# How to prove this theorem on diagonalizable operator on finite dimensional Vector space.

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?

• No, you are asked to prove an equivalence and you attemped only one direction. The implication you started with is correct though. $W_i$ is T-invariant because these spaces are generated by eigenvectors. So, include your attempt for the converse. – EpsilonDelta Jun 8 at 13:08