Non-degenerate bilinear forms and invertible matrices

Question

I have the following question:

Show that a symmetric bilinear form is nondegenerate if and only if its matrix in a basis is invertible

Ok so both directions "if non-degenerate then the matrix is invertible" and "if matrix is invertible then the form is nondegenerate" have to be proven for this.

For the first direction. If the bilinear form is non-degenerate its null space is $\left\{0\right\}$, so for every $v \neq 0$ there exists a $v'$ such that $\langle v, v' \rangle \neq 0$ and so there are no zero eigenvalues, I'm pretty sure this somehow implies the determinant of the matrix is also non-zero, as it is the product of the eigenvalues and hence invertible (but not sure how to show this).

For the other direction, if the matrix is invertible, then the determinant is non-zero and, if my previous assumption is correct, this implies there are no zero eigenvectors, which implies the null space is $\left\{0\right\}$ and hence non-degenerate?

Thanks for any guidance, the notes I have on this topic are pretty poor so I'm having a hard time fully understanding everything. Thanks again.

Answer 1

Let me assume that you have a bilinear form $q : V \times V \to F$, where $V$ is a finite-dimensional vector space over the field $F$. Fix a basis $\beta = \{e_1,\dotsc,e_n\}$ of $V$, which defines the corresponding dual basis $\beta^\ast = \{e_1^\ast,\dotsc,e_n^\ast\}$ of $V^\ast$.

First, check that $q$ defines a linear transformation $Q : V \to V^\ast$ by $Q(v) := q(v,\cdot)$ satisfying $\ker Q = \ker q$. Now, since for any $\phi \in V^\ast$, $\phi = \sum_k \phi(e_k) e_k^\ast$, $$ Q(e_k) = \sum_j Q(e_k)(e_j) e_j^\ast = \sum_j q(e_k,e_j) e_j^\ast = \sum_j q(e_j,e_k)e_j^\ast $$ for each $k$, and hence the matrix $[Q]_{\beta^\ast\beta}$ of $Q$ with respect to the bases $\beta$ and $\beta^\ast$ is precisely the matrix $[q]_\beta$ of $q$ with respect to $\beta$. As a result, your matrix of interest $[q]_\beta$ is invertible if and only if the linear transformation $Q$ is invertible.

So, let's collect what we know:

1. $[q]_\beta = [Q]_{\beta^\ast\beta}$, so that $[q]_\beta$ is invertible if and only if $Q$ is invertible.
2. $\ker q = \ker Q$, so that $q$ is non-degenerate if and only if $Q$ is injective.
3. $\dim V = \dim V^\ast$, so that by the rank-nullity theorem, $Q : V \to V^\ast$ is injective if and only if it is surjective.

What can you therefore conclude?

Answer 2

Here is one more, a simple one.

Non-degeneracy of the bilinear form f(x,y) in a finite dimensional vector space V is for $ x,y \in V $ defined by the condition

$ \big( f(x,y)=0 \; \forall \, x \big) \; \Leftrightarrow \; y=\mathit{0} $.

Let M be the matrix representing the symmetric bilinear form f, i.e. $ f(x,y) = x^T M y \, \big( = y^T M x \big) $.

Obviously, we have $ \big( x^T M y = 0 \; \forall \, x \big) \; \Leftrightarrow \; My=\mathit{0} $.

"$ \; \Rightarrow \; $" : Let f be non-degenerate. Then

$ My=\mathit{0} \; \Leftrightarrow \; y=\mathit{0} $,

and in particular, the implication $ \, My=\mathit{0} \; \Rightarrow \; y=\mathit{0} \, $ means that M is invertible.

"$ \; \Leftarrow \; $" : On the contrary, let M be invertible. Then from the above we have

$ \big( f(x,y)=0 \; \forall \, x \big) \; \Rightarrow \; y=\mathit{0} $.

Obviously, $ y=\mathit{0} \; \Rightarrow \; My=\mathit{0} $, so we also get

$ \big( f(x,y)=0 \; \forall \, x \big) \; \Leftarrow \; y=\mathit{0} $.

Thus, f is non-degenerate.

Answer 3

I also proposed the question, but (I think) solved it myself before reading any of this. We know that any bilinear form can be represented as $X^T * (n \times n \text{ matrix}) * Y$ where $X^T$ and $Y$ are $n$-dimensional vector spaces.

The definition I have of a nonDegenerate bilinear form is that $B ( X, Y) = 0$ for all $X\Longrightarrow Y = 0$ ; somewhat convoluted no?

Prove the contrapositive -- There exists a non-zero vector $Y$ such that for all $X, B (X, Y) = 0$

Start with non-zero $Y$ and extend it to a new basis for a vector space. This makes $Y$ equal the column vector $( 1, 0, . . . , 0 )$.

Now consider the row vector $X^T$ * matrix of $B$ in the new basis, to get $B ( X , Y)$ to be zero, all $X$ must transform to a row vector with 0 in the first position. This implies that the matrix has a zero row, hence not invertible.

Answer 4

Shortly after Branimir posted his solution I did realise a proof, and I thought I'd post it as an answer.

The nullspace of the bilinear form is the set of vectors $v$ such that the coordinate of $v$ is a solution to the homogeneous equation $AX=0$, where $A$ is the matrix of the symmetric form w.r.t a basis. The form corresponds to the product $X^TAY$ so if $Y$ is a vector such that $AY = 0$, then this implies also that $X^TAY = 0$ for all $X$ and thus $Y$ is in the null space. Now suppose that $AY \neq 0$, then $AY$ has at least one non-zero coordinate, so one of the products is not zero, so $Y$ is not a null vector. So we see that if a form is non-degenerate then the only solution to $AX=0$ is when $X=0$ which implies that $A$ is invertible.

To do the other direction, if $A$ is invertible, it can be row reduced to the identity matrix, which implies that the only solution to $AX = 0$ is when $X=0$, that is the nullspace of the form is $\left\{0\right\}$ and hence the form is non-degenerate.