Geometric multiplicity of eigenvalues in a diagonal block matrix

Question

I'm trying to prove that the geometric multiplicity of an eigenvalue in a diagonal block matrix is the sum of the geometric multiplicities of the eigenvalue with respect to every block. I know that if I have a diagonal block matrix with $k$ blocks and I take the associated endomorphism(fixed a basis) $f:V \to V$, then $V=W_1 \oplus ... \oplus W_k$, where $W_i$ is a $f$-invariant subspace $\forall i$. So I can consider the induced endomorphisms on each $W_i$(let's call it $f_i$), and clearly the eigenspace of and eigenvalue $\lambda$ with respect to $f_i$ is $V_\lambda \cap W_i$, where $V_\lambda$ is the autospace of $\lambda$ with respect to $f$. Clearly these "induced eigenspaces" are still in direct sum, so I can use Grassman relation to obtain: $$\sum_{i=1}^{k} \dim(V_\lambda \cap W_i)=\dim((V_\lambda \cap W_1)\oplus \cdots \oplus (V_\lambda \cap W_k))$$ The first member is the sum of the geometric multiplicities with respect to every block, so I have to prove that: $$(V_\lambda \cap W_1)\oplus \cdots \oplus (V_\lambda \cap W_k)=V_\lambda$$ I'm having some troubles with this last step, could you help me please?

Answer 1

We must show that $V_{\lambda} \subseteq (V_\lambda \cap W_1)\oplus \cdots \oplus (V_\lambda \cap W_k)$. So, suppose that $x \in V_\lambda$.

Because $x = W_1 \oplus \cdots \oplus W_k$, there exist $x_j \in W_j$ (for $j = 1,\dots,k$) such that $x = x_1 + \cdots + x_k$. Because $f(x) = \lambda x$, we have $$ f_1(x_1) + \cdots + f_k(x_k) = f(x_1 + \cdots + x_k) = \lambda(x_1 + \cdots + x_k) = \lambda x_1 + \cdots + \lambda x_k. $$ Because $f_j(x_j) \in W_j$ for each $j$ and because $W_1 \oplus \cdots \oplus W_k$ is a direct sum, we have $$ f_1(x_1) + \cdots + f_k(x_k) = \lambda x_1 + \cdots + \lambda x_k \implies f_j(x_j) = \lambda x_j, \quad j = 1 ,\dots,k. $$ So, it is indeed the case that $x \in (V_\lambda \cap W_1)\oplus \cdots \oplus (V_\lambda \cap W_k)$.