Non-degeneracy of bilinear form and invertible matrix

Question

Let $B$ be a symmetric bilinear form on a vector space $V$, finite dimensional over $F$. Let $\{v_1,\cdots,v_n\}$ be a basis. Then the non-degeneracy implies that the matrix $[a_{ij}]=[B(v_,v_j)]$ is invertible.

Suppose $v_i$'s satisfy condition that $B(v_i,v_i)\neq 0$ for all $i$.

Is it true that the matrix $[b_{ij}]$ with $b_{ij} = \frac{B(v_i,v_j)}{B(v_j,v_j)}$ is also non-singular?

Answer 1

Yep. The matrix $B$ is obtained from $A$ by repeatedly multiplying the $j$-th column of $A$ by the non-zero scalar $\frac{1}{B(v_j,v_j)}$. Since column operations don't change the non-singularity of a matrix, the matrix $B$ is also non-singular.