Let $\mathbb{k}$ be a field. Let $V$ be a finite-dimensional $\mathbb{k}$-vector space. Let $\varphi:V\times V\to \mathbb{k}$ ($\operatorname{char}\mathbb{k}\neq 2$) is bilinear (symmetric or skew-symmetric) or sesquilinear (hermitian or skew-hermitian) function. Let $W\subset V$ is a subspace. Show that $V=W+W^{\perp}$ if and only if $\varphi$ is non-degenerate on $W$.
My approach:
$\Leftarrow$ If $\varphi$ is non-degenrate on $W$ it means that $\ker \varphi |_W=\{0\}$, i.e. $W\cap W^{\perp}=\{0\}$. But also one can show that $\dim V\leq \dim W+\dim W^{\perp}$. Since $\dim(W+W^{\perp})=\dim W+\dim W^{\perp}-\dim (W\cap W^{\perp})=\dim W+\dim W^{\perp}\geq \dim V$. Since $W+W^{\perp}\subset V$ then it follows that $\dim(W+W^{\perp})\leq \dim V$ and these two inequalities give us that $\dim V=\dim(W+W^{\perp})$ and it follows that $V=W+W^{\perp}$.
$\Rightarrow$ In this case literally I have no ideas. I took $x\in \ker \varphi |_W$ and I am trying to show that $x=0$ but do not know how to do that.
Can anyone show to me the proof, please?
I have spent a whole day trying to do that.
Actually when I was trying to prove the $\Rightarrow$ direction I have noticed that if $\varphi$ is non degenerate then result follows quite easily.
It is enough to show that $\ker \varphi|_W=\{0\}$ or $W\cap W^{\perp}=\{0\}$.
Let $x\in W\cap W^{\perp}$. Then for any $v\in V$ we have $v=a+b$ with $a\in W, b\in W^{\perp}$ and $$\varphi(v,x)=\varphi(a,x)+\varphi(b,x)=0+0=0.$$
In particular, $\varphi(e_j,x)=0$ for $1\leq j \leq n$ where $\{e_1,\dots,e_n\}$ is some basis of $V$. Let $b_{ij}=\varphi(e_i,e_j)$ and we can write it matrix form $BX=0$, where $X=(x_1,\dots,x_n)^t$ column of coordinates of $x$. Since $\operatorname{rk} B=n$ then $B$ is invertible and $x=0$. We are done.