Solutions To Mathematics Textbooks/Algebra (9780132413770)/Chapter 4

Exercise 1.1

Let $f(M):F^{m\times n}\to F^{l\times p}$ such that $f(M)=AMB$ where $A\in F^{l\times m},B\in F^{n\times p}$. Then for $M,N\in F^{m\times n}$ and $\alpha ,\beta \in F$ it holds $f(\alpha M+\beta N)=A(\alpha M+\beta N)B=\alpha AMB+\beta ANB=\alpha f(M)+\beta f(N)$ , so $f$ is a linear transformation.

Exercise 1.3

The matrix $A\in F^{m\times n}$ is a linear mapping $A:F^n\to F^m$. Let $𝖡=\{v_1,\dots ,v_n\}$ be a basis for $F^n$. Then the space $\text{im}A=\text{Span}(A(v_1),\dots ,A(v_n))$ has dimension at most $\min (n,m)$. Then, using the dimension formula we have $\dim (\text{im}A)+\dim (\ker A)=\dim F^n=n$, so rearranging we get $\dim (\ker A)=n-\dim (\text{im}A)\ge n-\min (m,n)\ge n-m$.

Exercise 1.4

Let $A\in F^{m\times n}$ be a matrix of rank 1. Then, the image of $A$ is a space spanned by a single vector, say $v\in F^n$, and $Av=w$ for some nonzero $w\in F^m$. We can assume that $v=(1,0,\dots ,0)$, since the vector is unique up to a scaling and change of basis. Then, the kernel of $A$ is given by the vectors $e_i$ for $i=2,\dots ,n$. Next, consider the matrix $B=\frac{1}{v^{t}v}w{v}^t$, so that $Bv=w$ as well, and $B{e}_i=0,i=2,\dots ,n$. It is easy to see that $A$ and $B$ describe the same linear transformation, so they are equal as matrices. The representation of $B$ is not unique, since we could scale one of the vectors $v$ and $w$ arbitrarily as long as we scale the other accordingly.

Exercise 1.5

a) It is very easy to see that performing the vector space operations coordinate-wise preserves the vector space structure in the product space.

b) Let $T(u,v)=u+v$. Then we have $T(u_1+u_2,v_1+v_2)=u_1+v_1+u_2+v_2=T(u_1,v_1)+T(u_2,v_1)$ and $T(au,av)=au+av=aT(u,v)$, so $T$ is a linear operator.

c) We have $\dim (\text{im}T)+\dim (\ker T)=\dim (U\times W)$ where $\ker T=\{(u,w)\in U\times W:u=-w\}=\{(v,-v):v\in U\cap W\}$ so $\dim (\ker T)=\dim (U\cap W)$. Furthermore, we have by definition that $\text{im}T=U\oplus W$, and $\dim (U\times W)=\dim U+\dim W$. Therefore, the dimension formula has the form $\dim (U\cap W)+\dim (U\oplus W)=\dim U+\dim W$.

Exercise 2.1

We can write $A=a_{11}{e}_{11}+a_{12}{e}_{12}+a_{21}{e}_{21}+a_{22}{e}_{22}$, $B=b_{11}{e}_{11}+b_{12}{e}_{12}+b_{21}{e}_{21}+b_{22}{e}_{22}$ and $M=m_{11}{e}_{11}+m_{12}{e}_{12}+m_{21}{e}_{21}+m_{22}{e}_{22}$. We have the following multiplication table

	$e_{11}$	$e_{12}$	$e_{21}$	$e_{22}$
$e_{11}$	$e_{11}$	$e_{12}$	$0$	$0$
$e_{12}$	$0$	$0$	$e_{11}$	$e_{12}$
$e_{21}$	$e_{21}$	$e_{22}$	$0$	$0$
$e_{22}$	$0$	$0$	$e_{21}$	$e_{22}$

where the first element of a column denotes the matrix that is multiplied from the right by the first element of a given row. Then, $AM=(a_{11}{m}_{11}+a_{21}{m}_{12})e_{11}+(a_{12}{m}_{11}+a_{22}{m}_{12})e_{12}+(a_{11}{m}_{21}+a_{21}{m}_{22})e_{21}+(a_{12}{m}_{21}+a_{22}{m}_{22})e_{22}$. For $AMB$ we have then in the given basis the form

$\left(\begin{array}{c}{a}_{11}{m}_{11}{b}_{11}+a_{21}{m}_{12}{b}_{11}+a_{11}{m}_{21}{b}_{12}+a_{21}{m}_{22}{b}_{12}\\ a_{12}{m}_{11}{b}_{11}+a_{22}{m}_{12}{b}_{11}+a_{12}{m}_{21}{b}_{12}+a_{22}{m}_{22}{b}_{12}\\ a_{11}{m}_{11}{b}_{21}+a_{21}{m}_{12}{b}_{21}+a_{11}{m}_{21}{b}_{22}+a_{21}{m}_{22}{b}_{22}\\ a_{12}{m}_{11}{b}_{21}+a_{22}{m}_{12}{b}_{21}+a_{12}{m}_{21}{b}_{22}+a_{22}{m}_{22}{b}_{22}\end{array}\right)$.

Exercise 2.3

The matrix with the given property satisfies the equation $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{c}x\\ x\end{array}\right)=\left(\begin{array}{c}x\\ 3x\end{array}\right)$. Solving this yields that the matrix has to have the form $\left(\begin{array}{cc}a& 1-a\\ c& 3-c\end{array}\right)$ for any $a,c\in ℝ.$