Dual space – Serlo

We have already seen the vector space of linear maps $\operatorname {Hom} _{K}(V,W)$ between two $K$-vector spaces $V$ and $W$. We will now consider the case where the vector space $W$ corresponds to the field $K$.

Motivation

Consider the following example: We want to buy apples and pears. An apple costs $$2$ and a pear $$3$. If $x\in \mathbb {R}$ is the number of apples and $y\in \mathbb {R}$ is the number of pears, how much do we have to pay in total? The formula for the total price is $2x+3y$. We can express this equation as $\mathbb {R}$-linear map

$P\colon \mathbb {R} ^{2}\to \mathbb {R} ,(x,y)\mapsto 2x+3y$

Let's assume that the prices increase by half. To get the formula that gives the new total price, we need to multiply the old formula by ${\frac {3}{2}}$. The formula that gives this price would then be ${\frac {3}{2}}(2x+3y)=3x+{\frac {9}{2}}y$. The corresponding linear map is

$Q\colon \mathbb {R} ^{2}\to \mathbb {R} ,(x,y)\mapsto 3x+{\frac {9}{2}}y.$

We thus recognize that $Q(x,y)={\frac {3}{2}}P(x,y)$. Suppose now that the price of apples increases by $$2$ and the price of pears by $$4$. We obtain the corresponding formula for the total price by adding $2x+4y$ to the original formula, i.e. $(2x+3y)+(2x+4y)=4x+7y$. This can be understood as the addition of linear maps. We define $R,S\colon \mathbb {R} ^{2}\to \mathbb {R}$ by $R(x,y)=2x+4y$ and $S(x,y)=4x+7y$. Then $(P+R)(x,y)=P(x,y)+R(x,y)=S(x,y)$ holds true. So in this example, we simpy added linear maps from $\mathbb {R} ^{2}$ to $\mathbb {R}$ and multiplied them by scalars.

The total price is indicated by linear maps from $\mathbb {R} ^{2}\to \mathbb {R}$. Such a map assigns a value, namely the price, to each vector. In other words, we can say that the mapping "measures" these vectors. This is why we call linear maps from $\mathbb {R} ^{2}$ to $\mathbb {R}$ linear measurement functions. We have seen above that sums and scalar multiples of such maps are again linear maps. In other words, linear combinations of linear maps are again linear maps. So also on the set of linear maps on $\mathbb {R} ^{2}$, we can find a vector space structure.

What about other vector spaces? Let's look at the $\mathbb {C}$-vector space $\mathbb {C} [x]_{\leq n}$ of complex polynomials of degree at most $n\in \mathbb {N}$. There are a number of simple measurement functions here. These can, for example, assign to a polynomial $p$ its value at a point $a\in \mathbb {C}$:

$\operatorname {eval} _{a}\colon \mathbb {C} [x]_{\leq n}\to \mathbb {C} ,p\mapsto p(a).$

Alternatively, we can assign to a polynomial the value of its derivative at the point $a\in \mathbb {C}$:

$D_{a}\colon \mathbb {C} [x]_{\leq n}\to \mathbb {C} ,p\mapsto p'(a).$

Since the coefficients of polynomials are scalars, we can use them to define further measurement functions. For example, for $p=a_{n}x^{n}+\ldots +a_{1}x+a_{0}$, consider the mappings $f,g\colon \mathbb {C} [x]_{\leq n}\to \mathbb {C}$ defined by $f(p)=a_{n}+\ldots +a_{1}$ and $g(p)=a_{0}$. Then $(f+g)(p)=f(p)+g(p)=a_{n}+\ldots +a_{1}+a_{0}=p(1)=:\operatorname {eval} _{1}(p)$. We can also see here that sums of measurement functions are again measurement functions.

In general, we can also consider the space of linear measurement functions $V\to K$ over an arbitrary $K$-vector space $V$. We will see that, as in the previous examples, this is a vector space. This space is called the dual space of $V$.

Definition

Definition (Dual space)

Let $V$ be a vector space over a field $K$. Then the space of linear mappings $V^{*}:=\operatorname {Hom} _{K}(V,K)$ between the $K$-vector spaces $V$ and $K$ is called the dual space of $V$.

The following theorem states that the dual space is a vector space.

Theorem ($V^{*}$ is a vector space)

Let $V$ be a vector space over a field $K$. Then $V^{*}$ with the two relations

${\begin{aligned}+\colon V^{*}\times V^{*}&\to V^{*}\\(f,g)&\mapsto f+g,{\text{ where }}(f+g)(v):=f(v)+g(v){\text{ for all }}v\in V,\end{aligned}}$

and

${\begin{aligned}\cdot \colon K\times V^{*}&\to V^{*}\\(\lambda ,f)&\mapsto \lambda \cdot f,{\text{ where }}(\lambda \cdot f)(v):=\lambda \cdot f(v){\text{ for all }}v\in V,\end{aligned}}$

a $K$-vector space.

Proof ($V^{*}$ is a vector space)

We know from the article on function spaces that for $K$-vector spaces $V$ and $W$, the set of linear maps $\operatorname {Hom} _{K}(V,W)$ is also a $K$-vector space. Since $K$ itself is a (1-dimensionsl) $K$-vector space, we know that for every $K$ vector space $V$, also $V^{*}=\operatorname {Hom} _{K}(V,K)$ is a $K$ vector space.

Examples of vectors in the dual space

Example (Characterization of $(\mathbb {R} ^{2})^{*}$)

The dual space of $\mathbb {R} ^{2}$ is the vector space of all linear maps from $\mathbb {R} ^{2}$ to $\mathbb {R}$. Each such linear map $f\in (\mathbb {R} ^{2})^{*}$ is given by multiplication with a (1x2) matrix, the representing matrix, and is therefore of the form

$f\colon \mathbb {R} ^{2}\to \mathbb {R} ,\quad {\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}a&b\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=ax+by$

for certain $a,b\in \mathbb {R}$. Thus, the elements in the dual space of $\mathbb {R} ^{2}$ are described by linear equations of the form $f(x,y)=ax+by$.

More generally, an element of $(\mathbb {R} ^{n})^{*}$ is represented by a (1xn) matrix ${\begin{pmatrix}a_{1}&\ldots &a_{n}\end{pmatrix}}$ or a linear equation of the form $f(x_{1},\ldots ,x_{n})=a_{1}x_{1}+\ldots +a_{n}x_{n}$ with coefficients $a_{i}\in \mathbb {R}$.

Example (Limit of convergent sequences)

Let $c$ be the space of convergent sequences $(x_{n})_{n\in \mathbb {N} }\subseteq \mathbb {R}$. Because sums and scalar multiples of convergent sequences are convergent sequences again, $c$ is a $\mathbb {R}$ vector space. You can read a proof of the vector space properties here.

We consider the mapping $f\colon c\to \mathbb {R} ,(x_{n})_{n\in \mathbb {N} }\mapsto \lim _{n\to \infty }x_{n}$, which sends a sequence to its limit value. For example, $f((1)_{n\in \mathbb {N} })=\lim _{n\to \infty }1=1$ or $f\left(\left({\tfrac {1}{n}}\right)_{n\in \mathbb {N} }\right)=\lim _{n\to \infty }({\tfrac {1}{n}})=0$. From the limit theorems we know that

${\begin{aligned}\lim _{n\rightarrow \infty }(a_{n}+b_{n})&=\lim _{n\to \infty }a_{n}+\lim _{n\to \infty }b_{n}{\text{ und }}\\\lim _{n\rightarrow \infty }\lambda \cdot a_{n}&=\lambda \cdot \lim _{n\to \infty }a_{n}\end{aligned}}$

applies to all convergent sequences $(a_{n})_{n\in \mathbb {N} },(b_{n})_{n\in \mathbb {N} }\in c$ and scalars $\lambda \in \mathbb {R}$. It follows that $f$ is a linear map and therefore $f\in c^{*}$ holds.

Example (Polynomial space and the evaluation mapping)

Let $K$ be a field. We consider the polynomial ring $K[X]$ as a $K$-vector space. For $\lambda \in K$ we define the mapping

${\begin{aligned}\operatorname {eval} _{\lambda }\colon K[X]&\to K,\\P&\mapsto P(\lambda ),\end{aligned}}$

which evaluates a polynomial at the position $\lambda$. For example, we have $\operatorname {eval} _{1}(x^{2}-1)=1^{2}-1=0$ and $\operatorname {eval} _{0}(x^{2}-1)=0^{2}-1=-1$.

By direct computation, e can verify that this mapping is $K$-linear, i.e. an element of $K[X]^{*}$:

For $P,Q\in K[X]$ and $k\in K$ we then have:

${\begin{aligned}\operatorname {eval} _{\lambda }(P+k\cdot Q)&=(P+k\cdot Q)(\lambda )\\&=P(\lambda )+k\cdot Q(\lambda )\\&=\operatorname {eval} _{\lambda }(P)+k\cdot \operatorname {eval} _{\lambda }(Q).\end{aligned}}$

Example (Derivative)

Let $C^{1}(\mathbb {R} )$ be the space of continuously differentiable functions $\mathbb {R} \to \mathbb {R}$. Let $x\in \mathbb {R}$ be fixed and consider the mapping

${\begin{aligned}\partial _{x}\colon C^{1}(\mathbb {R} )&\to \mathbb {R} ,\\f&\mapsto f'(x)\end{aligned}}$

which sends a differentiable function to its derivative at the point $x$. For example, for $x=0$, the value of the mapping in $f(t)=t^{2}-1$ is given by

$\partial _{0}(f)=f'(0)=(2t)|_{t=0}=0.$

We verify by direct computation that the mapping $\partial _{x}$ (for fixed $x\in \mathbb {R}$) is linear: For $f,g\in C^{1}(\mathbb {R} )$ and $\lambda \in \mathbb {R}$ we have

${\begin{aligned}\partial _{x}(f+\lambda g)&=(f+\lambda g)'(x)\\&=f'(x)+\lambda g'(x)\\&=\partial _{x}(f)+\lambda \partial _{x}(g).\end{aligned}}$

This follows from the properties of the derivative. So $\partial _{x}$ is an element of $C^{1}(\mathbb {R} )^{*}$.

Example (Integral)

Let $C^{0}([0,1])$ be the space of continuous functions $[0,1]\to \mathbb {R}$. Consider the mapping

${\begin{aligned}I\colon C^{0}([0,1])&\to \mathbb {R} ,\\f&\mapsto \int _{0}^{1}f(x)dx\end{aligned}}$

which sends a continuousfunction on $[0,1]$ to its integral over this interval. As an example, for $f(x)=x^{2}-1$,

$I(f)=\int _{0}^{1}x^{2}-1dx=[{\frac {1}{3}}x^{3}-x]_{x=0}^{1}=({\frac {1}{3}}-1)-0=-{\frac {2}{3}}.$

We verify by direct calculation that the mapping $I$ is linear: For $f,g\in C^{0}([0,1])$ and $\lambda \in \mathbb {R}$ the following applies

${\begin{aligned}I(f+\lambda g)&=\int _{0}^{1}f(x)+g(x)dx\\&=\int _{0}^{1}f(x)dx+\int _{0}^{1}g(x)dx\\&=I(f)+\lambda I(g).\end{aligned}}$

This follows from the properties of the integral. So $I$ is an element of $C^{0}([0,1])^{*}$.

Dual Basis

We now know what the dual space $V^{*}$ of a $K$-vector space $V$ is: It consists of all linear maps from $V$ to $K$. Intuitively, we can understand these maps as linear maps that measure vectors from $V$. This is why we sometimes call elements of the dual space $V^{*}$ "(linear) measurement functions" in this article.

Motivated by this intuitive notion of "measurements", we ask ourselves: Is there a subset $M\subseteq V^{*}$ of measurement functions that can be used to uniquely determine vectors? In other words, is there a subset $M$ so that we can find a measurement function $f\in M$ with $f(v)\neq f(w)$ for every choice of vectors $v,w\in V$ with $v\neq w$?

Let's first consider what this means using an example:

Example (Unique determination of vectors using measurement functions)

Let us consider $V=\mathbb {R} ^{2}$. Then the dual space $V^{*}$ is the vector space of all linear maps $\mathbb {R} ^{2}\to \mathbb {R}$. Consider the linear maps $f,g,h\in V^{*}$ with

$f(x,y)=2x-y,\quad g(x,y)={\frac {1}{2}}y-x,\quad h(x,y)=y.$

If $M=\{f\}$, we cannot use these functions to determine vectors uniquely: For $v=(1,1)$ and $w=(0,-1)$, we have $v\neq w$, but $f(v)=2-1=1=0-(-1)=f(w)$.

Even with the measurement functions in $M'=\{f,g\}$, the vectors $(1,1)$ and $(0,-1)$ cannot be distinguished: We also have $g(1,1)=-{\frac {1}{2}}=g(0,-1)$.

However, if we consider the subset of measurement functions $M''=\{f,h\}$ instead, then vectors in $\mathbb {R} ^{2}$ are uniquely determined by the measurements in $M''$: Let $v=(x,y)$ and $w=(x',y')$ be any vectors with $v\neq w$. Assume that $f(v)=f(w)$ and $h(v)=h(w)$ apply. From $h(v)=h(w)$ we obtain $y=y'$. Together with $2x-y=f(v)=f(w)=2x'-y'$, we would then also get $2x=2x'$, i.e. $x=x'$. This would mean that $v=w$, which is a contradiction to our assumption. Therefore, $f(v)\neq f(w)$ or $h(v)\neq h(w)$ (or both) applies. Hence, for each choice of different vectors in $v,w\in \mathbb {R} ^{2}$, at least one of the two measurements in $M''$ provides different values for $v$ and $w$. Vectors are therefore uniquely determined by the measurements in $M''$.

In sumary, our question is: Does there exist a subset $M\subseteq V^{*}$ such that $v,w\in V$ applies to all vectors: If $f(v)=f(w)$ applies to all measurements $f\in M$, then $v=w$ must be true.

We will first try to answer this question in $K^{n}$.

Measurement functions for unique determination of vectors

A vector $v=(v_{1},\ldots ,v_{n})\in K^{n}$ is uniquely determined by its entries $v_{i}$. If we select measurement functions from $(K^{n})^{*}$ in such a way that their values provide us with the entries of a vector, then we have ensured that a vector is already uniquely determined by these values. Let us therefore consider the following mappings for $i\in \{1,\ldots ,n\}$

$f_{i}\colon K^{n}\to K,\quad x=(x_{1},\ldots ,x_{n})\mapsto x_{i}.$

You can check that the maps $f_{i}$ are linear. In addition, $f_{i}(v)=v_{i}$ holds for every $i$. The map $f_{i}$ therefore provides the $i$-th entry of vectors in $K^{n}$. A vector $v\in K^{n}$ is already uniquely determined by the values of $f_{i}$: Suppose we have vectors $v=(v_{1},\ldots ,v_{n})$ and $w=(w_{1},\ldots ,w_{n})$ in $K^{n}$ with equal function values among the $f_{i}$, i.e., with $f_{i}(v)=f_{i}(w)$ for all $i$. Then $v_{i}=f_{i}(v)=f_{i}(w)=w_{i}$ applies for all $i$ and therefore $v=w$. Thus, if $v,w\in K^{n}$ with $f_{i}(v)=f_{i}(w)$ for all $i$, then $v=w$ follows.

It is also intuitively clear that we cannot omit any of the measurement functions $f_{i}$ in order to uniquely determine a vector by its measurement values. For example, if we omit $f_{j}$, $j\in \{1,\ldots ,n\}$, then for

$v=(0,\ldots ,0)\quad {\text{ und }}\quad w=(0,\ldots ,0,\underbrace {1} _{j{\text{-th position}}},0,\ldots ,0)$

we may have $f_{i}(v)=0=f_{i}(w)$ for all measurement functions with $i\neq j$, but nevertheless $v\neq w$. The measurement functions $f_{i}$ with $i\neq j$ therefore no longer uniquely determine a vector.

So the $f_{i}$ with $i=1,\ldots n$ form a set of measurement functions that uniquely determine vectors from $K^{n}$. Further, they are minimal because we cannot omit any of the functions.

Can we generalize this to a general vector space $V$? In $K^{n}$ we have used the fact that a vector $v=(v_{1},\ldots ,v_{n})\in K^{n}$ is uniquely determined by its entries $v_{i}$. Now, the $v_{i}$ are precisely the coordinates of $v$ with respect to the standard basis $\{e_{1},\ldots ,e_{n}\}\subseteq K^{n}$:

$v=v_{1}\cdot e_{1}+\ldots +v_{n}\cdot e_{n}.$

In a general vector space $V$, we do not have a standard basis. However, as soon as we have chosen any basis $B$, we can speak of the coordinates of a vector with respect to $B$ in the same way as in $K^{n}$. Just as in $K^{n}$ with the standard basis, in $V$ with the selected basis $B$, a vector $v\in V$ is uniquely determined by its coordinates with respect to $B$. As soon as we have chosen a basis, we can try to proceed in the same way as in $K^{n}$.

In the following, we assume that $V$ is finite-dimensional, i.e. $\dim V=n<\infty$. Let $B=\{b_{1},\ldots ,b_{n}\}$ be a basis of $V$. Then every vector $v\in V$ is of the form

$v=a_{1}\cdot b_{1}+\ldots +a_{n}\cdot b_{n}$

with uniquely determined coordinates $a_{1},\ldots ,a_{n}\in K$. Analogous to $K^{n}$, we now define the linear measurement functions for $i\in \{1,\ldots ,n\}$ in $V^{*}$

$f_{i}\colon V\to K,\quad v=a_{1}\cdot b_{1}+\ldots +a_{n}\cdot b_{n}\mapsto a_{i}.$

One of the measurement functions $f_{i}$ therefore determines the $i$-th coordinate of vectors with respect to the basis $B$. Thus,

$v=f_{1}(v)\cdot b_{1}+f_{2}(v)\cdot b_{2}+\ldots +f_{n}(v)\cdot b_{n}.$

for every vector $v\in V$.

Warning

Note that the definition of $f_{i}$ depends on the selected basis $B$.

Since vectors in $V$ are already uniquely determined by their coordinates, they are also already uniquely determined by the values of $f_{i}$. In other words, for all $v,w\in V$ we have

$f_{1}(v)=f_{1}(w),\,f_{2}(v)=f_{2}(w),\,\ldots ,\,f_{n}(v)=f_{n}(w)\implies \underbrace {\sum _{i=1}^{n}f_{i}(v)\cdot b_{i}} _{=v}=\underbrace {\sum _{i=1}^{n}f_{i}(w)\cdot b_{i}} _{=w}\implies v=w.$

For the same reason as with $K^{n}$, none of the $f_{i}$ can be omitted: If the $j$-th measurement function $f_{j}$, $i\in \{1,\ldots ,n\}$, is missing, then any two vectors for which only the $j$-th coordinate with respect to $B$ differs, can no longer be distinguished.

Question: Which two vectors can you choose here?

We choose an example analogous to $K^{n}$ and set

$v=0\cdot b_{1}+\ldots +0\cdot b_{j-1}+1\cdot b_{j}+0\cdot b_{j+1}+\ldots +0\cdot b_{n}=b_{j}$

and

$w=0\cdot b_{1}+\ldots +0\cdot b_{n}=0_{V}.$

Then $f_{i}(v)=0_{K}=f_{i}(w)$ holds for all $i\in \{1,\ldots ,j-1,j+1,\ldots ,n\}$, but nevertheless $v\neq w$. If the $j$-th measurement function is omitted, then vectors are no longer uniquely determined by the function values of $f_{i}$.

The measurement functions form a basis

Let $V$ be a vector space with a fixed basis $B=\{b_{1},\ldots ,b_{n}\}$ and let the $f_{i}$ be defined as above. If you want to determine vectors uniquely using the values of $f_{i}$, you cannot do without any of the $f_{i}$. The reason for this is that the result of a measurement $f_{j}(v)$ (the $j$-th coordinate of $v$ with respect to $B$) cannot be deduced from the other measurements. That means, we cannot represent any of the measurement functions $f_{j}$ as a linear combination of the other $f_{i}$ ($i\neq j$). In other words, the measurement functions $f_{i}$ are linearly independent.

On the other hand, the values of $f_{i}$ already tell us everything there is to know about a vector $v\in V$: Its coordinates with respect to the selected basis $B$. Can all other measurement functions from $V^{*}$ therefore be combined from $f_{1},\ldots ,f_{n}$? Any measurement function $g\colon V\to K$ from $V^{*}$ is already uniquely determined by its values on the basis vectors $b_{1},\ldots ,b_{n}$ according to the principle of linear continuation. For $i\in \{1,\ldots ,n\}$, let $\lambda _{i}=g(b_{i})\in K$ be these values. Furthermore, $f_{i}(b_{i})=1$ and $f_{i}(b_{j})=0$ apply for $j\neq i$ and all $i\in \{1,\ldots ,n\}$. By inserting the $b_{i}$ we obtain that

$g=\lambda _{1}\cdot f_{1}+\ldots +\lambda _{n}\cdot f_{n}$

assume the same values on the basis vectors. According to the principle of linear continuation, the two linear maps are therefore identical. Thus, every $g\in V^{*}$ can be written as a linear combination of $f_{i}$. In other word, the measurement functions $f_{i}$ form a generating system of $V^{*}$.

Hence, $\{f_{1},\ldots ,f_{n}\}\subseteq V^{*}$ is a basis of the dual space and we can prove the following theorem:

Theorem (Existence of a dual basis)

Let $V$ be a finite dimensional vector space and $B=\{b_{1},\ldots ,b_{n}\}$ a basis of $V$. Then there exists a unique basis $B^{*}=\{f_{1},\ldots ,f_{n}\}$ of $V^{*}$ such that

$f_{i}(b_{j})={\begin{cases}1&{\text{if }}i=j\\0&{\text{else}}\end{cases}}$

is true for all $i,j\in \{1,\ldots ,n\}$.

Proof (Existence of a dual basis)

Proof step: Existence and uniqueness of the $f_{i}$.

According to the principle of linear continuation, the linear maps $f_{i}$ exist and are uniquely determined by their values on the basis vectors of $V$.

Proof step: The $f_{i}$ are linearly independent.

Let $\lambda _{1},\ldots ,\lambda _{n}\in K$ with $\sum _{i=1}^{n}\lambda _{i}f_{i}=0_{V^{*}}$. Let further $i\in \{1,\ldots ,n\}$. Because $f_{i}(b_{i})=1$ and $f_{j}(b_{i})=0$ for $j\neq i$, we obtain the following by plugging in $b_{i}$:

$0_{K}=0_{V^{*}}(b_{i})=(\sum _{i=1}^{n}\lambda _{i}f_{i})(b_{i})=\sum _{i=1}^{n}\lambda _{i}f_{i}(b_{i})=\lambda _{i}.$

Because $i\in \{1,\ldots ,n\}$ was arbitrary, we conclude $\lambda _{1}=\ldots =\lambda _{n}=0_{K}$.

Proof step: The $f_{i}$ form a generating system.

Let $f\in V^{*}$ be arbitrary. For $i\in \{1,\ldots ,n\}$ we define $\lambda _{i}=f(b_{i})\in K$ and set $g=\sum _{i=1}^{n}\lambda _{i}f_{i}$. Then, proceeding as in the proof of linear independence, we obtain

$g(b_{i})=(\sum _{i=1}^{n}\lambda _{i}f_{i})(b_{i})=\sum _{i=1}^{n}\lambda _{i}f_{i}(b_{i})=\lambda _{i}$

for each $i\in \{1,\ldots ,n\}$. Because $f(b_{i})=g(b_{i})$ applies to all $i$ and because a linear map is already uniquely determined by the images of its basis vectors, we have $f=g\in \operatorname {span} \{f_{1},\ldots ,f_{n}\}$. The $f_{i}$ therefore form a generating system.

We call the uniquely determined basis $B^{*}$ the dual basis with respect to $B$ and denote its basis vectors by $b_{i}^{*}=f_{i}$.

Definition (Dual basis)

Let $V$ be a finite dimensional vector space with basis $B=\{b_{1},\ldots ,b_{n}\}$. The uniquely determined basis $B^{*}=\{b_{1}^{*},\ldots ,b_{n}^{*}\}$ with

$b_{i}^{*}(b_{j})={\begin{cases}1&{\text{if }}i=j\\0&{\text{else}}\end{cases}}$

is called the dual basis of $B$.

Warning

Note that $B^{*}$ depends on the basis chosen for $V$. Furthermore, you cannot "dualize" individual vectors from $V$, but only entire bases.

What happens in the infinite dimension?

Above, we only considered the case $\dim V<\infty$. Can we proceed analogously if $V$ is infinite dimensional? To define the measurement functions $f_{i}$, we must first choose a basis of $V$. Let $B=\{b_{i}\mid i\in I\}\subseteq V$ be a basis of $V$, where $I$ is an (infinite) index set. The principle of linear continuation also applies in infinite dimensions: For given values $\lambda _{i}\in K$, $i\in I$, there is exactly one linear map $f\colon V\to K$ with $f(b_{i})=\lambda _{i}$ for all $i\in I$. Just as in the finite-dimensional case, we can therefore define the map $f_{i}\colon V\to K$ for $i\in I$ using the rule

$f_{i}(b_{j})={\begin{cases}1,&j=i\\0,&j\neq i\end{cases}}.$

We can then show that $\{f_{i}\mid i\in I\}$ is also a linearly independent subset of $V^{*}$ in infinite dimensions. The proof is analogous to the proof of linear independence in the theorem on the dual basis.

However, in infinitely many dimensions, $\{f_{i}\mid i\in I\}$ cannot be a generating system of $V^{*}$: One can consider the function

$h\colon V\to K,\quad b_{i}\mapsto 1{\text{ for all }}i\in I,$

which assumes the value 1 on all basis vectors. This function cannot be represented as a finite linear combination of $f_{i}$.

So in infinitely many dimensions, the "dual basis" $\{f_{i}\mid i\in I\}$ is not a basis of the dual space.

Exercises

Exercise (Determining dual basis vectors and their kernels)

Let $V$ be a finite-dimensional vector space and let $v\in V$ with $v\neq 0$. Show that there exists an $f\in V^{*}$ with $f(v)\neq 0$.

When deriving the dual basis, we were guided by the idea that vectors in $V$ should be distinguishable by "measurements" in $V^{*}$. In this exercise, we will convince ourselves that this is true: We can always find a measurement $f\in V^{*}$ for which $f(0_{V})=0$ (this applies to every linear mapping), but $f(v)\neq 0$. We may therefore find an element in the dual space with which we can distinguish $v$ and the zero vector.

How to get to the proof? (Determining dual basis vectors and their kernels)

We have to construct a linear map $f\colon V\to K$. This map is one element of $V^{\ast }$. According to the principle of linear continuation, we can construct linear maps by specifying what they do on a basis. To use this principle, it is convenient to have a basis of $V$. Even more convenient is to have a basis of $V$ that contains $v$ as a basis vector.

We can construct such a basis using the basis completion theorem, which tells us that $V$ has a basis $b_{1},\dots ,b_{n}$ with $b_{1}=v$. Using the principle of linear continuation, we can thus construct a linear map that does not send $b_{1}=v$ to $0$. For example, we can choose that $f\colon V\to K$, which sends all $b_{1}$ to $1$ and $b_{i}$ for $i=2,\dots ,n$ to $0$.

This is exactly the dual basis vector $b_{1}^{\ast }$ in the dual basis to $b_{1},\dots ,b_{n}$.

Solution (Determining dual basis vectors and their kernels)

According to the Basis completion theorem, there exists a basis $B=\{b_{1},\dots ,b_{n}\}$ with $b_{1}=v$. From the definition of the dual basis we obtain that the dual basis vector $b_{1}^{\ast }$ of $B^{\ast }$ has the property $b_{1}^{\ast }(v)=b_{1}^{\ast }(b_{1})=1\neq 0$. Thus $f=b_{1}^{\ast }$ fulfills the desired condition.

Exercise (Determining the dual basis)

1. Consider the basis $B_{1}=\left\{{\begin{pmatrix}2\\0\\1\end{pmatrix}},{\begin{pmatrix}0\\1\\1\end{pmatrix}},{\begin{pmatrix}1\\1\\2\end{pmatrix}}\right\}$ of $\mathbb {R} ^{3}$. Determine the basis $B_{1}^{*}=\{v_{1}^{*},v_{2}^{*},v_{3}^{*}\}$ which is dual to $B_{1}$, that is, for $1\leq i\leq 3$ determine the explicit form of the function
   $v_{i}^{*}\colon \mathbb {R} ^{3}\to \mathbb {R} ,\quad {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto v_{i}^{*}({\begin{pmatrix}x\\y\\z\end{pmatrix}}).$
2. Consider the basis $B_{2}=\{t^{3}+t^{2},t^{2},t^{2}-t,1\}$ of $\mathbb {R} [t]_{\leq 3}$. Determine the basis dual to $B_{2}$ $B_{2}^{*}=\{p_{1}^{*},\ldots ,p_{4}^{*}\}$, i.e. for $1\leq i\leq 4$ determine the explicit form of the function
   $p_{i}^{*}\colon \mathbb {R} [t]_{\leq 3}\to \mathbb {R} ,\quad a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}\mapsto p_{i}^{*}(a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}).$
3. Consider the basis $B_{3}=\left\{{\begin{pmatrix}1&0\\0&0\end{pmatrix}},{\begin{pmatrix}0&1\\1&0\end{pmatrix}},{\begin{pmatrix}0&1\\-1&0\end{pmatrix}},{\begin{pmatrix}1&1\\1&1\end{pmatrix}}\right\}$ of $\mathbb {C} ^{2\times 2}$. Determine the basis $B_{3}^{*}=\{M_{1}^{*},\ldots ,M_{4}^{*}\}$ dual to $B_{3}$ , i.e. for $1\leq i\leq 4$ etermine the explicit form of the function
   $M_{i}^{*}\colon \mathbb {C} ^{2\times 2}\to \mathbb {C} ,\quad {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto M_{i}^{*}({\begin{pmatrix}a&b\\c&d\end{pmatrix}}).$

Solution (Determining the dual basis)

Solution sub-exercise 1:

Set $v_{1}=(2,0,1)^{T}$, $v_{2}=(0,1,1)^{T}$ and $v_{3}=(1,1,2)^{T}$. We are looking for linear maps $v_{1}^{*},v_{2}^{*},v_{3}^{*}\colon \mathbb {R} ^{3}\to \mathbb {R}$ whose values we only know on the basis vectors $v_{i}$. We must define $v_{i}^{*}((x,y,z)^{T})$ for general $x,y,z\in \mathbb {R}$.

By definition of the dual basis, we already know the function values of each $v_{i}^{*}$ on the basis vectors in $B_{1}$. Applying the principle of linear continuation, we can determine all function value: Because $B_{1}$ is a basis, there are coordinates $a,b,c\in \mathbb {R}$ for each $(x,y,z)^{T}\in \mathbb {R} ^{3}$ such that $(x,y,z)^{T}=av_{1}+bv_{2}+cv_{3}$. With the help of linearity we get

$v_{i}^{*}((x,y,z)^{T})=v_{i}^{*}(av_{1}+bv_{2}+cv_{3})=av_{i}^{*}(v_{1})+bv_{i}^{*}(v_{2})+cv_{i}^{*}(v_{3}).$

We know the values $v_{i}^{*}(v_{j})$ by definition of the dual basis. We therefore only need to determine the coordinates of any vector $(x,y,z)^{T}$ with respect to $B_{1}$. Then we can write out the $v_{i}^{*}$.

Proof step: Determining the coordinates of any vector $(x,y,z)^{T}$ with respect to $B_{1}$

We want to determine the coordinates with respect to $B_{1}$ of any vector $(x,y,z)^{T}$. Let $x,y,z\in \mathbb {R}$. We write

${\begin{pmatrix}x\\y\\z\end{pmatrix}}=x{\begin{pmatrix}1\\0\\0\end{pmatrix}}+y{\begin{pmatrix}0\\1\\0\end{pmatrix}}+z{\begin{pmatrix}0\\0\\1\end{pmatrix}}=xe_{1}+ye_{2}+ze_{3}.$

The coordinates of $(x,y,z)^{T}$ with respect to the standard basis $B_{st}=\{e_{1},e_{2},e_{3}\}$ are therefore simply $x$, $y$ and $z$. If we write $k_{B_{st}}$ for the coordinate map, this means

$k_{B_{st}}({\begin{pmatrix}x\\y\\z\end{pmatrix}})={\begin{pmatrix}x\\y\\z\end{pmatrix}}.$

We can convert these into coordinates $a,b,c$ with respect to $B_{1}$ by multiplying the coordinate vector of $B_{st}$ from the left by the basis transition matrix $T_{B_{1}}^{B_{st}}$ that implements the transfer from $B_{st}$ to $B_{1}$. Then

${\begin{pmatrix}a\\b\\c\end{pmatrix}}=T_{B_{1}}^{B_{st}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}.$

In order to determine the basis transition matrix $T_{B_{1}}^{B_{st}}$, we calculate the coordinates of the standard basis vectors $e_{1},e_{2},e_{3}$ with respect to $B_{1}$. These form the columns of $T_{B_{1}}^{B_{st}}$.

We start with $e_{1}$: We are looking for $a_{1},b_{1},c_{1}\in \mathbb {R}$ such that

$a_{1}v_{1}+b_{1}v_{2}+c_{1}v_{3}=a_{1}{\begin{pmatrix}2\\0\\1\end{pmatrix}}+b_{1}{\begin{pmatrix}0\\1\\1\end{pmatrix}}+c_{1}{\begin{pmatrix}1\\1\\2\end{pmatrix}}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$

For this we hae to solve the linear system

${\begin{aligned}2a_{1}+c_{1}&=1\\b_{1}+c_{1}&=0\\a_{1}+b_{1}+2c_{1}&=0\end{aligned}}$

which yields $a_{1}=1$, $b_{1}=1$ and $c_{1}=-1$. In the same way, we determine the coordinates $a_{2}=1,b_{2}=3,c_{2}=-2$ of $e_{2}$ with respect to $B_{1}$ and the coordinates $a_{3}=-1,b_{3}=-2,c_{3}=2$ of $e_{3}$ with respect to $B_{1}$. Then

$T_{B_{1}}^{B_{st}}={\begin{pmatrix}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{pmatrix}}={\begin{pmatrix}1&1&-1\\1&3&-2\\-1&-2&2\end{pmatrix}}.$

Note: We could also have solved all three systems at once by summarizing the "right-hand sides" column by column, i.e. by taking the inverse of ${\begin{pmatrix}2&0&1\\0&1&1\\1&1&2\end{pmatrix}}$ . This makes sense, because this matrix is the basis transition matrix from $B_{1}$ to the standard basis. Its inverse is therefore the matrix $T_{B_{1}}^{B_{st}}$ that transitions from $B_{st}$ to $B_{1}$.

The coordinates of $(x,y,z)^{T}$ with respect to $B_{1}$ are therefore

$T_{B_{1}}^{B_{st}}k_{B_{st}}({\begin{pmatrix}x\\y\\z\end{pmatrix}})={\begin{pmatrix}1&1&-1\\1&3&-2\\-1&-2&2\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}x+y-z\\x+3y-2z\\-x-2y+2z\end{pmatrix}}.$

Of course, it is also okay to guess the coordinates of $(x,y,z)^{T}$ with respect to $B_{1}$ by looking closely without solving systems of equations.

Proof step: Result for $v_{1}^{*},v_{2}^{*},v_{3}^{*}$

We can now write any $(x,y,z)^{T}$ as

${\begin{pmatrix}x\\y\\z\end{pmatrix}}=(x+y-z)v_{1}+(x+3y-2z)v_{2}+(-x-2y+2z)v_{3}.$

Using linearity of $v_{i}^{*}$ and the definition of the dual basis, we obtain

$v_{1}^{*}((x,y,z)^{T})=(x+y-z)\underbrace {v_{1}^{*}(v_{1})} _{=1}+(x+3y-2z)\underbrace {v_{1}^{*}(v_{2})} _{=0}+(-x-2y+2z)\underbrace {v_{1}^{*}(v_{3})} _{=0}=x+y-z.$

In the same way, we calculate $v_{2}^{*}((x,y,z)^{T})=x+3y-2z$ and $v_{3}^{*}((x,y,z)^{T})=-x-2y+2z$. In total, we have therefore determined the three basis vectors of the dual basis:

${\begin{aligned}v_{1}^{*}&\colon \mathbb {R} ^{3}\to \mathbb {R} ^{3},\quad {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto x+y-z,\\v_{2}^{*}&\colon \mathbb {R} ^{3}\to \mathbb {R} ^{3},\quad {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto x+3y-2z,\\v_{3}^{*}&\colon \mathbb {R} ^{3}\to \mathbb {R} ^{3},\quad {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto -x-2y+2z.\end{aligned}}$

Solution sub-exercise 2:

We know what the map $p_{i}^{*}$ does with the basis vectors $p_{i}\in B_{2}$. To find out how $p_{i}^{*}$ acts on a general vector $a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}$, we can express it in the basis $B_{2}$ via linear combination:

${\begin{aligned}&a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}\\&{\color {OliveGreen}\left\downarrow {\text{ expand with }}a_{3}t^{3}-a_{3}t^{3}{\text{ and }}-a_{1}t^{2}+a_{1}t^{2}\right.}\\=&a_{3}(t^{3}+t^{2})-a_{3}t^{2}+a_{2}t^{2}+a_{1}(t-t^{2})+a_{1}t^{2}+a_{0}\\&{\color {OliveGreen}\left\downarrow {\text{sort by basis vectors }}p_{1},p_{2},p_{3},p_{4}\right.}\\=&a_{3}(t^{3}+t^{2})+(a_{1}+a_{2}-a_{3})t^{2}-a_{1}(t^{2}-t)+a_{0}\\=&a_{3}p_{4}+(a_{1}+a_{2}-a_{3})p_{3}-a_{1}p_{2}+p_{1}\end{aligned}}$

This allows us to calculate the desired functions. For $p_{1}^{*}$ we have

${\begin{aligned}&p_{1}^{*}(a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0})\\=&p_{1}^{*}(a_{3}p_{4}+(a_{1}+a_{2}-a_{3})p_{3}-a_{1}p_{2}+p_{1})\\&{\color {OliveGreen}\left\downarrow p_{1}^{*}{\text{ is linear.}}\right.}\\=&a_{3}\underbrace {p_{1}^{*}(p_{4})} _{=0}+(a_{1}+a_{2}-a_{3})\underbrace {p_{1}^{*}(p_{3})} _{=0}-a_{1}\underbrace {p_{1}^{*}(p_{2})} _{=0}+\underbrace {p_{1}^{*}(p_{1})} _{=1}\\=&1\end{aligned}}$

For $p_{2}^{*}$ we get

${\begin{aligned}&p_{2}^{*}(a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0})\\=&p_{2}^{*}(a_{3}p_{4}+(a_{1}+a_{2}-a_{3})p_{3}-a_{1}p_{2}+p_{1})\\&{\color {OliveGreen}\left\downarrow p_{2}^{*}{\text{ is linear.}}\right.}\\=&a_{3}\underbrace {p_{2}^{*}(p_{4})} _{=0}+(a_{1}+a_{2}-a_{3})\underbrace {p_{2}^{*}(p_{3})} _{=0}-a_{1}\underbrace {p_{2}^{*}(p_{2})} _{=1}+\underbrace {p_{2}^{*}(p_{1})} _{=0}\\=&-a_{1}\end{aligned}}$

So the function of $p_{3}^{*}$ is

${\begin{aligned}&p_{3}^{*}(a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0})\\=&p_{3}^{*}(a_{3}p_{4}+(a_{1}+a_{2}-a_{3})p_{3}-a_{1}p_{2}+p_{1})\\&{\color {OliveGreen}\left\downarrow p_{3}^{*}{\text{ is linear.}}\right.}\\=&a_{3}\underbrace {p_{3}^{*}(p_{4})} _{=0}+(a_{1}+a_{2}-a_{3})\underbrace {p_{3}^{*}(p_{3})} _{=1}-a_{1}\underbrace {p_{3}^{*}(p_{2})} _{=0}+\underbrace {p_{3}^{*}(p_{1})} _{=0}\\=&a_{1}+a_{2}-a_{3}\end{aligned}}$

For $p_{4}^{*}$ we get

${\begin{aligned}&p_{4}^{*}(a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0})\\=&p_{4}^{*}(a_{3}p_{4}+(a_{1}+a_{2}-a_{3})p_{3}-a_{1}p_{2}+p_{1})\\&{\color {OliveGreen}\left\downarrow p_{4}^{*}{\text{ is linear.}}\right.}\\=&a_{3}\underbrace {p_{4}^{*}(p_{4})} _{=1}+(a_{1}+a_{2}-a_{3})\underbrace {p_{4}^{*}(p_{3})} _{=0}-a_{1}\underbrace {p_{4}^{*}(p_{2})} _{=0}+\underbrace {p_{4}^{*}(p_{1})} _{=0}\\=&a_{3}\end{aligned}}$

In summary, we obtain the following functions

${\begin{aligned}&p_{1}^{*}\colon \mathbb {R} [t]_{\leq 3}\to \mathbb {R} ,\quad a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}\mapsto 1\\&p_{2}^{*}\colon \mathbb {R} [t]_{\leq 3}\to \mathbb {R} ,\quad a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}\mapsto -a_{1}\\&p_{1}^{*}\colon \mathbb {R} [t]_{\leq 3}\to \mathbb {R} ,\quad a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}\mapsto a_{1}+a_{2}-a_{3}\\&p_{1}^{*}\colon \mathbb {R} [t]_{\leq 3}\to \mathbb {R} ,\quad a_{3}t^{3}+a_{2}t^{2}+a_{1}t+a_{0}\mapsto a_{3}\\\end{aligned}}$

Solution sub-exercise 3:

We know the values of each $M_{i}^{*}$ when applied to the basis vectors $M_{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},M_{2}={\begin{pmatrix}0&1\\1&0\end{pmatrix}},M_{3}={\begin{pmatrix}0&1\\-1&0\end{pmatrix}},M_{4}={\begin{pmatrix}1&1\\1&1\end{pmatrix}}$ and want to find the value for any matrix $A={\begin{pmatrix}a&b\\c&d\end{pmatrix}}$. To do this, we express $A$ as a linear combination of $M_{i}$:

${\begin{aligned}{\begin{pmatrix}a&b\\c&d\end{pmatrix}}&=a{\begin{pmatrix}1&0\\0&0\end{pmatrix}}+b{\begin{pmatrix}0&1\\0&0\end{pmatrix}}+c{\begin{pmatrix}0&0\\1&0\end{pmatrix}}+d{\begin{pmatrix}0&0\\0&1\end{pmatrix}}\\&=aM_{1}+b{\frac {1}{2}}(M_{2}+M_{3})+c{\frac {1}{2}}(M_{2}-M_{3})+d(M_{4}-M_{1}-M_{2})\\&=(a-d)M_{1}+({\frac {b}{2}}+{\frac {c}{2}}-d)M_{2}+({\frac {b}{2}}-{\frac {c}{2}})M_{3}+dM_{4}.\end{aligned}}$

Using the definition of the dual basis and the linearity of $M_{i}^{*}$, we can now specify the solution: We have $M_{i}^{*}(M_{j})=0$ for $i\neq j$ and $M_{i}^{*}(M_{i})=1$, so the following applies

${\begin{aligned}M_{1}^{*}&\colon \mathbb {C} ^{2\times 2}\to \mathbb {C} ,\quad {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto a-d,\\M_{2}^{*}&\colon \mathbb {C} ^{2\times 2}\to \mathbb {C} ,\quad {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto {\frac {b}{2}}+{\frac {c}{2}}-d,\\M_{3}^{*}&\colon \mathbb {C} ^{2\times 2}\to \mathbb {C} ,\quad {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto {\frac {b}{2}}-{\frac {c}{2}},\\M_{4}^{*}&\colon \mathbb {C} ^{2\times 2}\to \mathbb {C} ,\quad {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto d,\\\end{aligned}}$

Exercise (Elements of the dual space and their kernel)

Let $V$ be an $n$-dimensional $K$-vector space and let $f,g\in V^{*}$. Show: If $\ker(f)=\ker(g)$, then there exists a $\lambda \in K$ with $g=\lambda f$.

How to get to the proof? (Elements of the dual space and their kernel)

For the elements $v$ in the kernel of $f$ and $g$, we have $g(v)=\lambda f(v)=0$ for all $\lambda \in K$. This means that the desired $\lambda$ only depends on the $v\in V$, which are not in the kernel of $f$ and $g$. To understand this in more detail, we first look at the dimension of the kernel. Using the dimension formula, we obtain

$\dim \ker(f)+\dim \operatorname {im} (f)=\dim V$

and therefore $\dim \ker(f)=n-\dim \operatorname {im} (f)$. Now $\operatorname {im} (f)$ is a subspace of $K$. Because $K$ is one-dimensional, we get that the dimension of the image of $f$ is either $0$ or $1$. Thus $\dim \ker(f)=n$ or $\dim \ker(f)=n-1$.

Now we have $\ker(f)=\ker(g)$. This means that they both have the same dimension. In case $\dim \ker(f)=\dim \ker(g)=n$, they have the same dimension as $V$. So in this case, $\ker(f)=\ker(g)=V$, so $f$ and $g$ are the zero map. Therefore, $f=g$ and we can choose $\lambda =1$.

It remains to consider the case $\dim \ker(f)=n-1$. Here, we actually have vectors in which $\lambda$ plays a role. To compare the maps, it makes sense to look at them as applied to a basis, since according to the principle of linear continuation we know that $f$ and $g$ are already completely determined by their behavior on a basis. It is useful to choose a basis of $V$ with respect to which we already know a lot about our maps $f$ and $g$. We already know what both do on $\ker(f)=\ker(g)$. Let $b_{1},\dots ,b_{n-1}$ be a basis of $\ker(f)=\ker(g)$. Then we can use the basis completion theorem to continue this basis to a basis $b_{1},\dots ,b_{n-1},b_{n}$ of $V$.

Since $b_{n}\not \in \ker(f)=\ker(g)$, we know that $f(b_{n})\neq 0$ and $g(b_{n})\neq 0$. Furthermore, we know that $f(b_{i})=g(b_{i})=0$ for $i=1,\dots ,n-1$. We now need a candidate for $\lambda$. Since $\lambda$ depends on elements from $V$ that are not mapped to $0$, it makes sense to use $b_{n}$ for the candidate. With $\lambda =g(b_{n})/f(b_{n})$ we get $g(b_{n})=\lambda f(b_{n})$.

To see whether $g(v)=\lambda f(v)$ is valid for all $v\in V$, by the principle of linear continuation, it is sufficient to check this on our basis $b_{1},\dots ,b_{n}$. We already know that the statement is true for $b_{n}$, as well as for $b_{i}$ with $i=1,\dots ,n-1$, since $g(b_{i})=0=\lambda \cdot 0=\lambda \cdot f(b_{i})$. This proves the statement.

Solution (Elements of the dual space and their kernel)

The function $f\colon V\to K$ is a linear map between two finite-dimensional vector spaces. From the dimension formula we get

$\dim \ker(f)+\dim \operatorname {im} (f)=\dim V$

Since the image $\operatorname {im} (f)$ is a subspace of $K$, we have $\dim \operatorname {im} (f)\leq \dim(K)=1$. Furthermore, $\dim V=n$ applies. We can therefore conclude

$n=\dim V=\dim \ker(f)+\dim \operatorname {im} (f)\leq \dim \ker(f)+1$

Therefore, $\dim \ker(f)\geq n-1$. On the other hand, $\dim \ker(f)\leq \dim V=n$, because the kernel $\ker(f)$ is a subspace of $V$. Hence, there are only two possibilities:

1. The dimension of $\ker(f)$ is $n$.
2. The dimension of $\ker(f)$ is $n-1$.

Similarly, we can conclude that the dimension of the kernel of $g$ is either $n$ or $n-1$.

We assume that $\ker(f)=\ker(g)$ and show that there is then a $\lambda \in K$ with $g=\lambda f$. For this, we consider the two cases $\dim \ker(f)=n$ and $\dim \ker(f)=n-1$ separately.

Fall 1: $\dim \ker(f)=n$

In this case, the kernel of $f$ is an $n$-dimensional subspace of the $n$-dimensional vector space $V$. Therefore, $\ker(f)=V$ and because of our assumption also $\ker(g)=V$. Therefore, for all $v\in V$, we have $f(v)=0$ and $g(v)=0$. This means $f$ and $g$ are both the zero map, i.e. $f=0=g$. This proves the statement with $\lambda =1$.

Fall 2: $\dim \ker(f)=n-1$

In this case, the dimension formula implies

$\dim \operatorname {im} (f)=\dim V-\dim \ker(f)=n-(n-1)=1$

Let $b_{1},\ldots ,b_{n-1}\in V$ be a basis of $\ker(f)$. Because $\ker(f)=\ker(g)$, it is also a basis of $\ker(g)$. Due to the basis completion theorem, we can complete $b_{1},\ldots ,b_{n-1}$ to a basis $b_{1},\ldots ,b_{n-1},b_{n}$ of $V$. We then define $\alpha :=f(b_{n})\in K$ and $\beta :=g(b_{n})\in K$. The vector $b_{n}$ is not in $\ker(f)$, therefore $\alpha \neq 0$. Define then $\lambda :={\tfrac {\beta }{\alpha }}$. We show that $g=\lambda f$. Because of the principle of linear continuation, it is sufficient to prove this equality on the basis of $b_{1},\ldots ,b_{n}$.

We first consider $b_{i}$ with $i\in \{1,\ldots ,n-1\}$. Since $b_{i}\in \ker(f)=\ker(g)$, we have that

$g(b_{i})=0=\lambda \cdot 0=\lambda f(b_{i}).$

For the basis vector $b_{n}$, we have

$g(b_{n})=\beta ={\frac {\beta }{\alpha }}\cdot \alpha =\lambda \cdot \alpha =\lambda f(b_{n}).$

So $g$ and $\lambda f$ agree when applied to any basis vector. Thus, $g=\lambda f$.

Exercise (Dual basis and hyperplanes)

Let $V$ be an $n$-dimensional $K$-vector space.

1. Let $f\in V^{*}$ with $f\neq 0$. Show that $\dim \ker(f)=n-1$ holds.
2. Let $U$ be an $n-1$-dimensional subspace of $V$. Show that there is an element $f\in V^{*}$ with $\ker(f)=U$.
3. Assuming that $K\neq \mathbb {F} _{2}$, is it true that the $f$ from sub-exercise 2 is uniquely determined by the subspace $U$?

An $n-1$-dimensional subspace of an $n$-dimensional vector space $V$ is also called a hyperplane in $V$. For example, the hyperplanes in $\mathbb {R} ^{3}$ are exactly the planes through the origin. The first part of the exercise thus shows that the kernel of a non-zero element in dual space is a hyperplane in $V^{*}$.

Solution (Dual basis and hyperplanes)

Solution sub-exercise 1:

We can use the dimension formula to relate the dimension of the kernel to the dimension of $V$:

$\dim _{K}\ker(f)=\dim _{K}V-\dim _{K}\operatorname {im} (f)=n-\dim _{K}\operatorname {im} (f).$

So we have shifted our problem to the calculation of $\dim _{K}\operatorname {im} (f)$. Now $\operatorname {im} (f)\subseteq K$, that is, $\dim _{K}\operatorname {im} (f)\leq \dim _{K}K=1$. This means that the dimension of $\operatorname {im} (f)$ is either $0$ or $1$.

We know that $f\neq 0$, so there is a $v\in V$ with $f(v)\neq 0$. This means that $\operatorname {in} (f)\neq 0$ and the dimension of $\operatorname {in} (f)$ cannot be $0$. Therefore, $\dim _{K}\operatorname {im} (f)=1$ and we get

$\dim _{K}\ker(f)=n-\dim _{K}\operatorname {im} (f)=n-1.$

Solution sub-exercise 2:

According to the principle of linear continuation, a linear mapping is determined by what it does on a basis. To be able to use this principle, we first choose a basis $B_{U}=\{b_{1},\dots ,b_{n-1}\}$ of $U$. The basis completion theorem then provides us with a vector $b_{n}\in V$, such that $B=\{b_{1},\dots ,b_{n}\}$ is a basis of $V$.

According to the principle of linear continuation, we can then define a candidate for the linear map $f\colon V\to K$ by saying what happens on a basis of $V$. The vectors $b_{1},\dots ,b_{n-1}$ are elements of $U$. Since $U$ is to be the kernel of $f$, we must require $f(b_{i})=0$ for $i=1,\dots ,n-1$. The last basis vector $b_{n}$ is not in $U$. This means that $b_{n}$ must not lie in the kernel of $f$. For example, that we can demand $f(b_{n})=1$. To summarize, we define $f\colon V\to K$ as the linear map with

$f(b_{i})={\begin{cases}0,&i=1,\dots ,n-1\\1,&i=n.\end{cases}}$

Since $U$ is generated by $b_{1},\dots ,b_{n-1}$, we have $U\subseteq \ker(f)$. We therefore only have to show that $\ker(f)\subseteq U$. For this, let $v\in \ker(f)$. Because $B$ is a basis of $V$, we find $\lambda _{1},\dots ,\lambda _{n}$ with $v=\lambda _{1}b_{1}+\dots +\lambda _{n}b_{n}$. Now we know that

${\begin{aligned}0=f(v)&=f(\lambda _{1}b_{1}+\dots +\lambda _{n}b_{n})\\&=\lambda _{1}{\underset {=0}{\underbrace {f(b_{1})} }}+\dots +\lambda _{n-1}{\underset {=0}{\underbrace {f(b_{n-1})} }}+\lambda _{n}{\underset {=1}{\underbrace {f(b_{n})} }}\\&=\lambda _{n}\end{aligned}}$

Hence $\lambda _{n}=0$ and $v=\lambda _{1}b_{1}+\lambda _{n-1}b_{n-1}\in U$. Therefore, we have $\ker(f)=U$.

Solution sub-exercise 3:

The mapping $f$ is not unique: We know that $f\neq 0$ because $U\neq V$. Therefore $v\in V$ exists with $f(v)\neq 0$. Because $K\neq \mathbb {F} _{2}$, there is an element $\lambda \in K$ with $\lambda \notin \{0,1\}$. Thus $\lambda f(v)\neq f(v)$. Now consider the linear map $g\colon V\to K;w\mapsto \lambda \cdot f(v)$. This map has the same kernel as $f$ because $g(w)=0$ if $\lambda f(w)=0$. This is the case if $f(w)=0$, since $\lambda \neq 0$.

Furthermore, $g\neq f$, because $f(v)\neq \lambda f(v)=g(v)$. The linear map from the second part is therefore not unique.

In the last task, we required $K\neq \mathbb {F} _{2}$ because we needed an element in the proof that is neither $0$ nor $1$. The field $\mathbb {F} _{2}$ only consists of the elements $0$ and $1$. This means that if we want to construct a linear map $f\colon V\to K$ that has an $n-1$-dimensional subspace $U$ as its kernel, then we must define it as

$f(v)={\begin{cases}0,&v\in U\\1,&v\not \in U\end{cases}}$

This map is linear amd it is the only way to have a linear map with kernel $U$. Thus, for $K=\mathbb {F} _{2}$ we arrive at a different result in the last sub-exercise: The map is then unique.

Exercise (Basis of the kernel of $v_{i}^{*}$)

Let $V$ be a $K$-vector space, $B=\{v_{1},\ldots ,v_{n}\}\subseteq V$ a basis and $B^{*}=\{v_{1}^{*},\ldots ,v_{n}^{*}\}\subseteq V^{*}$ is the base dual to $B$. Show: For each $i\in \{1,\ldots ,n\}$ it holds true that

$\ker(v_{i}^{*})=\operatorname {span} \{v_{1},\ldots ,v_{i-1},v_{i+1},\ldots ,v_{n}\}.$

In particular, $B\setminus \{v_{i}\}$ is a basis of $\ker(v_{i}^{*})$.

Solution (Basis of the kernel of $v_{i}^{*}$)

By definition of the dual basis, $v_{i}^{*}(v_{j})=0$ holds for all $j\neq i$. Therefore, $v_{j}\in \ker(v_{i}^{*})$ applies for all $j\neq i$ and since the kernel is a subspace, we have

$\operatorname {span} \{v_{1},\ldots ,v_{i-1},v_{i+1},\ldots ,v_{n}\}\subseteq \ker(v_{i}^{*}).$

Since $v_{i}^{*}(v_{i})=1$ holds, $v_{i}^{*}$ is not the zero mapping. With the previous exercise, we conclude $\dim \ker(v_{i}^{*})=n-1$. Since the $v_{1},\ldots ,v_{n}$ are linearly independent, we have $\dim \operatorname {span} \{v_{1},\ldots ,v_{i-1},v_{i+1},\ldots ,v_{n}\}=n-1$, and since this span is contained in the kernel of $v_{i}^{*}$, the two subspaces are equal.

Exercise

Consider the basis

$B=\{v_{1},v_{2},v_{3}\}=\left\{{\begin{pmatrix}2\\1\\0\end{pmatrix}},{\begin{pmatrix}0\\2\\1\end{pmatrix}},{\begin{pmatrix}1\\2\\1\end{pmatrix}}\right\}$

of $\mathbb {R} ^{3}$.

1. For $B$ determine the dual basis $B^{*}=\{v_{1}^{*},v_{2}^{*},v_{3}^{*}\}$ with $v_{i}^{*}\colon \mathbb {R} ^{3}\to \mathbb {R}$ for $i=1,2,3$.
2. Determine the kernel $\ker(v_{i}^{*})$ and draw it in $\mathbb {R} ^{3}$ for $i=1,2,3$.

Solution

Solution sub-exercise 1:

The matrix of a linear map $f\colon \mathbb {R} ^{3}\to \mathbb {R}$ with respect to $\{e_{1}\}$. of the canonical bases $\{e_{1},e_{2},e_{3}\}$ of $\mathbb {R} ^{3}$ and $\{1\}$ of $\mathbb {R}$ is the uniquely determined matrix ${\begin{pmatrix}a&b&c\end{pmatrix}}$ with

$f({\begin{pmatrix}x\\y\\z\end{pmatrix}})={\begin{pmatrix}a&b&c\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}=ax+by+cz$

for all $(x,y,z)^{T}\in \mathbb {R} ^{3}$.

We are looking for the formula of the linear maps $v_{i}^{*}\colon \mathbb {R} ^{3}\to \mathbb {R}$, $i=1,2,3$. That means, we determine the three corresponding representative matrices ${\begin{pmatrix}a_{i},b_{i},c_{i}\end{pmatrix}}$ with respect to the canonical bases. By definition of the dual basis, the following should hold

${\begin{pmatrix}a_{1}&b_{1}&c_{1}\end{pmatrix}}{\begin{pmatrix}2\\1\\0\end{pmatrix}}=1,\quad {\begin{pmatrix}a_{1}&b_{1}&c_{1}\end{pmatrix}}{\begin{pmatrix}0\\2\\1\end{pmatrix}}=0,\quad {\begin{pmatrix}a_{1}&b_{1}&c_{1}\end{pmatrix}}{\begin{pmatrix}1\\2\\1\end{pmatrix}}=0$

and the same for $i=2,3$. If we summarize these equations in matrix form, we get

${\begin{pmatrix}a_{1}&b_{1}&c_{1}&\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{pmatrix}}{\begin{pmatrix}2&0&1\\1&2&2\\0&1&1\end{pmatrix}}={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}$

We must therefore determine an inverse of the matrix on the left-hand side of the equation, which has the basis vectors in $B$ as columns.

The inverse is

${\begin{pmatrix}2&0&1\\1&2&2\\0&1&1\end{pmatrix}}^{-1}={\begin{pmatrix}0&1&-2\\-1&2&-3\\1&-2&4\end{pmatrix}}={\begin{pmatrix}a_{1}&b_{1}&c_{1}&\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{pmatrix}}.$

The rows are the desired dual basis vectors. We therefore have

${\begin{aligned}v_{1}^{*}&\colon \mathbb {R} ^{3}\to \mathbb {R} ^{3},\quad {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto {\begin{pmatrix}0&1&-2\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}=y-2z,\\v_{2}^{*}&\colon \mathbb {R} ^{3}\to \mathbb {R} ^{3},\quad {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto {\begin{pmatrix}-1&2&-3\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}=-x+2y-3z,\\v_{3}^{*}&\colon \mathbb {R} ^{3}\to \mathbb {R} ^{3},\quad {\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto {\begin{pmatrix}1&-2&4\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}=x-2y+4z.\end{aligned}}$

Solution sub-exercise 2:

From the previous exercise we know that $\ker(v_{1}^{*})=\operatorname {span} \{v_{2},v_{3}\}$, $\ker(v_{2}^{*})=\operatorname {span} \{v_{1},v_{3}\}$ and $\ker(v_{3}^{*})=\operatorname {span} \{v_{1},v_{2}\}$. Plotted in $\mathbb {R} ^{3}$, we obtain a plane spanned by the two vectors in $\mathbb {R} ^{3}$.

Instead of using the previous exercise, we can also calculate the kernels of the matrices $v_{i}^{*}$:

Proof step: $\ker(v_{1}^{*})$

The kernel of $v_{1}^{*}$ contains all $(x,y,z)^{T}\in \mathbb {R} ^{3}$ with $v_{1}^{*}((x,y,z)^{T})=y-2z=0$, i.e., with $y=2z$. So the following holds:

$\ker(v_{1}^{*})=\left\{{\begin{pmatrix}a\\2b\\b\end{pmatrix}}\mid a,b\in \mathbb {R} \right\}=\operatorname {span} \left\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\2\\1\end{pmatrix}}\right\}.$

Note that $(1,0,0)^{T}=v_{3}-v_{2}$, so the result for the kernel is the same as in the previous exercise.

Proof step: $\ker(v_{2}^{*})$

The kernel of $v_{2}^{*}$ contains all $(x,y,z)^{T}\in \mathbb {R} ^{3}$ with $v_{2}^{*}((x,y,z)^{T})=-x+2y-3z=0$, i.e., with $x=2y-3z$. So the following holds:

$\ker(v_{2}^{*})=\left\{{\begin{pmatrix}2a-3b\\a\\b\end{pmatrix}}\mid a,b\in \mathbb {R} \right\}=\operatorname {span} \left\{{\begin{pmatrix}2\\1\\0\end{pmatrix}},{\begin{pmatrix}-3\\0\\1\end{pmatrix}}\right\}.$

Here, also $(-3,0,1)^{T}=v_{3}-2v_{1}$, so the result is the same as in the previous exercise.

Proof step: $\ker(v_{3}^{*})$

The kernel of $v_{3}^{*}$ contains all $(x,y,z)^{T}\in \mathbb {R} ^{3}$ with $v_{3}^{*}((x,y,z)^{T})=x-2y+4z=0$, i.e. with $x=2y-4z$. So the following holds:

$\ker(v_{2}^{*})=\left\{{\begin{pmatrix}2a-4b\\a\\b\end{pmatrix}}\mid a,b\in \mathbb {R} \right\}=\operatorname {span} \left\{{\begin{pmatrix}2\\1\\0\end{pmatrix}},{\begin{pmatrix}-4\\0\\1\end{pmatrix}}\right\}.$

Because $(-4,0,1)^{T}=v_{2}-2v_{1}$, this agrees with the result determined in the previously exercise.

Exercise (Dual map)

Let $f\colon V\to W$ be a linear map. We define the map

$f^{*}\colon W^{*}\to V^{*},\quad g\mapsto f^{*}(g):=g\circ f.$

1. Show that $f^{*}$ is linear.
2. Show: $(\operatorname {id} _{V})^{*}=\operatorname {id} _{V^{*}}$ and $(g\circ f)^{*}=f^{*}\circ g^{*}$ for linear maps $f\colon V\to W$ and $g\colon W\to X$.
3. Show: If $f$ is surjective, then $f^{*}$ is injective.
4. Show: If $f$ is injective, then $f^{*}$ is surjective.
5. Show: If $f$ is bijective, then $f^{*}$ is bijective and the inverse is given by $(f^{*})^{-1}=(f^{-1})^{*}$.

$f^{*}$ is called the dual mapping with respect to $f$. By definition, the dual map therefore receives linear mappings from $W$ to $K$ as input and turns them into linear mappings from $V$ to $K$. This is achieved by precomposition with $f$. A mapping $W{\overset {g}{\to }}K$ therefore becomes $V{\overset {f}{\to }}W{\overset {g}{\to }}K$. In words, $f^{*}$ can be described as "execute $f$ first".

Solution (Dual map)

Solution sub-exercise 1:

For more clarity in the proof, we write $\boxplus _{V}$ or $\boxplus _{W}$ for the addition of linear maps in $V^{*}$ or $W^{*}$ and $+$ for the addition in the vector space $K$. We also write $\boxdot _{V}$ or $\boxdot _{W}$ for the scalar multiplication in $V^{*}$ or $W^{*}$ and $\cdot$ for the scalar multiplication in $K$.

Let $g,h\in W^{*}$ and $\lambda \in K$. We have to show that

$f^{*}(g\boxplus _{W}h)=f^{*}(g)\boxplus _{V}f^{*}(h)\quad {\text{ and }}\quad f^{*}(\lambda \boxdot _{W}g)=\lambda \boxdot _{V}f^{*}(g)$

We must therefore prove the equality of elements in $V^{*}$, i.e., of maps $V\to K$. To do this, we show

$f^{*}(g\boxplus _{W}h)(v)=(f^{*}(g)\boxplus _{V}f^{*}(h))(v)$ and $f^{*}(\lambda \boxdot _{W}g)(v)=(\lambda \boxdot _{V}f^{*}(g))(v)$

for all $v\in V$.

Proof step: $f^{*}(g\boxplus _{W}h)=f^{*}(g)\boxplus _{V}f^{*}(h)$

Let $v\in V$. Then

${\begin{aligned}f^{*}(g\boxplus _{W}h)(v)&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f^{*}\right.}\\[0.3em]&=((g\boxplus _{W}h)\circ f)(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\circ \right.}\\[0.3em]&=(g\boxplus _{W}h)(f(v))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus _{W}\right.}\\[0.3em]&=g(f(v))+h(f(v))\\[0,3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\circ \right.}\\[0.3em]&=(g\circ f)(v)+(h\circ f)(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f^{*}\right.}\\[0.3em]&=f^{*}(g)(v)+f^{*}(h)(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxplus _{V}\right.}\\[0.3em]&=(f^{*}(g)(v)\boxplus _{V}f^{*}(h))(v).\\[0.3em]\end{aligned}}$

Because $v\in V$ was arbitrary, this shows the equality of the maps $f^{*}(g\boxplus _{W}h)$ and $f^{*}(g)\boxplus _{V}f^{*}(h)$.

Proof step: $f^{*}(\lambda \boxdot _{W}g)=\lambda \boxdot _{V}f^{*}(g)$

Let $v\in V$. Then

${\begin{aligned}f^{*}(\lambda \boxdot _{W}g)(v)&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f^{*}\right.}\\[0.3em]&=((\lambda \boxdot _{W}g)\circ f)(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\circ \right.}\\[0.3em]&=(\lambda \boxdot _{W}g)(f(v))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot _{W}\right.}\\[0.3em]&=\lambda \cdot g(f(v))\\[0,3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\circ \right.}\\[0.3em]&=\lambda \cdot (g\circ f)(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f^{*}\right.}\\[0.3em]&=\lambda \cdot (f^{*}(g))(v)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\boxdot _{V}\right.}\\[0.3em]&=(\lambda \boxdot _{V}f^{*}(g))(v).\\[0.3em]\end{aligned}}$

Because $v\in V$ was arbitrary, this shows the equality of the maps $f^{*}(\lambda \boxdot _{W}g)$ and $\lambda \boxdot _{V}f^{*}(g)$.

Solution sub-exercise 2:

We show $(\operatorname {id} _{V})^{*}(g)=g$ for all $g\in V^{*}$. It then follows that $(\operatorname {id} _{V})^{*}$ is the identity on $V^{*}$. Let $g\in V^{*}$. By definition of the dual map, we have

$(\operatorname {id} _{V})^{*}(g)=g\circ \operatorname {id} _{V}=g.$

Since $g\in V^{*}$ was arbitrary, the statement is shown.

Now let $f\colon V\to W$ and $g\colon W\to X$. Then $g\circ f\colon V\to X$ applies, i.e. $(g\circ f)^{*}\colon X^{*}\to V^{*}$. Furthermore, $f^{*}\colon W^{*}\to V^{*}$ and $g^{*}\colon X^{*}\to W^{*}$ and therefore $f^{*}\circ g^{*}\colon X^{*}\to V^{*}$. To show the equality of the maps $(g\circ f)^{*}=f^{*}\circ g^{*}$, we show that $(g\circ f)^{*}(k)=(f^{*}\circ g^{*})(k)$ holds for all $k\in X^{*}$. So if $k\in X^{*}$, then we get

${\begin{aligned}(g\circ f)^{*}(k)&=\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}(g\circ f)^{*}\right.}\\[0.3em]&=k\circ (g\circ f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{associativity of }}\circ \right.}\\[0.3em]&=(k\circ g)\circ f\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}g^{*}\right.}\\[0.3em]&={\color {OliveGreen}\underbrace {\color {black}g^{*}(k)} _{\in W^{*}}\color {black}\circ f}\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}f^{*}\right.}\\[0.3em]&=f^{*}(g^{*}(k))\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{definition of }}\circ \right.}\\[0.3em]&=(f^{*}\circ g^{*})(k).\end{aligned}}$

Because $k\in X^{*}$ was arbitrary, the statement is shown.

Solution sub-exercise 3:

Let $f\colon V\to W$ be surjective. We want to show that $f^{*}\colon W^{*}\to V^{*}$ is injective. Due to the linearity of $f^{*}$, it is sufficient to show that $\ker(f^{*})=\{0_{W^{*}}\}$. Let $g\in W^{*}$ with $f^{*}(g)=0_{V^{*}}$. This means that $g$ maps from $W$ to $K$ and $f^{*}(g)=g\circ f$ is the zero mapping from $V$ to $K$. We want to conclude that $g$ is the zero mapping in $W^{*}$, i.e. that $g(w)=0_{K}$ for all $w\in W$. For this, let $w\in W$ be arbitrary. Because $f$ is surjective, there exists an $v\in V$ with $f(v)=w$. It follows that

$g(w)=g(f(v))=(g\circ f)(v)=f^{*}(g)(v)=0_{V^{*}}(v)=0_{K}.$

Because $w\in W$ was arbitrary, we conclude $g=0_{W^{*}}$.

Solution sub-exercise 4:

Let $f\colon V\to W$ be injective. We want to show that $f^{*}\colon W^{*}\to V^{*}$ is surjective. So let $g\in V^{*}$ be arbitrary. This means that $g$ is a linear map from $V$ to $K$. We want to define a map $h\in W^{*}$ from $W$ to $K$ such that $f^{*}(h)=h\circ f=g$.

Because $f$ is injective, the restriction of $f$ to the image $f(V)$ of $f$ is an isomorphism. We denote this restriction by ${\tilde {f}}\colon V\to f(V)$. Then ${\tilde {f}}^{-1}\colon f(V)\to V$ and the following holds

${\tilde {f}}^{-1}\circ f={\tilde {f}}^{-1}\circ {\tilde {f}}=\operatorname {id} _{V}.$

Because $g$ is defined on $V$, we can define and obtain $h:=g\circ {\tilde {f}}^{-1}$:

$f^{*}(h)=h\circ f=(g\circ {\tilde {f}}^{-1})\circ f=g\circ ({\tilde {f}}^{-1}\circ f)=g\circ \operatorname {id} _{V}=g.$

Because $g\in V^{*}$ was arbitrary, the surjectivity of $f^{*}$ is shown.

Solution sub-exercise 5:

If $f\colon V\to W$ is bijective, then it follows from the previous two sub-exercises that $f^{*}$ is also bijective. We calculate that $(f^{-1})^{*}$ is the inverse of $f$: From sub-exercise 2 we get

$f^{*}\circ (f^{-1})^{*}=(f^{-1}\circ f)^{*}=\left(\operatorname {id} _{V}\right)^{*}=\operatorname {id} _{V^{*}}.$

Analogously, one can show $(f^{-1})^{*}\circ f^{*}=\operatorname {id} _{W^{*}}$.