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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.

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  • $\begingroup$ Hint: $T$ induces a linear map $W\to W$. $\endgroup$ – Wojowu Mar 6 at 14:40
  • $\begingroup$ @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 $T$. $\endgroup$ – CornerSolution Mar 6 at 15:59
  • $\begingroup$ That's perfectly right. $\endgroup$ – Wojowu Mar 6 at 16:01
  • $\begingroup$ Ah, great! So simple in the end. I should've asked for help much sooner. Thanks again! $\endgroup$ – CornerSolution Mar 6 at 16:09

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