# $V = W_1\oplus W_2\oplus… \oplus W_s$ we must have $W = (W_1 \cap W )\oplus (W_2\cap W)\oplus … (W_s \cap W)$?

Let $T$ be a linear operator on an $n-$dimensional vector space $V$ ($n > 1$) and $W$ is a $k-$dimensional $(0 < k < n)$, $T-$invariant subspace. Show that if $T$ has $n$ distinct eigenvalues, then for any $T-$invariant direct sum decomposition of $V = W_1\oplus W_2\oplus... \oplus W_s$ we must have $W = (W_1 \cap W )\oplus (W_2\cap W)\oplus ..... \oplus(W_s \cap W)$

I'm completely clueless about how to even attack this one. Any help is appreciated.

• What does the minus sign mean between vector spaces here? – Tobias Kildetoft Jun 23 '16 at 6:39
• Sorry, edited... – dragoboy Jun 23 '16 at 6:41

This is a bit tricky, I am not aware of a reasonably direct argument. The only line of argument I know is as follows: By the assumption on the eigenvalues, $V$ has to be the direct sum of the $n$ eigenspaces for $T$. Next, one shows that any $T$-invariant subspace of $V$ is the direct sum of some of these eigenspaces. From this the result follows easily, since any of the $n$ eigenspaces must be contained in exactly one of the $W_s$. (If you prefer toe argue in terms of matrices, the main argument is that any matrix which commutes with a diagonal matrix with all eigenvalues different is itself diagonal.)