Prove bilinear form is nondegenerate

Question

This is from Linear Algebra, an Introductory Approach - Charles Curtis

Let $V$ be a finitely generated vector space over $F$ with basis $\{v_1, ..., v_n\}$. Let $A = (\alpha_{ij})$ be a fixed $n$ by $n$ matrix over $F$. For any $v=\epsilon_1v_1+...+\epsilon_nv_n$ and $u=\eta_1v_1+...+\eta_nv_n$ define the bilinear form $B(u,v) := \sum^{n}_{i=1}\sum^{n}_{j=1} \alpha_{ij}\epsilon_{i}\eta_{j}$. Show that $B$ is non-degenerate if and only if $rank(A) = n$.

I am having trouble getting this one anywhere useful. I think one way to start would be noting that $rank(A)=n$ gives $ker(A)={0}$, which seems to me like it oughta connect to the non-degeneracy condition.

Answer 1

So your bilinear form is non-degenerate if there does not exist a vector $v \neq 0$ such that $B(v,w) = 0$ for any vector $w \in V$. In other words, there doesn't exist a vector that sends all $V$ to zero.

Consider a bilinear form $B$ and matrix $A$. If $A$ is non-invertible, then the kernel of $A$ is non-trivial. Suppose $k$ is a vector in the kernel of $A$. Then, because

$$ \sum^n_{i=1} \sum^n_{j=1} \alpha_{ij}\epsilon_i\eta_j = \sum^n_{j=1} \eta_j \sum^n_{i=1} \alpha_{ij}\epsilon_i $$

You know that each term $ \sum^n_{i=1} \alpha_{ij}\epsilon_i $ in the summation goes to zero because $k$ is in the kernel of $A$. Then, for any $\eta \in V$,

$$ \sum^n_{i=1} \sum^n_{j=1} \alpha_{ij}\epsilon_i\eta_j = \sum^n_{i=1} \sum^n_{j=1} 0\eta_j = 0 $$

Which would imply the existence of a non zero vector $k \in V$ which sends $V$ to zero.

Similarly, if $B$ is degenerate then there exists a vector $\epsilon$ such that

$$ \sum^n_{i=1} \sum^n_{j=1} \alpha_{ij}\epsilon_i\eta_j = 0 $$

for any $\eta \in V$. However going by our previously equality we can observe that $\sum^n_{j=1} \eta_j \sum^n_{i=1} \alpha_{ij}\epsilon_i$ is actually an exact expression of,

$$ [\eta]^T[A][\epsilon] $$

over basis $\mathcal{B} = \{v_i\}$. If we examine $\eta = v_i $ for each $i$, we see that each term of $[A][\epsilon] $, must be zero for them all to be sent to zero, so either $\epsilon$ is the zero vector, or it is in the kernel of $A$. Because $\epsilon $ is assumed to be non-zero, it means that $A$ is non-invertible and we have an if and only if relation.

Answer 2

$\Rightarrow)$ To see that $rank(A)=n$, we have to proof that the columns of $A$ are linearly independant. So we notice that the $j$-th column is $(\alpha_{ij})_i$. If we have $\lambda_1,\dots,\lambda_n\in F$ such that $\sum_{j=1}^n\lambda_j(\alpha_{ij})_i=0$, then $(\sum_{j=1}^n\lambda_j\alpha_{ij})_i=0$, which means that for each $i=1,\dots,n$, $\sum_{j=1}^n\lambda_j\alpha_{ij}=0$. So if we put $u=\sum_{j=1}^n\lambda_jv_j$, the equality above is: $B(u,v_i)=\sum_{j=1}^n\lambda_j\alpha_{ij}=0$ for all $i=1,\dots,n$. As $B(u,\_)$ is null for all the elements in a basis, it must be the linear transformation zero, but as $B$ is non-degenerate, this means that $u$ must be zero. Therefore, $\lambda_j=0$ for each $j=1,\dots,n$. Meaning that the columsn are linearly independant, so $rank(A)=n$.

$\Leftarrow)$ If $B$ is degenerate, then exist $u\in V$ non zero such that $B(u,v)=0$ for all $v\in V$. In particular $B(u,v_i)=0$ for all $i=1,\dots,n$. If $u=\sum_{j=1}^n\lambda_jv_j$ with $\lambda_1,\dots,\lambda_n\in F$ with at least one diferent of zero, then $0=B(u,v_i)=\sum_{j=1}^n\lambda_j\alpha_{ij}$ for all $i=1,\dots,n.$. From this, we get that $(\sum_{j=1}^n\lambda_j\alpha_{ij})_i=0$. Therefore $\sum_{j=1}^n\lambda_j(\alpha_{ij})_i=0$, so the columns $\{(\alpha_{ij})_i\}_j$ are not linearly independat, so the $rank(A)<n$.