Showing that a bilinear form is non degenerate

Question

Given a finite dimensional vector space $V$ over $F$ and a fixed matrix $(\alpha_{ij})=A \in M_n(F)$ and the bilinear form on $V \times V$ by $B(u,v)=\sum_{i=1}^{n} \sum_{j=1}^{n} \alpha_{ij} \zeta_i \eta_j$, where $\zeta_i$ and $\eta_j$ are the coordinates of $u,v$, I want to show that if the rank of $A$ is $n$, then $B$ is non-degenerate.

I am absolutely stuck. :( I cannot seem to get a hold on this problem. I would really appreciate a hint in the right direction. The only thing that I think is in the right direction is that if $rank(A)=n$ then $n(A)=\lbrace 0 \rbrace$ and the columns form a basis of $V$.... Also, that $B(u,v) \equiv u^TAv$.

Answer 1

The definition of non degenerate bilinear form is that if there exist $u$ such that for all $v$, $u^TAv = 0$ then $u=0$.

Choose $v$ as vectors in the canonical basis such as $(1,0,\cdots , 0)$, $(0,1,\cdots , 0)$ , etc. and you will find that it means that $$u^TA_1 = 0$$ $$u^TA_2 = 0$$ $$ \cdots $$ $$u^TA_n = 0$$ Where the $A_i$s are the column vectors of $A$. Treat this as a system of linear equation and the fact that $rank(A) = n$ implies that the only solution is $u = 0$.

Answer 2

Take $v\in V$, $v\ne0$, then also the coordinate vector $\eta$ of $v$ is non-zero, and $A\eta\ne0$ (as $\ker(A)=\{0\}$).

Hence there is an index $k$ such that $(A\eta)_k$, the $k$-th entry of $A\eta$, is non-zero. Now set $u$ the be the $k$-th basis element. Then the coordinate vector of $u$ is equal to $e_k=\pmatrix{0&\dots&0&1&0&\dots&0}^T$, i.e., $\zeta_i = \delta_{ik}$, and $$ B(u,v) = \sum_{i,j}\alpha_{ij}\zeta_i \eta_j = \sum_j \alpha_{kj}\eta_j = (A\eta)_k\ne0, $$ and the bilinear form is non-degenerate.

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The proof is much simple if $K=\mathbb R$ or $\mathbb C$. Then set $\zeta:=(A\eta)^H$ implying $B(u,v) = \|A\eta\|^2\ne0$.

Answer 3

$B$ non-degenerate means that $B(u,v) = 0$ for all $v \in V$ implies that $u = 0$. Equivalently, this means that if $u \neq 0$, then there is some $v$ such that $B(u,v) \neq 0$. But $B(u,v) = u^T A v$. So suppose $A$ has rank $n$. Then if $u \neq 0$, it follows that $u^T A \neq 0$ (basically $A$ having full rank means that it has trivial kernel). Hence there is some $v$ such that $u^T A v \neq 0$ (basically $u^T A$ is a row vector with a nonzero component, say at entry i. So let $v = (0,..., 1, ..., 0)^T$ where the $1$ is at entry $i$) .