The proof of theorem 2 of section 58 (Jordan Form) of Halmos' Finite-Dimensional Vector Spaces says the following (this is near the end of the first paragraph in page 114):
Since the only proper values of $A$ on $M_j$ is $\lambda_j$, and since $A$ on $N_j$ doesn't have the proper value $\lambda_j$ (that is, $A - \lambda_j$ is invertible on $N_j$), it follows that the dimension of $M_j$ is exactly $m_j$ and that each of the subspaces $M_j$ is disjoint from the span of all others.
I'm wondering why that last statement is true:
and that each of the subspaces $M_j$ is disjoint from the span of all others
I can't follow it. On my own, I managed to prove that any two $M_i$ and $M_j$, with $i \ne j$, are disjoint. However, I couldn't manage to prove that $M_j$ is disjoint from the span of all others, which is the sum of all those other subspaces.
Can anyone shed any light on this? I'm sort of stuck on this for about three days.
Here’s one way to show that $M_j$ is disjoint from the span of the other $M_i$.
Suppose we have $v_1+\cdots+v_d=0$, where $v_i\in M_i$. Let $A_i=A-\lambda_iI$ and $r_i$ such that $A_i^{r_i}v_i=0$. Set $S=A_2^{r_2}\cdots A_d^{r_d}$. Then $Sv_i=0$ for $i\ge 2$, so $Sv_1=0$.
Now, suppose that $v_1\ne0$. Consider the sequence $$ v_1,\, A_dv_1,\dots,A_d^{r_d}v_1,\,A_{d-1}A_d^{r_d}v_1,\dots,\, A_{d-1}^{r_{d-1}}A_d^{r_d}v_1,\dots,\, A_2^{r_2}\cdots A_d^{r_d}v_1=Sv_1. $$ $M_1$ is invariant under each of the $A_i$ in this sequence, so each vector in this sequence is an element of $M_1$. Since $Sv_1=0$, there is a vector $w\ne 0$ in this sequence such that $A_iw=0$ for some $i\ge2$, i.e., $w\in \ker A_i$, but this is impossible since the eigenspaces are pairwise disjoint. Therefore, $v_1=0$ and, by symmetry, all of the $v_i$ are $0$.
There’s probably a more straightforward proof, but I’d have to take a look at Halmos’ proof again to find it.