MHB Bilinear Form Non-Degenerate on a Subspace.

[caffeinemachine]

I am trying to prove the following standard result:Let $V$ be a finite dimensional vector space over a field $F$ and $f:V\times V\to F$ be a symmetric bilinear form on $V$. Let $W$ be a subspace of $V$ such that $f$ is non-degenerate on $W$.
Then
$$V=W\oplus W^\perp$$(Here $W^\perp=\{v\in V:f(w,v)=0\text{ for all } w\in W\}$).Here is what I tried:The bilinear form gives us a map $L_f:V\to V^*$ defined as
$$(L_fu)v=f(u,v),\quad\forall u,v\in V$$
Let $W^0$ denote the annihilator of $W$.
We show that
$$v\in W^\perp \text{ if and only if } L_fv\in W^0$$Let $v\in W^\perp$.
Then $(L_fv)w=f(v,w)=0$ for all $w\in W$. Therefore $L_fv\in W^0$.
Now say $L_fv\in W^0$ for some $v\in V$.
Then $(L_fv)w=0$ for all $w\in W$, giving $f(v,w)=0$ for all $w\in W$.
Therefore $v\in W^\perp$.Also, it is clear that $W\cap W^\perp=0$.Since $\dim W^0=\dim V-\dim W$, we would be done if we could show that $\dim W^\perp \geq \dim W^0$.From the observation done above, it is natural to consider the map $T:W^\perp\to W^0$ defined as $Tv=L_fv$ for all $v\in W^\perp$.
We just need to show that $T$ is surjective.
But here I am stuck.Can somebody help.Thanks.

 

[Fernando Revilla]

We need not to use the concept of dual space. As $V$ is finite dimensional, if $B_W=\{e_1,\ldots,e_r\}$ is a basis of $W$ and $B=\{e_1,\ldots,e_r,\ldots,e_n\}$ a basis of $V$ then, $f(x,y)=X^TAY$ with $A=[a_{ij}]$ symmetric, $\text{rank}(A)=n$ and $X,$ $Y$ coordinates of $x,$ $y$ with respect to $B.$ Then, $$x\in W^{\perp}\Leftrightarrow \left \{ \begin{matrix} f(x.e_1)=0\\\ldots\\f(x.e_r)=0\end{matrix}\right.\Leftrightarrow \left \{ \begin{matrix} a_{11}x _1+\ldots +a_{n1}x_n=0\\\ldots\\a_{1r}x_1+\ldots +a_{nr}x_n=0\end{matrix}\right.\Leftrightarrow MX=0.$$ But $\text{rank}(M)=r$ because $A$ has maximum rank so, $\dim W^{\perp}=n-r.$ This implies $\dim W+\dim W^{\perp}=\dim V.$ As $W\cap W^{\perp}=\{0\},$ we conclude $V=W\oplus W^{\perp}.$

 

[caffeinemachine]

We need not to use the concept of dual space. As $V$ is finite dimensional, if $B_W=\{e_1,\ldots,e_r\}$ is a basis of $W$ and $B=\{e_1,\ldots,e_r,\ldots,e_n\}$ a basis of $V$ then, $f(x,y)=X^TAY$ with $A=[a_{ij}]$ symmetric, $\text{rank}(A)=n$ and $X,$ $Y$ coordinates of $x,$ $y$ with respect to $B.$ Then, $$x\in W^{\perp}\Leftrightarrow \left \{ \begin{matrix} f(x.e_1)=0\\\ldots\\f(x.e_r)=0\end{matrix}\right.\Leftrightarrow \left \{ \begin{matrix} a_{11}x _1+\ldots +a_{n1}x_n=0\\\ldots\\a_{1r}x_1+\ldots +a_{nr}x_n=0\end{matrix}\right.\Leftrightarrow MX=0.$$ But $\text{rank}(M)=r$ because $A$ has maximum rank so, $\dim W^{\perp}=n-r.$ This implies $\dim W+\dim W^{\perp}=\dim V.$ As $W\cap W^{\perp}=\{0\},$ we conclude $V=W\oplus W^{\perp}.$

Here we cannot assume that $\text{rank}(A)=n$ because $f$ is not necessarily degenerate on the whole of $V$. The hypothesis merely demands the non-degeneracy of $f$ on $W$.

Also, what does $M$ denote in your answer?

 

[Fernando Revilla]

Here we cannot assume that $\text{rank}(A)=n$ because $f$ is not necessarily degenerate on the whole of $V$.

Sorry, I didn't notice it, but it does not matter, if $f_{W}:W\times W\to K$ is not degenerate then, the matrix of $f_W$ with respect to $B_W=\{e_1,\ldots,e_r\}$ is $$A_W=\begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1r}\\ a_{21} &a_{22} & \ldots & a_{2r} \\ \vdots&&&\vdots \\ a_{r1} & a_{r2} &\ldots & a_{rr}\end{bmatrix}\text{ with }\det( A_W)\neq 0.\qquad (*)$$

Also, what does $M$ denote in your answer?

Denotes the matrix of the cooresponding system, that is $M=\begin{bmatrix} a_{11} & a_{21} & \ldots & a_{n1}\\ \vdots&&&\vdots \\ a_{1r} & a_{2r} &\ldots & a_{nr}\end{bmatrix},$ and $\text{rank}(M)=r$ by $(*).$

 

[caffeinemachine]

Sorry, I didn't notice it, but it does not matter, if $f_{W}:W\times W\to K$ is not degenerate then, the matrix of $f_W$ with respect to $B_W=\{e_1,\ldots,e_r\}$ is $$A_W=\begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1r}\\ a_{21} &a_{22} & \ldots & a_{2r} \\ \vdots&&&\vdots \\ a_{r1} & a_{r2} &\ldots & a_{rr}\end{bmatrix}\text{ with }\det( A_W)\neq 0.\qquad (*)$$
Denotes the matrix of the cooresponding system, that is $M=\begin{bmatrix} a_{11} & a_{21} & \ldots & a_{n1}\\ \vdots&&&\vdots \\ a_{1r} & a_{2r} &\ldots & a_{nr}\end{bmatrix},$ and $\text{rank}(M)=r$ by $(*).$

This solves the problem I was originally trying to solve. Thanks.

But can you say for sure that $T$, defined in the last paragraph of my first post, is necessarily surjective?

 

[Fernando Revilla]

But can you say for sure that $T$, defined in the last paragraph of my first post, is necessarily surjective?

Yes, $T: W^{\perp}\to W^{0}$ is a linear map, injective and $\dim W^{\perp}=\dim W^{0}$ so, $T$ is surjective.