# If a Bilinear Form is Non-Degenerate on a Subspace $W$, then $V=W\oplus W^\perp$.

$\newcommand{\range}{\text{image}}\newcommand{\ann}{\text{Ann}}\newcommand{\set}[1]{\{#1\}}$

Problem: Let $V$ be a finite dimensional vector space over a field $F$ and $f$ be a symmetric bilinear form on $V$. Let $W$ be a subspace of $V$. If $f$ is non-degenerate on $W$, then $V=W\oplus W^\perp$.

Below I provide a proof of this. My problem is that my proof doesn't seem to make use of the fact that $f$ is symmetric. Is the above theorem also true if we strike out the word symmetric? If yes, then one may not read my proof and just leave a comment. If not then can one check where have I used the symmetry of $f$ in the argument?

Definitions and Notations:

Given a bilinear form $f$ on a (finite dimensional) vector space $V$, we define $R_f:V\to V^*$ as $(R_fv)(u)=f(u, v)$ for all $u, v\in V$. Clearly $R_f$ is a linear map.

For a subspace $W$ of $V$, we write $f|W$ to denotes the restriciton of $f$ to $W\times W$. It can be seen that $f|W$ is a bilinear form on $W$. We define $W^\perp$ as $\set{v\in V: f(w, v)=0 \ for \ all \ w\in W}$. Also, we write $\ann W$ to denote the annihilator of $W$.

We say that $f$ is non-degenerate on a subspace $W$ of $V$ is $R_{f|W}:W\to W^*$ has rank $\dim W$.

The Purported Proof:

It is clear that $f$ is non-degenerate on $W$ if and only if $W\cap W^\perp=\set{0}$. So it suffices to show that $\dim W+\dim W^{\perp}=\dim V$. To this end, first we note the simple fact that given $v\in V$ we have \begin{equation*} v\in W^\perp\ \iff\ R_fv\in \ann W \tag{1} \end{equation*}

Let $\pi:V^*\to V^*/\ann W$ be the canonical projection map and conider the liner map $\pi\circ R_f:V\to V^*/\ann W$.

By the Rank-Nullity Theorem we have

$$\begin{array}{rcl} \dim V &=& \dim \ker (\pi\circ R_f) + \dim \range(\pi\circ R_f)\\ \\ &\leq& \dim \ker (\pi\circ R_f) + \dim (V^*/\ann W)\\ \\ &=& \dim \ker (\pi\circ R_f) + \dim W \end{array}$$

Note that from (1) we know that $\ker (\pi\circ R_f)=W^\perp$. Therefore $\dim V\leq \dim W^\perp+\dim W$. But since $W\cap W^\perp=\set{0}$, we also have $\dim V\geq \dim W+\dim W^\perp$. So we must have $\dim V=\dim W+\dim W^\perp$ and we are done.

• Yes, symmetry of the bilinear form is irrelevant. For the reason your proof exposes. Commented May 23, 2015 at 8:52
• Thank you. I asked this because in Hoffman and Kunze's book on Linear Algebra, there is an exercise which asks to prove it but provides symmetry of the form in the hypothesis. Commented May 23, 2015 at 9:10
• Well, there's one thing I missed due to it being before my morning tea: without symmetry, you generally have ${}^\perp W \neq W^\perp$, the left and right "orthogonal complement"s differ. But, if a bilinear form is non-degenerate on $W$, then both, $W^\perp$ and ${}^\perp W$ are (algebraic) complements of $W$. Commented May 23, 2015 at 9:25

For symmetric or skew-symmetric forms $(u,v)=0 \implies (v,u) = 0$. But you are right, the statement does not need any of them, just decide what $\perp$ means. The nice thing is that if non-degeneracy holds for one map, it holds for the flip one ( since, if a matrix is nonsingular, then its transpose also is). Here is a quick proof to have in your toolbox ( from Milnor&Husemoller's book Symmetric bilinear forms).
Take $v \in V$. The map $W \to F$, $u \mapsto (u,v)$ is a linear functional so it must be the dot product with a unique $w$ from $W$. Hence we have $v = w + (v - w)= w + w'$. Note that $w' \in W^{\perp}$.
If $W$ contains a null vector, the statement (that the intersection of a subspace and it complement is trivial) is false. For example, a nondegenerate real quadratic form that is indefinite contains contains nonzero vectors that are null. If such a vector is in $W$, it will be in the intersection with the complement.
• This is not true. Take the $2$-dimensional hyperbolic space (quadratic form $x^2-y^2$) with "null" vector $(1,1)$. If you take as $W$ the space spanned by $w$, then the restriction of the form to $W$ degenerates, so the theorem does not apply. But if you take the whole space as $W$, then its orthogonal complement is $\{0\}$, and does not contain $w$. Commented Jan 22, 2022 at 3:13