I'm building up to the study of Jordan normal form, and need to prove that a vector space can be decomposed as a direct sum of generalised eigenspaces of a linear map. The proof in my notes goes:
$$ \mathrm{characteristic\ polynomial} = \prod_{i} (x – \lambda_i)^{n_i}$$
Define
$$ h_j(x) = \prod_{i \neq j} (x – \lambda_i)^{n_i} $$
And consider the subspaces $W_j = \mathrm{im}(h_j(A))$, where $A$ is my linear map. The proof then goes on to show that any vector space can be decomposed as a direct sum of the $W_j$, using the fact the $h_j$ are coprime polynomials and Euclid's algorithm and so on.
My question is: how does this show that the space is a direct sum of generalised eigenspaces? Presumably we need to show that $W_j$ is the $\lambda_j-$generalised eigenspace $V_{\lambda_j}$. I managed to show that $W_{\lambda_j} \subseteq V_{\lambda_j}$ using the Cayley-Hamilton theorem:
$$(A – \lambda_j I)^{n_j} h_j(A) = \mathrm{cp}(A) = 0 $$
And hence any vector in the image of $h_j(A)$ is annihilated by $(A – \lambda_j I)^{n_j}$. However what I can't show is the opposite: that $V_{\lambda_j} \subseteq W_{\lambda_j}$. Thanks.
Best Answer
First, note that $A-\lambda_j I$ is invertible on $W_{\lambda_i}$ for $i\neq j$. Indeed, this follows from the fact that the polynomial $x-\lambda_j$ is invertible mod $(x-\lambda_i)^{n_i}$ and $(A-\lambda_i)^{n_i}$ annihilates $W_{\lambda_i}$.
Now suppose you have $v\in V_{\lambda_j}$, and write $v=\sum v_i$ where $v_i\in W_{\lambda_j}$. Then for some $n$, $0=(A-\lambda_j I)^nv=\sum (A-\lambda_j I)^n v_i$. Furthermore, it follows from the definition of $W_{\lambda_i}$ that $v_i\in W_{\lambda_i}$ implies $(A-\lambda_j I)^n v_i\in W_{\lambda_i}$. Since $V$ is the direct sum of the $W_{\lambda_i}$, $\sum (A-\lambda_j I)^n v_i=0$ now implies each $(A-\lambda_j I)^n v_i$ must be $0$. For $i\neq j$, this implies $v_i=0$, since $A-\lambda_j I$ is invertible on $W_{\lambda_i}$. Thus $v=v_j\in W_{\lambda_j}$.