Equality of eigenspace and generalized eigenspace when endomorphism is diagonalizable. Easier solution?

Question

If $V$ is a finite-dimensional vector space and $A\in \text{End}(V)$ is an endomorphism. Define $$ M_\lambda = \left\{ x \in V \mid \exists k > 0 \colon (A-\lambda)^k x = 0 \right\} $$ and $$ V_\lambda = \left\{ x \in V \mid A x = \lambda x \right\}$$.

The idea is to show that if $A$ is diagonalizable, then $V_\lambda = M_\lambda$.

---

I have found a solution which goes as follows: trivially $V_\lambda \subset M_\lambda$. Now, if $A$ is diagonalizable, we have a reduction of $V$ as a direct sum $$ V = \oplus_{\lambda \in \sigma (A)} V_\lambda $$ where $\sigma(A)$ is the spectrum of $A$ (the set of eigenvalues).

Likewise, we also have the reduction $$ V = \oplus_{\lambda \in \sigma(A)} M_{\lambda} $$ if the minimal polynomial of $A$ splits which it does since $A$ is diagonalizable. Now, considering that $\dim V_\lambda \leq \dim M_\lambda$, taking degrees of both sides we must have that $\dim V_\lambda = \dim M_\lambda$ for all $\lambda \in \sigma(A)$, so $V_\lambda = M_\lambda$ for all $\lambda$.

---

My question is: is there a more direct approach to this problem? I am wondering whether one can take some $x \in M_\lambda$ and show directly that $x \in V_\lambda$?

Answer 1

Clearly $0 \in M_\lambda$ and $0 \in V_\lambda$, so suppose that $x \in M_\lambda$ where $x \neq 0$. By definition, there exists some $k > 0$ such that $(A - \lambda)^k x = 0$, and assume $k$ is smallest of such $k$'s. We can write $Ax = \lambda x + (A - \lambda) x$ and so $$(A - \lambda)^k Ax = \lambda (A - \lambda)^k x + (A - \lambda)^{k+1}x = 0,$$ meaning that $M_\lambda$ is $A$-invariant. By $A$-invariance of $M_\lambda$, if $A$ is diagonalizable then there exists a basis $\{v_j\}_{j = 1}^{m}$ of $M_\lambda$ and scalars $\{\mu_j\}_{j = 1}^{m}$ such that $Av_j = \mu_j v_j$ for each $j$ and $m = \dim(M_\lambda)$. Then $x = \sum_{j = 1}^{m} c_j v_j$ for some scalars $c_j$. Now, $$0 = (A - \lambda)^k x = \sum_{j = 1}^{m} c_j (A - \lambda)^k v_j = \sum_{j = 1}^{m} c_j (\mu_j - \lambda)^k v_j.$$ Since the $v_j$'s are linearly indepenedent, it follows that $c_j (\mu_j - \lambda)^k = 0$ for each $1 \leq j \leq m$. Since $x \neq 0$, there exist some $j$'s such that $c_j \neq 0$. For each such $j$, $(\mu_j - \lambda)^k = 0$, which implies $\mu_j - \lambda = 0$. It follows that $k = 1$ so $x \in V_\lambda$.

Answer 2

You don't need $V$ to be finite-dimensional.

Assume wlog $\lambda=0$. Since $A$ is diagonalizable, so is its restriction $B$ to the invariant subspace $M_0$. Let $\mu$ be some eigenvalue of $B$. Any associated eigenvector $x\ne0$ satisfies both $$A(x)=\mu x\quad\text{and}\quad A^k(x)=0$$ for some integer $k>0$. We derive successively $\mu^kx=0$, then $\mu x=0$, and finally $\mu=0$, so $B$ is a diagonalizable operator whose unique eigenvalue is $0$, which means that $B=0$, in other words $M_0\subseteq V_0$.