# Abstract Nonsense

## Halmos Sections 34 and 35:Products and Polynomials (Pt. II)

Point of post: This post is a continuation of

6.

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

a) Prove that the set of all $B\in\text{End}\left(\mathscr{V}\right)$ for which $AB=\bold{0}$ is a subspace of $\text{End}\left(\mathscr{V}\right)$

b) Show that by a suitable choice of $A$ the dimension of the subspace described in a) can be $0,n$ or $n^2$.

c) Can every subspace of $\text{End}\left(\mathscr{V}\right)$ be gotten this way?

Proof:

a) We must merely note that if $B,B'$ both annihilate $A$, then for any $\alpha,\beta\in F$ we have that $A\left(\alpha B+\beta B'\right)=\alpha AB+\beta AB'=\bold{0}$. Thus, it follows that the set of all such $B$ is a subspace of $\text{End}\left(\mathscr{V}\right)$.

b) Let $T_0\in\text{End}\left(\mathscr{V}\right)$. It’s fairly easy to show that $\left\{v\in\mathscr{V}:T(v)=\bold{0}\right\}$ is a subspace of $\mathscr{V}$. Thus, we may find a basis $\{x_1,\cdots,x_m\}$ for it and extend this to a basis $\{x_1,\cdots,x_m,x_{m+1},\cdots,x_n\}$ for $\mathscr{V}$. We claim that

$\mathscr{B}=\left\{T_{i,j}:i\in[n]\text{ and }j\in[m]\right\}$

is a basis for $\mathscr{G}=\left\{T\in\text{End}\left(\mathscr{V}\right):T_0T=\bold{0}\right\}$. To see this we note that evidently $\mathscr{B}$ is linearly independent and $\text{span }\mathscr{B}\subseteq\mathscr{G}$, so it remains to show the reverse inclusion. To do this, suppose that $T\notin\text{span }\mathscr{B}$ but $T\in\mathscr{G}$. We note that since $T\notin\text{span }\mathscr{B}$ that

$\displaystyle T=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_{i,j}T_{i,j}$

where there is some $i_0\in[n]\text{ and }j_0\notin[m]$ such that $\alpha_{i,j}\ne 0$.  Note then that

$\displaystyle T(x_{i_0})=\sum_{j=1}^{n}\alpha_{i_0,j}T_{i_0,j}(x_{i_0})=\sum_{j=1}^{n}\alpha_{i_0,j}x_j$

and thus by assumption that $T\in\mathscr{G}$ we see that

$\displaystyle 0=T_0\left(T(x_{i_0})\right)=\sum_{j=1}^{n}\alpha_{i_0,j}T_0(x_j)=\sum_{j=m+1}^{n}\alpha_{i_0,j}T_0(x_j)=T\left(\sum_{j=m+1}^n x_j\right)$

and thus

$\displaystyle \sum_{j=m+1}^{n}\alpha_{i_0,j}x_j\in\ker T$

but by construction this is only possible if

$\displaystyle \sum_{j=m+1}^{n}\alpha_{i_0,j}x_j=\bold{0}$

and since $\{x_{m+1},\cdots,x_n\}$ is a linearly independent set this implies that $\alpha_{i_0,m+1}=\cdots=\alpha_{i_0,n}=0$ and so in particular $\alpha_{i_0,j_0}=0$ which is contradictory to our assumption. Thus, we see that if $T\notin\text{span }\mathscr{B}$ then $T\notin\mathscr{G}$ so that

$\text{End}\left(\mathscr{V}\right)-\text{span }\mathscr{B}\subseteq\text{End}\left(\mathscr{V}\right)-\mathscr{G}$

which is equivalent to

$\text{span }\mathscr{B}\supseteq\mathscr{G}$

from where it follows that $\text{span }\mathscr{B}=\mathscr{G}$. Therefore

$\dim_F\mathscr{G}=nm$

From this it’s fairly easy to find examples of transformations for which $\dim_F\mathscr{G}=0,n,n^2$ and furthermore it’s clear that the only possible dimensions of $\mathscr{G}$ are $0,n,2n,3n,\cdots,(n-1)n,n^2$.

c) No. Consider $\mathbb{R}^4$. Then, there does not exist a $T_0\in\text{End}\left(\mathbb{R}^4\right)$ such that $\left\{T\in\text{End}\left(\mathbb{R}^4\right):T_0T=\bold{0}\right\}$ has dimension $5$ since $5$ is not an integral multiple of $4$.

7.

Problem: Let $A\in\text{End}\left(\mathscr{V}\right)$ where $\mathscr{V}$ is an $n$-dimensional $F$-space and consider the map

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

Prove that $T_A$ is a linear transformation. Can every linear transformation on that space be obtained in this manner?

Proof: Almost by definition we see that

$T_A\left(\alpha X+\beta Y\right)=A\left(\alpha X+\beta Y\right)=\alpha A X+\beta A Y=\alpha T_A\left(X\right)+\beta T_A\left(Y\right)$

To prove the second part we let

$\Phi:\text{End}\left(\mathscr{V}\right)\to\text{End}\left(\text{End}\left(\mathscr{V}\right)\right):A\mapsto T_A$

This is evidently linear since

\begin{aligned}\Phi\left(\alpha A+\beta B\right)(X) &=T_{\alpha A+\beta B}\left(X\right)\\ &=\left(\alpha A+\beta B\right)X\\ &=\alpha A X+\beta B X\\ &=\alpha\left(\Phi\left(A\right)\right)(X)+\beta \left(\Phi\left(B\right)\right)(X)\end{aligned}

And since $X\in\text{End}\left(\mathscr{V}\right)$ was arbitrary it follows that $\Phi\left(\alpha A+\beta B\right)=\alpha \Phi\left(A\right)+\beta\Phi\left(B\right)$. Note then that $\Phi$ is also injective since $T_A=T_B$ implies that

$A=A\mathbf{1}=T_A\left(\mathbf{1}\right)=T_B\left(\mathbf{1}\right)=B\mathbf{1}=B$

and thus if $\Phi$ were surjective we’d have that

$\text{End}\left(\mathscr{V}\right)\cong\text{End}\left(\text{End}\left(\mathscr{V}\right)\right)$

but note that $\dim_F\text{End}\left(\mathscr{V}\right)=n^2$ and $\dim_F\text{End}\left(\text{End}\left(\mathscr{V}\right)\right)=n^4$. Thus, if $n\ne 1$ it follows that $\text{End}\left(\text{End}\left(\mathscr{V}\right)\right)\not\cong\text{End}\left(\mathscr{V}\right)$ and so $\Phi$ is not surjective, if $n=1$ then the above reasoning leads to a dimension argument that $\Phi$ must be surjective. Thus, it follows that

$\text{End}\left(\mathscr{V}\right)=\left\{T_A:A\in\text{End}\left(\mathscr{V}\right)\right\}$

if and only if $n=1$.

References:

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