# Abstract Nonsense

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

Point of post: In this post I will complete the problems listed at the end of sections 37 and 38 of Halmos.

Remark: For those who are just interested in the solutions to Halmos and haven’t read my side-along postings you will probably need to see the series of posts for which this and this are the first posts for notation.

1.

Problem: Let $A\in\text{End}\left(\mathbb{C}_n[x]\right)$ be defined by $\left(A(p)\right)(x)=p(x+1)$, and let $\mathcal{B}$ be the canonical ordered basis $(1,x,\cdots,x^{n-1})$. Find $\left[A\right]_{\mathcal{B}}$.

Proof: We note that for each $j\in[n-1]$ we have that

$\displaystyle A\left(x^j\right)=(x+1)^j=\sum_{i=0}^{j}{n\choose j} x^{i}=\sum_{i=0}^{n-1}{j\choose i}x^i$

where we’ve appealed to the convention that $\displaystyle {k\choose \ell}=0$ if $\ell>k$. Thus

$\displaystyle \left[A\right]_{\mathcal{B}}=\begin{pmatrix}{1 \choose 1} & {2\choose 1} & {3 \choose 1} & \cdots & {n\choose 1}\\ {1 \choose 2} & {2 \choose 2} & {3\choose 2} & \cdots & {n\choose 2}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ {1 \choose n-1} & {2 \choose n-1} & {3 \choose n-1} & \cdots & {n\choose n-1}\\ {1 \choose n} & {2\choose n} & {3 \choose n} & \cdots & {n\choose n}\end{pmatrix}$

2.

Problem: Find the matrix of the operation of the operation of conjugation on $\mathbb{C}$, considered as a real vector space with respect to the canonical ordered basis $\mathcal{B}=(1,i)$.

Proof: Let $T$ be the conjugation operator. We see then that

$T(1)=1+0i$

and

$T(i)=01+(-1)i$

so that

$\left[T\right]_{\mathcal{B}}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$

3.

Problem: Let $\pi\in S_n$ then compute $\left[A\right]_{\mathcal{B}}$ where $\mathcal{B}=(e_1,\cdots,e_n)$ (where $\{e_1,\cdots,e_n\}$ is the canonical basis for $\mathbb{C}^n$) and $A((\zeta_1,\cdots,\zeta_n))=(\zeta_{\pi(1)},\cdots,\zeta_{\pi(n)})$.

Proof: It’s fairly easy to see that

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

4.

Problem: Let $P=\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}$ and consider $T\in\text{End}\left(\text{Mat}_2(\mathbb{R})\right)$ given by $T(X)=PX$. Find $\left[T\right]_{\mathcal{B}}$ where

$\mathcal{B}=\left(\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}\right)$

Proof: We note that

$T\left(\begin{pmatrix}1 &0\\ 0 &1\end{pmatrix}\right)=\begin{pmatrix}1 & 0\\1 & 0\end{pmatrix}$

$T\left(\begin{pmatrix}0 &1\\ 0 & 0\end{pmatrix}\right)=\begin{pmatrix}0 &1\\ 0 & 1\end{pmatrix}$

$T\left(\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}\right)=\begin{pmatrix}1 & 0\\ 1 & 0\end{pmatrix}$

and

$T\left(\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}\right)=\begin{pmatrix}0 & 1\\ 0 & 1\end{pmatrix}$

So that

$\left[T\right]_{\mathcal{B}}=\begin{pmatrix}1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\end{pmatrix}$

5.

Problem: Let $\mathscr{V}$ be an $n$-dimensional $F$-space and $P\in\text{End}\left(\mathscr{V}\right)$. Let then

$T:\text{End}\left(\mathscr{V}\right)\to\text{End}\left(\mathscr{V}\right):X\mapsto PX$

Under what conditions is $T$ invertible?

Proof: The claim is that $T$ is an isomorphism if and only if $P$ is an isomorphism. To see that this condition is sufficient we note that if $P$ is an isomorphism and $T(X)=T(Y)$ then $PX=PY$ and thus since $P$ is an isomorphism we may conclude that $X=Y$. Thus, $T$ is a monomorphism, but by prior theorem we may then conclude that $T$ is an isomorphism.

To see that $P\in\text{GL}\left(\mathscr{V}\right)$ is a necessary condition we note that if $T$ is an isomorphism then $T$ is, in particular, an endomorphism. Thus, there exists some $X_0\in\text{End}\left(\mathscr{V}\right)$ such that $\mathbf{1}=T(X_0)=PX_0$. But, the existence of a right inverse for $P$ implies that $P$ is an endomorphism, and thus since $\mathscr{V}$ is finite dimensional we may conclude that $P$ is an isomorphism.

6.

Problem: Prove that if $I,J$ and $K$ are the complex matrices

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

respectively. Show that $I^2=J^2=K^2=-I_2$, $IJ=-JI=K$, $JK=-KJ=I$, and $KI=-IK=J$.

Proof: This is purely computational, and not even cleverly so. That said, I would like to remark that this is a group under matrix multiplication which is isomorphic to the quaternions.