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!

  • $\begingroup$ 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

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.


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.