# Restriction operator $T_W$ is diagonalizable if T is diagonalizable.

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?

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.
Well check that $$m_T(T_W) = 0$$. Therefore by minimality, $$m_{T_W}\mid m_T$$
• why must $m_T(T_W) = 0$ ?
• 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 $P$, $P(T)(x) = P(T_W)(x)$) Oct 13, 2019 at 9:55