# Is an invariant subspace always spanned by a set of generalized eigenvectors?

I have been trying to find an answer to this question for some time and haven't had any luck. Let me state the question formally:

Suppose $$T$$ is a linear mapping $$T:V \rightarrow V$$, where $$V$$ is an $$n$$-dimensional vector space, $$n<\infty$$. Suppose $$W\subset V$$ is an invariant subspace of $$V$$, i.e., $$W$$ is a subspace such that $$T(W) \subseteq W$$, with $$\dim(W) = k$$. Is it always the case that we can find $$k$$ generalized eigenvectors of $$T$$, $$v_1,\ldots,v_k$$, such that $$\text{span}(v_1,\ldots,v_k) = W$$?

It seems to me that this must be true, but I've had no luck proving it on my own. I've searched around and found lots of questions/answers about the "converse" question. For example, questions about proving that a generalized eigenspace (or the direct sum of several generalized eigenspaces) is $$T$$-invariant. That's not what I'm after here.

Any help would be appreciated.

• Hint: $T$ induces a linear map $W\to W$. – Wojowu Mar 6 at 14:40
• @Wojowu Thanks for the reply. So would an argument be something like: (1) $T$ induces a linear map $\tilde{T}: W \rightarrow W$. (2) Thus, $W$ is spanned by $k$ generalized eigenvectors of $\tilde{T}$. (3) A generalized eigenvector of $\tilde{T}$ must also be a generalized eigenvector of $T$. (4) Therefore $W$ is spanned by $k$ generalized eigenvectors of $T$. – CornerSolution Mar 6 at 15:59
• That's perfectly right. – Wojowu Mar 6 at 16:01
• Ah, great! So simple in the end. I should've asked for help much sooner. Thanks again! – CornerSolution Mar 6 at 16:09