# If $T \colon V\to V$ is normal and $W\subseteq V$ is $T$-invariant then $T_W$ is normal

Let $T \colon V\to V$ be a linear normal operator on a finite-dimensional vector space $V$. Suppose $W\subseteq V$ is a $T$-invariant subspace of $V$. I must prove that $T_W$ (the restriction of $T$ to $W$) is also normal.

My attempt: I proved that the minimal polynomial of $T_W$ divides the minimal polynomial of $T$ (for any linear map, not just normal). I then deduced that if $T$ is diagonalizable then $T_W$ is also diagonalizable. The problem is that I need to prove that $T_W$ is unitarily diagonalizable in order to deduce that $T_W$ is normal. But I fail to do this.

• Do you know that $T$ is normal if and only if $\Vert T^\ast x \Vert = \Vert Tx\Vert$ for all $x$? – PhoemueX Jun 19 '15 at 11:03
• @PhoemueX - yes. – user249293 Jun 19 '15 at 12:33
• Does that condition also hold for $T_W$ instead of $T$ if it holds for $T$? (you should verify that $(T_W)^\ast = T^\ast |_W$) – PhoemueX Jun 19 '15 at 12:50

As far as I see you already know that $T$ is diagonalizable but you cannot prove that the eigenspaces $V_{\lambda}$ of $T$ are orthogonal, isn'it ?. Here is a proof that they are indeed orthogonal.

First of all notice that $ker(T) = ker(T^*)$. This follows form the already notice fact that $\|Tx\| = \|T^*x\|$. Notice also that each $V_{\lambda}$ is $T^*$-invariant. Indeed, since $T$ and $T^*$ commute between them then $T^*$ commute with $T - \lambda Id$ hence $V_{\lambda} := ker(T - \lambda Id)$ is preserved by $T^*$.

Next, notice that the restriction of $T^*$ to each $V_{\lambda}$ is bijective. Indeed, since $ker(T^*) \cap V_{\lambda} = \{ 0 \}$ and each $V_{\lambda}$ has finite dimension it follows that each restriction is bijective.

Now we are ready to show that $V_{\lambda}$ is perpendicular to $V_{\beta}$ if $\lambda \neq \beta$. First suppose that $\lambda = 0$ and $\beta \neq 0$. Let $v \in V_0$ and $w \in V_{\beta}$. Since $T^*$ is bijective on $V_{\beta}$ there is $w' \in V_{\beta}$ such that $w = T^* w'$ so $$v.w = v. T^*w' = Tv . w' = 0.w' = 0$$ So $V_0 \perp V_{\beta}$ as we wanted to show.

If both $\lambda,\beta \neq 0$ then notice that $$T^* - \bar{\lambda} Id : V_{\beta} \to V_{\beta}$$ is injective hence bijective. To check this use that $\| (T^* - \bar{\lambda} Id)v \| = \| (T - \lambda Id)v \|$ for all $v \in V$. Finally, let $v \in V_{\lambda}$ and $w \in V_{\beta}$ arbitrary vectors. Since $(T^* - \bar{\lambda} Id)$ is bijective there is $w' \in V_{\beta}$ such that $(T^* - \bar{\lambda} Id)w' = w$ then $$v.w = v.(T^* - \bar{\lambda} Id)w' = (T - \lambda Id)v.w' = 0.w' = 0$$

So all eigenspaces are orthogonal between them and $T$ is unitarily diagonalizable as you wanted to show.

Summing up a key tool is that $\| T^* v \| = \|T v \|$ for all $v \in V$ iff $T$ is normal as user PhoemueX commented. The problem is that it is not clear if $W$ is also $T^*$-invariant, isn'it ?.