Diagonalizable transformation restricted to an invariant subspace is diagonalizable

Question

Suppose $V$ is a vector space over $\mathbb{C}$, and $A$ is a linear transformation on $V$ which is diagonalizable. I.e. there is a basis of $V$ consisting of eigenvectors of $A$. If $W\subseteq V$ is an invariant subspace of $A$ (so $A(W)\subseteq W$), show that $A|_W$ is also diagonalizable.

I tried supposing $A$ has distinct eigenvalues $\lambda_1,\ldots,\lambda_m$, with $V_i=\{v\in V: Av=\lambda_i v\}$. Then we can write $V=V_1\oplus\cdots\oplus V_m,$ but I'm not sure whether it is true that

$$W=(W\cap V_1)\oplus\cdots\oplus (W\cap V_m),.$$

If it is true, then we're done, but it may be wrong.

Answer 1

Theorem. A linear transformation is diagonalizable if and only if its minimal polynomial splits and has no repeated factors.

Proof. This follows by examining the Jordan canonical form, since the largest power of $(x-\lambda)$ that divides the minimal polynomial is equal to the size of the largest block of corresponding to $\lambda$ of the Jordan canonical form of the linear transformation. (Use the fact that every irreducible factor of the characteristic polynomial divides the minimal polynomial, and that the characteristic polynomial must split for the linear transformation to be diagonalizable to argue that you can restrict yourself to linear transformations with Jordan canonical forms). QED

Theorem. Let $A$ be a linear transformation on $V$, and let $W\subseteq V$ be an $A$-invariant subspace. Then the minimal polynomial of the restriction of $A$ to $W$, $A|_{W}$, divides the minimal polynomial of $A$.

Proof. Let $B=A|_{W}$, and let $\mu(x)$ be the minimal polynomial of $A$. Since $\mu(A)=0$ on all of $V$, the restriction of $\mu(A)$ to $W$ is $0$; but $\mu(A)|_{W} = \mu(A|_{W}) = \mu(B)$. Since $\mu(B)=0$, then the minimal polynomial of $B$ divides $\mu(x)$. QED

Corollary. If $A$ is diagonalizable, and $W$ is $A$-invariant, then the restriction of $A$ to $W$ is diagonalizable.

Proof. The minimal polynomial of $A$ splits and has no repeated factors; since the minimal polynomial of $A|_W$ divides a polynomial that splits and has no repeated factors, it follows that it itself has no repeated factors and splits. Thus, the restriction of $A$ to $W%$ is also diagonalizable. QED

Answer 2

This theorem is true for arbitrary $V$ (over an arbitrary field $\mathbb{F}$).

We can prove the following

Lemma. If $v_1 + v_2 + \cdots + v_k \in W$ and each of the $v_i$ are eigenvectors of $A$ corresponding to distinct eigenvalues, then each of the $v_i$ lie in $W$.

Proof. Proceed by induction. If $k = 1$ there is nothing to prove. Otherwise, let $w = v_1 + \cdots + v_k$, and $\lambda_i$ be the eigenvalue corresponding to $v_i$. Then:

$$Aw - \lambda_1w = (\lambda_2 - \lambda_1)v_2 + \cdots + (\lambda_k - \lambda_1)v_k \in W.$$

By induction hypothesis, $(\lambda_i - \lambda_1)v_i \in W$, and since the eigenvalues $\lambda_i$ are distinct, $v_i \in W$ for $2 \leq i \leq k$, then we also have $v_1 \in W$. $\quad \square$

Now each $w \in W$ can be written as a finite sum of nonzero eigenvectors of $A$ with distinct eigenvalues, and by the Lemma these eigenvectors lie in $W$. Then we have $W = \bigoplus_{\lambda \in F}(W \cap V_{\lambda})$ as desired (where $V_{\lambda} = \{v \in V\mid Av = \lambda v\}$).

Answer 3

Here is a minor variation of Zorn's very nice argument. I'll use Zorn's notation:

Let $w=v_1 + v_2 + \cdots + v_k$ be in $W$, each $v_i$ being a $\lambda_i$-eigenvector of $A$, and the $\lambda_i$ being distinct.

It suffices to check that each $v_i$ is in $W$.

But this is clear since

$$v_i=\left(\prod_{j\neq i}\ \frac{A-\lambda_j\,I}{\lambda_i-\lambda_j}\right)(w)\quad.$$