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Let $T$ be a linear operator on a finite-dimensional vector space $V$. Deduce that if the characteristic polynomial of $T$ splits, then any non-trivial $T$-invariant subspace of $V$ contains an eigenvector of $T.$

Let $W$ be a $T$-Invariant subspace. $W\neq\{0\}$($\because$ Given that $W$ is non-trivial). The characteristic polynomial of $T$ restricted to $W$ divides the characteristic polynomial of $T$. I am not able to proceed further. I need to prove $\exists v\neq 0, v\in W: T(v)=\lambda v$. Please help me.

Michael Hardy
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1 Answers1

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Hint: Since the characteristic polynomial of $T$ splits, so does characteristic polynomial of $T|W$.

lhf
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    characteristic polynomial of $T|{W}$ splits. Let $W_1$ be the matrix representation $T|{W}$ of the standard basis. $\det(W_1-tI)=0$ has solution. so, there exists $v\in W: T(v)=\lambda v$. right? –  Jun 03 '18 at 15:02
  • Am I correct?can you please tell me? –  Jun 04 '18 at 01:29