Question: let $T$ be linear operator on finite dimensional Vector space $V$ then $T$ is diagonalizable if and only if $V$ is direct sum of one dimensional $T$-invariant subspaces.
My attempt: if $T$ is diagonalizable linear operator on $n$ dimensional Vector space $V$ then $V$ has basis say $\beta =\{v_1,...,v_n\}$ consisting of an eigenvectors of $T$.
Taking $W_i=span\{v_i\}$ where $i=1,..,n$ then
$W_i ∩Wj=\{0\}$ for $i≠j$ and $V=W_1+...+W_n$.
Hence $V=W_1⊕...⊕W_n$ i.e. $V$ is direct sum of one dimensional $T$ invariant subspaces.
I had skip to prove $W_i$ are $T$ invariant but I think it's obvious by definiition of $W_i$.
Is above proof is valid? In fact I saw the above question/ theorem first time. Is the statement of above theorem is correct?
Please help me.