2
$\begingroup$

The question is: Given a diagonizable linear operator $T$ on vector space $V$, there exists a non zero subspace of $V$, $W$, such that $W$ is $T$ invariant. Prove that there exist eigenvectors of $T$, $v_1, v_2,... v_k$ such that $W = Span{(v_1, v_2,... v_k)}$.

I saw this question on here and the answer given, but I wanted to know if there was a different way to solve it.

Is this question not as easy as saying: $W$ is a subspace of $V$. There exist eigenvetors that span $V$, therefore some subset of those eigenvectors spans any subspace of $V$, including $W$. I realize there's probably a mistake in my logic, but where?

Thanks for your time.

$\endgroup$
  • $\begingroup$ Is $V$ finite-dimensional? $\endgroup$ – Roland Feb 11 '14 at 10:23
  • 1
    $\begingroup$ Say the matrix of $T$ is $\begin{pmatrix} 3 & 0 \\ 0 & 1\end{pmatrix}$. The eigenvectors are (up to multiplicative constants) $(1,0)$ and $(0,1)$. $W = \mathbb{R}\cdot (1,1)$ is a subspace, but not spanned by eigenvectors (of course $W$ is not invariant, but that's the point). $\endgroup$ – Daniel Fischer Feb 11 '14 at 10:24
  • $\begingroup$ I now realized that the fault in my thinking was this: there exist eigenvectors that span $U$, such that $W<=U$, but not necessarily $W=U$. $\endgroup$ – ILoveLev Feb 11 '14 at 11:05
0
$\begingroup$

I'm missing a couple of details here, but...

Since $W$ is invariant under $T$, $T|_W$, the restriction of $T$ to $W$, makes sense. It should then follow (somehow) from the fact that $T$ is diagonalisable on $V$ that $T|_W$ is diagonalisable on $W$. Since $T|_W$ is diagonalisable on $W$, there exists a basis $(v_1, \ldots, v_k)$ of $W$ consisting of eigenvectors of $T|_W$. In particular, span$(v_1, \ldots, v_k) = W$, and $(v_1, \ldots, v_k)$ are also eigenvectors of $T$.

Any help with the critical step is appreciated.

$\endgroup$
0
$\begingroup$

To give a little bit more detail on 0-0-0's answer: Let $V$ be an $\mathbb{F}$-vector space of finite dimension.

We first prove the following theorem:


Theorem: Let $\phi\in\mathsf{Hom}(V,V)$, $U\subseteq V$ invariant under $\phi$. Let $\phi|_U:U\to U$ be the restriction of $\phi$ to $U$. Then $q_{\phi|_U}$ divides $q_\phi$ where $q_\phi$ is the minimal polynomial of $\phi$, similar for $\phi|_U$.

Proof: Let $q_\phi$ be the minimal polynomial, i.e. $q_\phi(\phi)=\mathbf{0}$ where $\mathbf{0}$ is the null-endomorphism. Now still $q_\phi(\phi)|_U=\mathbf{0}$. It is straightforward to check, as $U$ is invariant, that $q_\phi(\phi)|_U=q_\phi(\phi|_U)=\mathbf{0}$. Now, per definition $q_{\phi|_U}(\phi|_U)=\mathbf{0}$ and as this minimal polynomial generates the ideal $$I(\phi|_U)=\{p\in\mathbb{F}[X]\mid p(\phi|_U)=\mathbf{0}\}$$ we have $q_{\phi|_U}\mid q_\phi$. $\Box$


As a corollary, we obtain the following:


Corollary: Let $\phi\in\mathsf{Hom}(V,V)$. If $\phi$ is diagonalizable and $U\subseteq V$ is invariant under $\phi$, then $\phi|_U:U\to U$ is diagonalizable.

Proof: $\phi$ is diagonalizable iff $$q_\phi=\prod_{i=1}^m(X-\lambda_i)$$ for pairwise distinct eigenvalues $\lambda_i$ of $\phi$. Now, by the above theorem, we have that $q_{\phi|_U}$ divides $q_\phi$, i.e. also $q_{\phi|_U}$ splits into distinct linear factors with algebraic multiplicity 1. Thus $\phi|_U$ is diagonalizable. $\Box$


Now let $U\subseteq V$ be a $\phi$-invariant subspace for $\phi\in\mathsf{Hom}(V,V)$ and $\mathrm{dim}(U)=k$. By the corollary, we have that $\phi|_U$ is diagonalizable, i.e. there exists an eigenbasis $B=(v_1,\dots,v_k)$ for $U$ s.t. $\phi|_U$ is represented by a diagonal matrix. Now, the vectors in this eigenbasis for $U$ are obviously also eigenvector w.r.t. $\phi$. Thus $U=\mathrm{span}(v_1,\dots,v_k)$ for these basis vectors as desired.

We can summarize this good with the following theorem:


Theorem: Let $\phi\in\mathsf{Hom}(V,V)$ be diagonalizable. Then a subspace $U\subseteq V$ with $\mathrm{dim}(U)=k$ is $\phi$-invariant iff $U=\mathrm{span}(v_1,\dots,v_k)$ for eigenvectors $v_i$.


The one direction we just proved, the other is a simple observation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.