Abstract Nonsense

Crushing one theorem at a time

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 - Posted by | Fun Problems, Halmos, Linear Algebra, Uncategorized | , , , , , ,

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