# Abstract Nonsense

## Halmos Sections 39 and 40: Invariance and Reducibility

Point of post: In this post we complete the problems at the end of sections 39 and 40 in Halmos.

1.

Problem: Find a finite dimensional vector space $\mathscr{V}$ and an endomorphism $T$ which leaves only $\{\bold{0}\}$ and $\mathscr{V}$ invariant.

Proof: Take $\mathbb{R}^2$ and the transformation identified with the matrix $\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$. So, let $\{\bold{0}\}<\mathscr{W}< \mathscr{V}$. Then, since $\dim_{\mathbb{R}}\mathbb{R}^2=2$ and we know that $\dim_{\mathbb{R}}\mathscr{W}=1$. So, let $\mathscr{W}=\text{span}\left\{\begin{pmatrix}\alpha\\ \beta\end{pmatrix}\right\}$. Then, for $\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$ to leave $\mathscr{W}$ invariant we must have that

$\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}\alpha\\ \beta\end{pmatrix}=\gamma\begin{pmatrix}\alpha\\ \beta\end{pmatrix}$

for some $\gamma\in\mathbb{R}-\{0\}$. Note though that

$\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}\alpha\\ \beta\end{pmatrix}=\begin{pmatrix}-\beta\\ \alpha\end{pmatrix}$

Comparing this though we see that if there existed such a $\gamma$ then

$\alpha=-\gamma\beta$

and

$\beta=\gamma\alpha$

Or,

$\beta=\gamma(-\gamma\beta)=-\gamma^2\beta$

But, this then implies that $\beta=-\beta$ or that $\beta=0$  and thus $\alpha=0$. But, then this implies that $\mathscr{W}=\{\bold{0}\}$ contradictory to assumption.

2.

Problem: Prove that the differentiation operator $D:\mathbb{C}_n[x]\to\mathbb{C}_n[x]$ leaves invariant the subspaces $\mathbb{C}_m[x]$ for every $m\in[n]$. Is there a complement of $\mathbb{C}_m[x]$ which reduces $D$?

Proof: The first part follows immediately since $\deg(p)\leqslant m\implies \deg\left(D(p)\right)\leqslant m-1\leqslant m$. For the second part we merely need notice that

3.

Problem: Prove that if $\mathscr{V}$ is an $F$-space and $\mathscr{W}_1,\mathscr{W}_2\leqslant\mathscr{V}$ and $T\in\text{End}\left(\mathscr{V}\right)$ is such that $\mathscr{W}_1,\mathscr{W}_2$ are invariant under $T$ prove that $\mathscr{W}_1+\mathscr{W}_2$ is invariant under $T$.

Proof: This follows immediately since

$T\left(\mathscr{W}_1+\mathscr{W}_2\right)=T\left(\mathscr{W}_1\right)+T\left(\mathscr{W}_2\right)\subseteq\mathscr{W}_1+\mathscr{W}_2$

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December 23, 2010 -

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