Diagonizable linear operator and invariant subspace question

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.

• Is $V$ finite-dimensional? – Roland Feb 11 '14 at 10:23
• 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). – Daniel Fischer Feb 11 '14 at 10:24
• I now realized that the fault in my thinking was this: there exist eigenvectors that span $U$, such that $W<=U$, but not necessarily $W=U$. – ILoveLev Feb 11 '14 at 11:05

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.

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.