Abstract Nonsense

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

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

7.

Problem:

a) Prove that if $T_1,T_2$ and $T_3$ are linear transformations on a two-dimensional vector space $\mathscr{V}$, then $\left(T_1T_2-T_2T_1\right)^2$ commutes with $T_3$.

b) Is the conclusion of a) true for higher dimensions?

Proof:

a) Maybe there is some clever way to do this, but right now I’m missing it.

Let’s first prove a lemma:

Lemma: Let $A,B\in\text{Mat}_2\left(F\right)$, then $\left(AB-BA\right)^2=\lambda I_2$ for some $\lambda$.

Proof: Let

$A=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$

and

$B=\begin{pmatrix}e & f\\ g & h\end{pmatrix}$

A messy but quick calculation shows that

$AB-BA=(bg-cf)\begin{pmatrix}1 & k_1\\ k_2 & -1\end{pmatrix}$

where $k_1,k_2$ are two constants. We see then that

$\left(AB-BA\right)^2=\left(bg-cf\right)^2(k_1k_2+1)I_2$

from where the conclusion follows. $\blacksquare$

Remark: The only thing that makes this problem manageable, is that the last equality held regardless of what $k_1,k_2$ are. Put more directly

$\begin{pmatrix}1 & m\\ n & -1\end{pmatrix}^2=(mn+1)I_2$

for any $m,n\in F$.

Note then that by this lemma, for any $A,B,C\in\text{Mat}_2\left(F\right)$ we have that

$\left(AB-BA\right)^2C=\lambda C=C\lambda=C\left(AB-BA\right)^2$

Thus, given any fixed ordered basis $\mathcal{B}$ for $\mathscr{V}$ we see that

\begin{aligned}\left[\left(T_1T_2-T_2T_1\right)^2\right]_{\mathcal{B}}\left[T_3\right]_{\mathcal{B}} &= \left(\left[T_1\right]_{\mathcal{B}}\left[T_2\right]_{\mathcal{B}}-\left[T_2\right]_{\mathcal{B}}\left[T_1\right]_{\mathcal{B}}\right)^2\left[T_3\right]_{\mathcal{B}}\\ &= \left[T_3\right]_{\mathcal{B}}\left(\left[T_1\right]_{\mathcal{B}}\left[T_2\right]_{\mathcal{B}}-\left[T_2\right]_{\mathcal{B}}\left[T_1\right]_{\mathcal{B}}\right)^2\\ &= \left[T_3\right]_{\mathcal{B}}\left[\left(T_1T_2-T_2T_1\right)\right]_{\mathcal{B}}\end{aligned}

From where it follows by previous theorem that $T_3\left(T_1T_2-T_2T_1\right)^2=\left(T_1T_2-T_3T_1\right)^2 T_3$.

b) The answer is no. Recall from a previous theorem that the center of $\text{Mat}_n\left(F\right)$ is the set of all multiplies of $I_n$. In particular, we see that if $T_1,T_2$ were to satisfy the desired conditions then $\left(T_1T_2-T_2T_1\right)^2$ would  be central (in the center). But, this would clearly imply that their induced matrix representations, under some arbitrary but fixed ordered basis $\mathcal{B}$, would have the property that $\left(\left[T_1\right]_{\mathcal{B}}\left[T_2\right]_{\mathcal{B}}-\left[T_2\right]_{\mathcal{B}}\left[T_1\right]_{\mathcal{B}}\right)^2$ is central, and thus a multiple of $I_n$. But, taking any $T_1$ and $T_2$ and ordered basis $\mathcal{B}$ such that

$\left[T_1\right]_{\mathcal{B}}=\begin{pmatrix}1 & 1 & \cdots & 1\\ 1 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1\end{pmatrix}$

and

$\left[T_2\right]_{\mathcal{B}}=\begin{pmatrix}1 & 1 & \cdots & 1\\ 1 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 0\end{pmatrix}$

can easily be shown to dissatisfy this condition if $n\geqslant 3$. Thus, the answer is no.

8.

Problem: Let $A\in\text{End}\left(\mathbb{C}^2\right)$ defined by $A\left((\zeta_1,\zeta_2)\right)=(\zeta_1+\zeta_2,\zeta_2)$. Prove that if a linear transformation $B$ commutes with $A$, then there exists a polynomial $p$ such that $B=p(A)$.

9.

Problem: For which of the following polynomials $p$ and matrices $A$ is it true that $p(a)=[0]$?

a) $p(t)=t^3-3t^2+3t-1$, $A=\begin{pmatrix}1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix}$

b) $p(t)=t^2-3t$, $A=\begin{pmatrix}1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1\end{pmatrix}$

c) $p(t)=t^3+t^2+t+1$, $A=\begin{pmatrix}1 & 1 & 0\\ 1 & 1 & 1\\ 0 & 1 & 1\end{pmatrix}$

d) $p(t)=t^3-2t$, $A=\begin{pmatrix}0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{pmatrix}$

Proof:

I’m not sure exactly how do ‘show’ the calculations, so I’ll just state the result and leave the rest to the reader.

a) This does satisfy $p(A)=[0]$

b) This does satisfy $p(A)=[0]$.

c) This does not, since $p(A)=4\begin{pmatrix}2 & 2 & 1\\ 2 & 3 & 2\\ 1 & 2 & 2\end{pmatrix}$

d) This does satisfy $p(A)=[0]$