Proof using self-adjoint property

Question

Let $V \;$ be an inner product space over a field $\Bbb{F}$ and let $T:V\to V$ be a self-adjoint linear map. Prove that $V = \operatorname{ker}(T)\oplus\operatorname{im}(T)$.

All I can think of is the definition of self-adjoint i.e. $T=T^*$ or $\left<T(v_1)|v_2\right>=\left<v_1|T^*(v_2)\right>\quad v_1,v_2\in V\quad$ and the identity $\operatorname{ker}{T}^∗ = \operatorname{im}{T}^⊥$.

Answer 1

Note that ${\rm dim}\ im(T) + {\rm dim}\ ker(T)=n$. So we have a claim : $$ im(T)\cap ker(T)=\{ 0\}$$

Assume that $Tx=v\neq 0\in im(T)\cap ker(T)$. Then $$ 0=( x,Tv)=(Tx,v)=(v,v) $$ Contradiction.