Exercises: Matrices – Serlo

Introduction

Exercise

Determine the $(4\times 4)$ matrix ${\mathcal {M}}=(m_{jk})$ whose entries satisfy:

$m_{jk}={\begin{cases}1&{\text{ for }}j=k,\\0&{\text{ for }}j=k-2,\\3&{\text{ for }}j=k+2,\\2&{\text{ else. }}\end{cases}}$

Solution

The matrix ${\mathcal {M}}$ is of the form ${\begin{pmatrix}m_{11}&m_{12}&m_{13}&m_{14}\\m_{21}&m_{22}&m_{23}&m_{24}\\m_{31}&m_{32}&m_{33}&m_{34}\\m_{41}&m_{42}&m_{43}&m_{44}\end{pmatrix}}$.

So the result is: ${\mathcal {M}}={\begin{pmatrix}1&2&0&2\\2&1&2&0\\3&2&1&2\\2&3&2&1\end{pmatrix}}$

Exercises on the vector space structure of matrices

Exercise (Derivation of matrix addition)

Let $L_{1},L_{2}:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be linear maps with

${\begin{aligned}&L_{1}{\begin{pmatrix}x\\y\end{pmatrix}}:={\begin{pmatrix}2x\\y-x\end{pmatrix}}&L_{2}{\begin{pmatrix}x\\y\end{pmatrix}}:={\begin{pmatrix}y\\x\end{pmatrix}}\end{aligned}}$

Determine the matrices of $A_{1},\,A_{2}$ with respect to the standard basis. How can you define $A_{1}+A_{2}$ such that the result corresponds to the matrix of $L_{1}+L_{2}$?

In this case, the standard basis corresponds to $E=\{e_{1},e_{2}\}$ with $e_{1}={\begin{pmatrix}1\\0\end{pmatrix}},e_{2}={\begin{pmatrix}0\\1\end{pmatrix}}$.

How to get to the proof? (Derivation of matrix addition)

Write the two maps in the same tabular form as used for $L$ above.

You can apply the same method to directly find the matrix representing $L_{1}+L_{2}$.

There is now a very obvious way to define the matrix addition. If you try this, you should get the right result.

Proof (Derivation of matrix addition)

We first determine $A_{1}$ by writing down the table and summarizing it as a matrix. For the map $L_{1}{\begin{pmatrix}x\\y\end{pmatrix}}:={\begin{pmatrix}2x\\y-x\end{pmatrix}}$ we have

$L_{1}(e_{1})={\begin{pmatrix}2\\-1\end{pmatrix}},L_{1}(e_{2})={\begin{pmatrix}0\\1\end{pmatrix}}$

This gives us

$A_{1}={\begin{pmatrix}2&0\\-1&1\end{pmatrix}}$

Now we do the same with $L_{2}{\begin{pmatrix}x\\y\end{pmatrix}}:={\begin{pmatrix}y\\x\end{pmatrix}}$ to obtain $A_{2}$:

$L_{2}(e_{1})={\begin{pmatrix}0\\1\end{pmatrix}},L_{2}(e_{2})={\begin{pmatrix}1\\0\end{pmatrix}}$

We conclude the table into a matrix

$A_{2}={\begin{pmatrix}0&1\\1&0\end{pmatrix}}$

We are now looking for the matrix of $L_{1}+L_{2}$:

${\begin{aligned}&(L_{1}+L_{2})(e_{1})=L_{1}(e_{1})+L_{2}(e_{1})={\begin{pmatrix}2\\-1\end{pmatrix}}+{\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}2\\0\end{pmatrix}}\\[0.3em]&(L_{1}+L_{2})(e_{2})=L_{1}(e_{2})+L_{2}(e_{2})={\begin{pmatrix}0\\1\end{pmatrix}}+{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}1\\1\end{pmatrix}}\\[0.3em]\end{aligned}}$

This results in the desired matrix

$A={\begin{pmatrix}2&1\\0&1\end{pmatrix}}$

We now want to define the addition of two matrices in such a way that $A_{1}+A_{2}=A$. Remember that we have already defined the vector addition in $\mathbb {R} ^{n}$ component by component - so this definition is a good first attempt. And indeed, with this rule we obtain

$A_{1}+A_{2}={\begin{pmatrix}2&0\\-1&1\end{pmatrix}}+{\begin{pmatrix}0&1\\1&0\end{pmatrix}}={\begin{pmatrix}2+0&0+1\\-1+1&1+0\end{pmatrix}}={\begin{pmatrix}2&1\\0&1\end{pmatrix}}=A$

Solution (Derivation of matrix addition)

If we define the addition of matrices as the addition of the respective components, we arrive at the desired result.

Let $L_{1}:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ be the above linear map with

$L_{1}{\begin{pmatrix}x\\y\end{pmatrix}}:={\begin{pmatrix}2x\\y-x\end{pmatrix}}$

Exercise (Derivation of scalar multiplication)

Determine the matrix $A_{1}$ with respect to the standard basis for the map $L_{1}$ and the matrix $M_{1}$ for the mapping $\beta \cdot L_{1}$. How can you define the multiplication of a matrix by a scalar such that $M_{1}=\beta \cdot A_{1}$?

Solution (Derivation of scalar multiplication)

We already know from the previous exercise that

$L_{1}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}2x\\y-x\end{pmatrix}}\,\Longrightarrow \,A_{1}={\begin{pmatrix}2&0\\-1&1\end{pmatrix}}$

If we now multiply $L_{1}$ by the scalar $\beta$, we get

${\begin{aligned}&(\beta \cdot L_{1}){\begin{pmatrix}1\\0\end{pmatrix}}=\beta \cdot L_{1}{\begin{pmatrix}1\\0\end{pmatrix}}=\beta \cdot {\begin{pmatrix}2\\-1\end{pmatrix}}={\begin{pmatrix}2\beta \\-\beta \end{pmatrix}}\\&(\beta \cdot L_{1}){\begin{pmatrix}0\\1\end{pmatrix}}=\beta \cdot L_{1}{\begin{pmatrix}0\\1\end{pmatrix}}=\beta \cdot {\begin{pmatrix}0\\1\end{pmatrix}}={\begin{pmatrix}0\\\beta \end{pmatrix}}\end{aligned}}$

Whence $M_{1}={\begin{pmatrix}2\beta &0\\-\beta &\beta \end{pmatrix}}$. Here you can quickly see that we can also define scalar multiplication element by element. Then

$\beta \cdot A_{1}=\beta \cdot {\begin{pmatrix}2&0\\-1&1\end{pmatrix}}={\begin{pmatrix}2\beta &0\\-\beta &\beta \end{pmatrix}}=M_{1}$

Exercises on matrix multiplication

Exercise (Derivation of matrix multiplication)

Let $K$ be a field and let $A,A'\in K^{2\times 2}$. Furthermore, let $f\colon K^{2}\to K^{2};x\mapsto A\cdot x$ and $g\colon K^{2}\to K^{2};x\mapsto A'\cdot x$. Let $B=\{(1,0)^{T},(0,1)^{T}\}$ be the standard basis of $K^{2}$. Describe $M_{B}^{B}(g\circ f)$ as a function of the entries of $A$ and $A'$.

Solution (Derivation of matrix multiplication)

We already know from the article on matrices of linear maps that $M_{B}^{B}(f)=A$ and $M_{B}^{B}(g)=A'$. If we write

$A={\begin{pmatrix}a&b\\c&d\end{pmatrix}}\quad {\text{und}}\quad A'={\begin{pmatrix}e&f\\g&h\end{pmatrix}}$

then

$f{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}}\quad {\text{und}}\quad g{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}ex+fy\\gx+hy\end{pmatrix}}$

Now we calculate:

${\begin{aligned}(g\circ f){\begin{pmatrix}x\\y\end{pmatrix}}&=g\left(f{\begin{pmatrix}x\\y\end{pmatrix}}\right)=g{\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}}\\[0.3em]&={\begin{pmatrix}e(ax+by)+f(cx+dy)\\g(ax+by)+h(cx+dy)\end{pmatrix}}\\[0.3em]&={\begin{pmatrix}(ae+cf)x+(be+df)y\\(ag+ch)x+(bg+dh)y\end{pmatrix}}\\[0.3em]&={\begin{pmatrix}ae+cf&be+df\\ag+df&bg+dh\end{pmatrix}}\cdot {\begin{pmatrix}x\\y\end{pmatrix}}\end{aligned}}$

Using the same argument as at the beginning of this solution, we now know that

$M_{B}^{B}(g\circ f)={\begin{pmatrix}ae+cf&be+df\\ag+df&bg+dh\end{pmatrix}}$

Exercise

We are given the matrix ${\mathcal {M}}={\begin{pmatrix}3&2&1\\2&3&1\\1&3&2\end{pmatrix}}$. Calculate the expression ${\mathcal {M}}^{0}+{\mathcal {M}}+{\frac {1}{2}}{\mathcal {M}}^{2}$.

Solution

We first consider each summand of the expression to be calculated individually:

${\mathcal {M}}^{0}={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}=I_{3},{\mathcal {M}}={\begin{pmatrix}3&2&1\\2&3&1\\1&3&2\end{pmatrix}}$

and bacause

${\mathcal {M}}^{2}={\begin{pmatrix}3&2&1\\2&3&1\\1&3&2\end{pmatrix}}\cdot {\begin{pmatrix}3&2&1\\2&3&1\\1&3&2\end{pmatrix}}={\begin{pmatrix}14&15&7\\13&16&7\\11&17&8\end{pmatrix}}$

we get

${\frac {1}{2}}{\mathcal {M}}^{2}={\begin{pmatrix}{\frac {1}{2}}\cdot 14&{\frac {1}{2}}\cdot 15&{\frac {1}{2}}\cdot 7\\{\frac {1}{2}}\cdot 13&{\frac {1}{2}}\cdot 16&{\frac {1}{2}}\cdot 7\\{\frac {1}{2}}\cdot 11&{\frac {1}{2}}\cdot 17&{\frac {1}{2}}\cdot 8\end{pmatrix}}={\begin{pmatrix}7&7.5&3.5\\6.5&8&3.5\\5.5&8.5&4\end{pmatrix}}$

Together, this results in:

${\begin{aligned}{\mathcal {M}}^{0}+{\mathcal {M}}+{\frac {1}{2}}{\mathcal {M}}^{2}&={\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}+{\begin{pmatrix}3&2&1\\2&3&1\\1&3&2\end{pmatrix}}+{\begin{pmatrix}7&7.5&3.5\\6.5&8&3.5\\5.5&8.5&4\end{pmatrix}}\\&={\begin{pmatrix}1+3+7&0+2+7.5&0+1+3.5\\0+2+6.5&1+3+8&0+1+3.5\\0+1+5.5&0+3+8.5&1+2+4\end{pmatrix}}\\&={\begin{pmatrix}11&9.5&4.5\\8.5&12&4.5\\6.5&11.5&7\end{pmatrix}}\end{aligned}}$

Exercise

Use matrix multiplication to prove the addition theorems for the cosine and the sine, i.e.,

$\cos(\alpha +\beta )=\cos(\alpha )\cdot \cos(\beta )-\sin(\alpha )\cdot \sin(\beta )$

$\sin(\alpha +\beta )=\sin(\alpha )\cdot \cos(\beta )+\cos(\alpha )\cdot \sin(\beta )$

Solution

We consider the rotation matrix ${\begin{pmatrix}\cos(\alpha +\beta )&-\sin(\alpha +\beta )\\\sin(\alpha +\beta )&\cos(\alpha +\beta )\end{pmatrix}}$ and remember that rotations in the plane can be understood as linear maps. Accordingly, it does not matter whether we rotate directly by an angle $\alpha +\beta$, or first by an angle $\alpha$ and then by $\beta$. Therefore, we have

${\begin{aligned}{\begin{pmatrix}\cos(\alpha +\beta )&-\sin(\alpha +\beta )\\\sin(\alpha +\beta )&\cos(\alpha +\beta )\end{pmatrix}}&={\begin{pmatrix}\cos(\alpha )&-\sin(\alpha )\\\sin(\alpha )&\cos(\alpha )\end{pmatrix}}\cdot {\begin{pmatrix}\cos(\beta )&-\sin(\beta )\\\sin(\beta )&\cos(\beta )\end{pmatrix}}\\&={\begin{pmatrix}\cos(\alpha )\cdot \cos(\beta )-\sin(\alpha )\cdot \sin(\beta )&-\cos(\alpha )\cdot \sin(\beta )-\sin(\alpha )\cdot \cos(\beta )\\\sin(\alpha )\cdot \cos(\beta )+\cos(\alpha )\cdot \sin(\beta )&\cos(\alpha )\cdot \cos(\beta )-\sin(\alpha )\cdot \sin(\beta )\end{pmatrix}}\end{aligned}}$

A comparison of the entries of the matrices provides the addition theorems to be shown.

Exercises on representative and basis change matrices

Exercise (Calculating coordinate vectors with respect to a basis)

Let $v=(3,2)^{T}\in \mathbb {R} ^{2}$. Calculate the coordinate vector of $v$ with respect to the basis $B=\{(1,-1)^{T},(1,1)^{T}\}$.

Solution (Calculating coordinate vectors with respect to a basis)

We wish to find out what the coordinate vector of $v$ looks like with respect to the basis $B=\{(1,-1)^{T},(1,1)^{T}\}$. This gives us a system of equations that needs to be solved:

$v={\begin{pmatrix}3\\2\end{pmatrix}}=a_{1}\cdot {\begin{pmatrix}1\\-1\end{pmatrix}}+a_{2}\cdot {\begin{pmatrix}1\\1\end{pmatrix}}$

We now have two equations. On the one hand

$3=a_{1}+a_{2}$

and on the other

$2=-a_{1}+a_{2}.$

Solving this linear system gives $a_{1}=0.5$ and $a_{2}=2.5$. This results in the following for the coordinate vector

$[v]_{B}={\begin{pmatrix}0.5\\2.5\end{pmatrix}}.$

Exercises on the rank of a matrix

Exercise

Determine the rank of the following matrix: ${\mathcal {M}}_{1}={\begin{pmatrix}1&2&0\\0&5&1\\0&10&2\end{pmatrix}}$

Solution

We transform the matrix ${\mathcal {M}}_{1}$ into row-level form and read off the rank of the matrix using the number of zero rows. We obtain:

${\begin{pmatrix}1&2&0\\0&5&1\\0&10&2\end{pmatrix}}{\overset {{\text{Add}}_{3,2}(-2)}{\rightsquigarrow }}{\begin{pmatrix}1&2&0\\0&5&1\\0&0&0\end{pmatrix}}$

We have created a zero row by converting to row-step form. The rank of our matrix is therefore ${\text{rank}}({\mathcal {M}}_{1})=2$.

In this case, the shorthand ${\text{Add}}_{3,2}(-2)$ indicates that we have added the third row of the matrix with $-2$ times the second row

Exercise

Determine the rank of the following matrix: ${\mathcal {M}}_{2}={\begin{pmatrix}1&2&0&3\\0&5&1&2\\3&16&1&-2\\0&4&0&-1\end{pmatrix}}$

Solution

We transform the matrix ${\mathcal {M}}_{2}$ into row-level form and read off the rank of the matrix using the number of zero rows. We obtain:

${\begin{alignedat}{3}&{\begin{pmatrix}1&2&0&3\\0&5&1&2\\3&16&1&-2\\0&4&0&-1\end{pmatrix}}&{\overset {{\text{Add}}_{3,4}(-4)}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&5&1&2\\3&0&1&2\\0&4&0&-1\end{pmatrix}}\\{\overset {{\text{Add}}_{3,1}(-3)}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&5&1&2\\0&-6&1&-7\\0&4&0&-1\end{pmatrix}}&{\overset {{\text{Mult}}_{3}({\frac {4}{6}})}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&5&1&2\\0&-4&{\frac {4}{6}}&-{\frac {28}{6}}\\0&4&0&-1\end{pmatrix}}\\{\overset {{\text{Add}}_{3,4}(1)}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&5&1&2\\0&0&{\frac {4}{6}}&-{\frac {17}{3}}\\0&4&0&-1\end{pmatrix}}&{\overset {{\text{Mult}}_{2}(-{\frac {4}{5}})}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&-4&-{\frac {4}{5}}&-{\frac {8}{5}}\\0&0&{\frac {4}{6}}&-{\frac {17}{3}}\\0&4&0&-1\end{pmatrix}}\\{\overset {{\text{Add}}_{4,2}(1)}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&4&{\frac {4}{5}}&{\frac {8}{5}}\\0&0&{\frac {4}{6}}&-{\frac {17}{3}}\\0&0&-{\frac {4}{5}}&-{\frac {13}{5}}\end{pmatrix}}&{\overset {{\text{Mult}}_{3}({\frac {6}{5}})}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&4&{\frac {4}{5}}&{\frac {8}{5}}\\0&0&{\frac {4}{5}}&-{\frac {34}{5}}\\0&0&-{\frac {4}{5}}&-{\frac {13}{5}}\end{pmatrix}}\\{\overset {{\text{Add}}_{4,3}(1)}{\rightsquigarrow }}&{\begin{pmatrix}1&2&0&3\\0&4&{\frac {4}{5}}&{\frac {8}{5}}\\0&0&{\frac {4}{5}}&-{\frac {34}{5}}\\0&0&0&-{\frac {47}{5}}\end{pmatrix}}&&\end{alignedat}}$

By converting to row-level form, we have thus shown that for ${\mathcal {M}}_{2}$ we have ${\text{rank}}({\mathcal {M}}_{2})=4$.

However, we could have seen much more quickly at this point that ${\text{rank}}({\mathcal {M}}_{2})=4$. It is also sufficient to show that the column vectors (or equivalently the row vectors) are linearly independent. In this example, we decide to use the column vectors and show their linear independence. Let $\lambda ,\mu ,\theta ,\phi \in \mathbb {R}$.

$0{\overset {!}{=}}\lambda \cdot {\begin{pmatrix}1\\0\\3\\0\end{pmatrix}}+\mu \cdot {\begin{pmatrix}2\\5\\16\\4\end{pmatrix}}+\theta \cdot {\begin{pmatrix}0\\1\\1\\0\end{pmatrix}}+\phi \cdot {\begin{pmatrix}3\\2\\-2\\-1\end{pmatrix}}$

This gives us the linear system:

${\begin{alignedat}{4}1\cdot \lambda &+2\cdot \mu &+0\cdot \theta &+3\cdot \phi &=0\\[0.3em]0\cdot \lambda &+5\cdot \mu &+1\cdot \theta &+2\cdot \phi &=0\\[0.3em]3\cdot \lambda &+16\cdot \mu &+1\cdot \theta &-2\cdot \phi &=0\\[0.3em]0\cdot \lambda &+4\cdot \mu &+0\cdot \theta &-1\cdot \phi &=0\end{alignedat}}$

with the unique solution $\lambda =\mu =\theta =\phi =0$, which shows the linear independence of the column vectors. However, the rank of a matrix describes the maximum number of linearly independent column vectors of the matrix. So ${\text{rank}}({\mathcal {M}}_{2})=4$.

The exercise therefore shows that it is not always advantageous to convert the matrix into row-level form in order to read off the rank of the matrix.

Exercises on matrix inversion

Exercise

Let ${\mathcal {M}}\in \mathbb {R} ^{n\times n}$ be invertible. Furthermore, let ${\mathcal {M}}^{3}={\mathcal {M}}$. Show that ${\mathcal {M}}$ is self-inverse, i.e., ${\mathcal {M}}={\mathcal {M}}^{-1}$.

Solution

Since ${\mathcal {M}}\in \mathbb {R} ^{n\times n}$ is invertible, there exists an ${\mathcal {M}}^{-1}\in \mathbb {R} ^{n\times n}$ with ${\mathcal {M}}\cdot {\mathcal {M}}^{-1}=I_{n}$. Thus, we can write

${\begin{aligned}{\mathcal {M}}^{3}&={\mathcal {M}}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{multiplication from the right by }}{\mathcal {M}}^{-1}\right.}\\[0.3em]{\mathcal {M}}^{3}\cdot {\mathcal {M}}^{-1}&={\mathcal {M}}\cdot {\mathcal {M}}^{-1}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\mathcal {M}}\cdot {\mathcal {M}}^{-1}=I_{n}\right.}\\[0.3em]{\mathcal {M}}^{2}&=I_{n}\\[0.3em]&\quad {\color {Gray}\left\downarrow \ {\text{multiplication from the right by }}{\mathcal {M}}^{-1}\right.}\\[0.3em]{\mathcal {M}}^{2}\cdot {\mathcal {M}}^{-1}&=I_{n}\cdot {\mathcal {M}}^{-1}\\[0.3em]{\mathcal {M}}&={\mathcal {M}}^{-1}\\[0.3em]\end{aligned}}$