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Let $T$ be a diagonalizable linear operator on the $n$-dimensional vector space $V$, and let $W$ be a subspace of $V$ which is invariant under $T$. Prove that the restriction operator $T_W$ is diagonalizable.

Since $T$ is diagonalizable, it's minimal polynomial must be of the form $m_T(x)=(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_r)$ where $\lambda_1,\lambda_2,..,\lambda_r$ are distinct. Now my claim is that minimal polynomial of restriction operator $(m_{T_W}(x))$ divides minimal polynomial of $T$. If that is the case then $m_{T_W}(x) = (x-\lambda_{a_1})(x-\lambda_{a_2})\cdots(x-\lambda_{a_k})$ where $k\in \mathbb{Z_{\ge0}}$ ,$a_1,a_2,..a_k \in$ $\{ 1,2,...,r\}$ are distinct. Hence $T_W$ will also be diagonalizable.

How do i prove that $m_{T_W}(x)$ must divide $ m_T(x) $ ? Is there a different way to solve this?

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2 Answers 2

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From Cayley's theorem, we have $m_T(T)=0$, since $T_W$ is the restriction of $T$ on $W$, $m(T_W)=0$ of course. Since $m_{T_W}$ is the minimal polynomial of $T_W$, we have $m_{T_W}\mid m_T$ by the definition of minimal polynomial.

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    $\begingroup$ Cayley's theorem is irrelevant here. $\endgroup$ Oct 9, 2019 at 10:10
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Well check that $m_T(T_W) = 0$. Therefore by minimality, $m_{T_W}\mid m_T$

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  • $\begingroup$ why must $m_T(T_W) = 0$ ? $\endgroup$
    – Sam
    Oct 13, 2019 at 8:27
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    $\begingroup$ Because $m_T(T_W) = m_T(T)_W$ : it immediately follows from.the definitions (take $x\in W$, then $T(x) = T_W(x)$ so by induction $T^k(x) = T_W^k(x)$ so for any polynomial $P$, $P(T)(x) = P(T_W)(x)$) $\endgroup$ Oct 13, 2019 at 9:55

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