# Abstract Nonsense

## Halmos Sections 37 and 38: Matrices and Matrices of Linear Transformations(Pt. III)

Point of post: This is a continuation of this post.

10.

Problem: Prove that if $A$ and $B$ are the complex matrices

$\begin{pmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\end{pmatrix}$

and

$\begin{pmatrix}i & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -i & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$

respectively, and if $C=AB-iBA$ then $C^3+C^2+C=[0]$.

Problem: This is just plain old computation, we leave this to the reader.

11.

Problem: IF $A,B\in\text{End}\left(\mathscr{V}\right)$ for some $n$-dimensional $F$-space $\mathscr{V}$, and if $AB=\bold{0}$ does it follow that $BA=\bold{0}$?

Proof: No, this is definitely not true. Take $A,B$ be the unique members of $\text{End}\left(\mathbb{R}^2\right)$ such that

$\left\{\begin{array}{c}A((1,0))=(0,1)\\ A((0,1))=(0,1)\end{array}\right\}\text{ and }\left\{\begin{array}{c}B((1,0))=(1,0)\\ B((0,1))=(0,0)\end{array}\right\}$

We see then that

$(BA)((0,1))=(BA)((1,0))=\bold{0}$

so that $BA=\bold{0}$. That said, $(AB)((1,0))=(0,1)$ so that $AB\ne \bold{0}$.

12.

Problem: What happens to the matrix of a linear transformation $T$ on a $n$-dimensional $F$-space $\mathscr{V}$ with respect to the ordered basis $\mathcal{B}=(x_1,\cdots,x_n)$ when the matrix is computed with respect to $\pi\mathcal{B}=(x_{\pi(1)},\cdots,x_{\pi(n)})$ for $\in\in S_n$?

Proof: The columns are permuted via $\pi$. Put more explicitly,

$\left[T\right]_{\pi\mathcal{B}}=\left(\begin{array}{c|c|c} & & \\ T(x_{\pi(1)}) & \cdots & T(x_{\pi(n)})\\ & & \end{array}\right)$

13.

Problem:

a) Suppose that $\mathscr{V}$ is an $n$-dimensional $F$-space with basis $\{x_1,\cdots,x_n\}$. Suppose that $\alpha_1,\cdots,\alpha_n\in F$ are distinct. If $A\in\text{End}\left(\mathscr{V}\right)$ such that $A(x_j)=\alpha_j x_j,\text{ }j\in[n]$ and $B\in\text{End}\left(\mathscr{V}\right)$ is such that $AB=BA$ prove then there exists scalars $\beta_1,\cdots,\beta_n$ such that $B(x_j)=\beta_j x_j$.

b) Prove that if $B\in\text{End}\left(\mathscr{V}\right)$ and $BA=AB$ for every $A\in\text{End}\left(\mathscr{V}\right)$ then $B=\beta\mathbf{1}$ for some $\beta\in F$.

Proof:

a) We note that for each $x_i$ there exists scalars $\beta_{i,j},\text{ }j\in[n]$ such that

$\displaystyle B(x_i)=\sum_{j=1}^{n}\beta_{i,j} x_j$

So,

$A\left(B(x_i)\right)=\sum_{j=1}^{n}\alpha_j \beta_{i,j}x_j$

But, this must be equal to

$\displaystyle B(A(x_i))=B(\alpha_i x_i)=\sum_{j=1}^{n}\alpha_i \beta_{i,j}x_j$

or, upon subtraction

$\displaystyle \sum_{j=1}^{n}\beta_{i,j}(\alpha_j -\alpha_i)x_j=\bold{0}$

But, since $\{x_1,\cdots,x_n\}$ is linearly independent this implies that $\beta_{i,j}(\alpha_j-\alpha_i)=0$ for $j\in[n]$. But, since $\alpha_j\ne \alpha_i$ for $j\ne i$ we may conclude that $\beta_{i,j}=0$ for $j\ne i$. Therefore, looking back we see that this implies that

$B(x_i)=\beta_{i,i}x_i$

Since $i\in[n]$ was arbitrary the conclusion follows.

b) See here.

14.

Problem: If $\{x_1,\cdots,x_k\}$ and $\{y_1,\cdots,y_k\}$ are linearly independent sets of vectors in an $n$-dimensional $F$-space $\mathscr{V}$ then there exists some $T\in\text{GL}\left(\mathscr{V}\right)$ such that $T(x_j)=y_j$ for $j=1,\cdots,k$

Proof: Extend $\{x_1,\cdots,x_k\}$ to a basis $\{x_1,\cdots,x_k,x_{k+1},\cdots,x_n\}$ and extend $\{y_1,\cdots,y_k\}$ to a basis $\{y_1,\cdots,y_k,y_{k+1},\cdots,y_n\}$. Then, let $T$ be the unique element of $\text{End}\left(\mathscr{V}\right)$ such that $T(x_\ell)=y_{\ell}$. Since this is a bijection between bases our previous characterization of isomorphisms let’s us conclude that $T\in\text{GL}\left(\mathscr{V}\right)$.

15.

Problem: If a matrix $A=[\alpha_{i,j}]\in\text{Mat}_n\left(F\right)$ is such that $\alpha_{i,i}=0,\text{ }i\in[n]$ then there exists $B=[\beta_{i,j}],C=[\gamma_{i,j}]\in\text{Mat}_n\left(F\right)$ such that $A=BC-CB$.

Proof: We may clearly assume that $F$ has infinitely distinct elements. Thus, choosing some countable subset $\left\{\beta_i\right\}_{i\in\mathbb{N}}$ we may define $\beta_{i,j}=\beta_i\delta_{i,j}$ (where $\delta_{i,j}$ is [as usual] the Kronecker delta symbol) and

$\displaystyle \gamma_{i,j}=\begin{cases}\frac{\alpha_{i,j}}{\beta_i-\beta_j} & \mbox{if}\quad i\ne j\\ 0 & \mbox{if}\quad i=j\end{cases}$

We note then that the general term of $BC$ is

$\displaystyle \sum_{r=1}^{n}\beta_i\delta_{i,r}\gamma_{r,j}=\beta_{i}\gamma_{i,j}=\begin{cases}\frac{\alpha_{i,j}}{\beta_i-\beta_j} & \mbox{if}\quad i\ne j\\ 0 & \mbox{if}\quad i=j\end{cases}$

and the general term of $CB$ is

$\displaystyle \sum_{r=1}^{n}\gamma_{i,r}\beta_r \delta_{r,j}=\gamma_{i,j}\beta_j=\begin{cases}\beta_j \frac{\alpha_{i,j}}{\beta_i-\beta_j} & \mbox{if}\quad i\ne j\\ 0 & \mbox{if} \quad i=j\end{cases}$

thus, the general term of $BC-CB$ is

$\begin{cases}\alpha_{i,j} & \mbox{if} \quad i\ne j\\ 0 & \mbox{if}\quad i=j\end{cases}=\alpha_{i,j}$

from where the conclusion follows.

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print