If $V$ is a finite dimensional inner product space over $\mathbb{C^{n}}$ and $T:V\rightarrow V$ is a normal linear transformation, how can I show that $V=\operatorname{ran}\left(T\right)\oplus\ker\left(T\right)$, using the fact that $\operatorname{ran}\left(T\right)^{\perp}=\ker\left(T^{\ast}\right)$, where $T^{\ast}$ is the adjoint of $T$?
Here is what I have done so far. Let $x\in\operatorname{ran}\left(T\right)\cap\ker\left(T\right)$. Since $T$ is normal, then \begin{align*} &Tx=0\\ \implies&T^{\ast}Tx=0\\ \implies&T^{\ast}Tx=Tx\\ \implies&TT^{\ast}x=Tx. \end{align*} This gives us that $\left\langle x,Tx\right\rangle=\left\langle x,TT^{\ast}x\right\rangle=\left\langle T^{\ast}x,T^{\ast}x\right\rangle=0$. Thus, $T^{\ast}x=0$, so $x\in\ker\left(T^{\ast}\right)$.
Since $\operatorname{ran}\left(T\right)^{\perp}=\ker\left(T^{\ast}\right)$, then $\operatorname{ran}\left(T\right)=\ker\left(T^{\ast}\right)^{\perp}$. Thus, $\operatorname{ran}\left(T\right)\cap\ker\left(T\right)=\ker\left(T^{\ast}\right)^{\perp}\cap\ker\left(T\right)$, so $\left\langle x,y\right\rangle$ for all $y\in\ker\left(T^{\ast}\right)$. Since $x\in\ker\left(T^{\ast}\right)$, then $\left\langle x,x\right\rangle=0$, so $x=0$. Thus, $\operatorname{ran}\left(T\right)\cap\ker\left(T\right)=\left\{0\right\}$.
Now I'm just stuck on trying to show that $\operatorname{ran}\left(T\right)+\ker\left(T\right)=V$, and I'm not sure where to go.
Note first (as you showed) that, for any $T$ (even if non-normal), $$\tag1 \ker T=\ker T^*T. $$ Indeed, if $Tx=0$, then $T^*Tx=0$. Conversely, if $T^*Tx=0$, then $$0=\langle x,T^*Tx\rangle=\langle Tx,Tx\rangle=\|Tx\|^2,$$ so $Tx=0$.
Using $(1)$ and that $T$ is normal, $$ \ker T=\ker T^*T=\ker TT^*=\ker T^*. $$ Then $$ (\ker T)^\perp=(\ker T^*)^\perp=\text{ran}\,(T). $$ The above shows that $\ker T$ and $\text{ran}\,T$ are the orthogonal complement of the other, so $V=\ker V\oplus \text{ran}\,T$.