Prove $f$ is diagonalizable iff $V=W \oplus Z$ where $W,Z \subseteq V$ are $f$ invariant

Question

Let $K$ be field algebraically closed, $f\in End(V)$ prove:

$f$ is diagonalizable $\iff$ $\forall W \subset V$ invarant under $f$ exist $Z \subset V$ invariant under $f$ such that $V=W\oplus Z$

i have olny idea with one directon namely assume that $f$ is diagonalizable then exist basis $A=\{v_1, .., v_n\}$ consisted with eigenvectors so choosing any subset of $A$ we have invariant subspace $W$ and we can pick $Z$ be taking the rest vectors from $A$ to take whole $V$, bu i'm not sure if it's works

I'm assuming $V$ is a vector space over $K$, is that right?. Are we only concerned with finite-dimensional vector spaces $V$ here? With your proof idea, you're choosing $W$, when you should be working with an arbitrary given $W \subseteq V$ that's invariant under $f$. — pjs36, Mar 26 '15 at 16:43
Yes V is vector space over K, yes only finite-dimensional. Can you show yours approach ? — Jessy, Mar 26 '15 at 17:11

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

This question has essentially been asked on this site before.

For the first direction, proof is here: Existence of T-invariant complement of T-invariant subspace when T is diagonalisable
For the second direction, proof is here: Proving a diagonal matrix exists for linear operators with complemented invariant subspaces

You can use the algebraic closedness of your field instead of the fundamental theorem of algebra. Otherwise the steps are the same.

Prove $f$ is diagonalizable iff $V=W \oplus Z$ where $W,Z \subseteq V$ are $f$ invariant

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