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Let $V$ be a finite-dimensional vector space and $T$ be a linear operator on $V$ whose minimal polynomial is a product of linear factors.

Prove that if $W$ is a proper $T$-invariant subspace of $V$, then there exists a vector $v\in V\backslash W$ such that $(T-cI)v\in W$ for some eigenvalue $c$ of $T$.

I have no idea how to attempt at all. Does anyone have ideas?

Thank you for your help!

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  • $\begingroup$ I think this could work: consider the the linear operator induced by $T$ on the quotient space $V/W$. $\endgroup$ – zipirovich Jul 31 '17 at 17:02
  • $\begingroup$ It might help to notice that $T$ is diagonalizable, and so can be expressed as a sum of projections onto its eigenspaces. Then again, this problem might be a step towards proving that. $\endgroup$ – amd Jul 31 '17 at 23:49
  • $\begingroup$ @amd Does 'minimal polynomial is a product of linear factors' mean that such minimal polynomial doesn't have multiple root? If does, then I cannot apply the argument you suggested. $\endgroup$ – bellcircle Aug 4 '17 at 13:07
  • $\begingroup$ @zipirovich I finally figured out the answer. Could you check if there is any error? $\endgroup$ – bellcircle Aug 4 '17 at 14:03
  • $\begingroup$ @amd I finally figured out the answer. Could you check if there is any error? $\endgroup$ – bellcircle Aug 4 '17 at 14:03
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Since the minimal polynomial factors into linear factors, $T$ has Jordan form.

Let $J$ be the Jordan form of $T$ and $\mathfrak B= \{v_i\}_{i=1}^{n}$ be the basis of $V$ s.t. $[T]_{\mathfrak B}=J$ where $\dim V=n$. Since $W<V$, there exists $v_i \in \mathfrak B$ which is not in $W$. Let $\lambda$ be the diagonal entry of the Jordan block corresponding to $v_i$. Then $T(v_i)$ is either $\lambda v_i$ or $\lambda v_i+v_{i+1}$. i.e. $(T-\lambda I)v_i=0$ or $v_{i+1}$. where $v_i$ and $v_{i+1}$ shares the common Jordan block.

In this way, for general $k \ge 0$, if $v_{i+k}\notin W$ then there are two cases:

Case 1: $(T-\lambda I)v_{i+k} =0$ or $(T-\lambda I)v_{i+k} =v_{i+k+1}\in W$

Case 2: $(T-\lambda I)v_{i+k}=v_{i+k+1} \notin W$

In the Case 1, $v_{i+k}$ and $\lambda$ are the desired vector and eigenvalue. In the Case 2, then $v_{i+1}\notin W$, then $(T-\lambda I)v_{i+k+1}=0$ or $v_{i+k+2}$, and determine whether this is the Case 1 or 2.

Since $V$ is finite dimensional, every Jordan block of $J$ has finite size. Therefore, this algorithm must terminate at the Case 1, and we can always find the vector and eigenvalue satisfying the condition of the question.

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