I need to prove this:
Let $V$ be a vector space. If $T\in L(V,V)$ is diagonalizable, then $V=\ker (T)\bigoplus \mathrm{Im}(T)$.
Well, if $T$ is a diagonalizable operator then there exists a basis $\beta=\{v_1,\dots,v_n\}$ for which matrix of $T$ is diagonal. Let the eigenvalues be $\lambda_1,\dots\lambda_n$ so $T(v_1)=\lambda_1v_1$ $\dots$ $T(v_n)=\lambda_nv_n$ and let $v\in\ker (T)\cap \mathrm{Im}(T)$ be arbitrary. Then $T(v)=0$ and $v=\sum_{i=1}^{n}c_iv_i$ so $0=\sum_{i=1}^{n}T(c_iv_i)$ so $\sum_{i=1}^{n}c_i\lambda_iv_i=0$ so $c_i\lambda_i=0$, what can I conclude later?