TITLE: Show that $T$ is diagonalizable iff $T_{W_i}$ is diagonalizable for all $i$.
QUESTION [1 upvotes]: I'm trying to prove the following theorem:
"Let $T$ be a linear operator on a finite-dimensional vector space $V$,
and let $W_1$, $W_2$,..., $W_k$ be $T-$invariant subspaces of $V$ such that $V =
W_1 \oplus W_2 \oplus···\oplus W_k$. Prove that $T$ is diagonalizable if and only if $T_{W_i}$
is diagonalizable for all $i$."
I've already proven the first implication, but I'm stuck on the latter. I'm trying to use the fact that the characteristic polynomial of $T$ can be written as the product of the characteristic polynomials of the restrictions $T_{W_i}$ (since the subspaces are $T-$invariant and direct sum of $V$), but I don't know if I can prove the theorem following this path.
I can also minimal polynomial and Cayley-Hamilton if need be. Can anyone tell me if I'm on the right track? Thank you in advance!

REPLY [1 votes]: There is an easier route. 
Since $T_{W_i}$ is diagonalizable always, for each $i$ we have a basis of $W_i$, $\{v_{1,i},v_{2,i},\dots,v_{n_i,i}\}$ and an operator $S_i$ (on $W_i$) s.t. $(T(v_{j,i}))$ is always a scalar multiple of $v_{j,i}$. 
Now, take the bases described for each of the $W_i$s and put them together. Then we get a basis for $\_\_\_\_\_$. 
Then, the matrix representation of $\_\_\_\_\_$ with respect to $\_\_\_\_\_$ is obviously diagonal.
FULL EXPLANATION:

 When we combine the bases for the $W_i$s, we get a basis for $V$. Under this basis, the matrix representation of $T$ is then clearly diagonal. 

note: if there is anything incorrect in my answer, feel free to edit or leave a comment
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