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

Exercise 1.1

Let $f(M):F^{m\times n}\rightarrow F^{l\times p}$ 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 $A:F^{n}\rightarrow F^{m}$. Let ${\mathsf {B}}=\{v_{1},\ldots ,v_{n}\}$ be a basis for $F^{n}$. Then the space ${\textrm {im}}~A={\textrm {Span}}(A(v_{1}),\ldots ,A(v_{n}))$ has dimension at most $\min(n,m)$. Then, using the dimension formula we have $\dim({\textrm {im}}~A)+\dim(\ker A)=\dim F^{n}=n$, so rearranging we get $\dim(\ker A)=n-\dim({\textrm {im}}~A)\geq n-\min(m,n)\geq n-m$.

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 $v=(1,0,\ldots ,0)$, 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 $i=2,\ldots ,n$. Next, consider the matrix $B={\frac {1}{v^{t}v}}wv^{t}$, so that $Bv=w$ as well, and $Be_{i}=0,i=2,\ldots ,n$. 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 $\dim({\textrm {im}}~T)+\dim(\ker T)=\dim(U\times W)$ 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 ${\textrm {im}}~T=U\oplus W$, 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

${\begin{pmatrix}a_{11}m_{11}b_{11}+a_{21}m_{12}b_{11}+a_{11}m_{21}b_{12}+a_{21}m_{22}b_{12}\\a_{12}m_{11}b_{11}+a_{22}m_{12}b_{11}+a_{12}m_{21}b_{12}+a_{22}m_{22}b_{12}\\a_{11}m_{11}b_{21}+a_{21}m_{12}b_{21}+a_{11}m_{21}b_{22}+a_{21}m_{22}b_{22}\\a_{12}m_{11}b_{21}+a_{22}m_{12}b_{21}+a_{12}m_{21}b_{22}+a_{22}m_{22}b_{22}\end{pmatrix}}$.

Exercise 2.3

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