Prove that if $B$ is symmetric then its matrix is symmetric.

Question

Let $V$ be a finite-dimensional vector space over the field $F,$ and let $B$ be a bilinear form on V.

$(a)$ $B$ is said to be symmetric if $B(v_1, v_2) = B(v_2, v_1)$ for all $v_1, v_2 \in V.$ Prove that if $B$ is symmetric then its matrix (with respect to any basis) is symmetric.

My trial:

Assume that $B$ is symmetric i.e.,$B(v_1, v_2) = B(v_2, v_1)$ for all $v_1, v_2 \in V.$ Now, I want to show that the matrix of $B$ is symmetric with respect to any basis, I know the general form of the matrix of $B$ with respect to the standard basis which is $ M = \begin{bmatrix} B(e_1, e_1) & B(e_1, e_2) \\ B(e_2, e_1) & B(e_2, e_2) \end{bmatrix}$ but then how can I prove the statement? could anyone help me please?

Answer 1

Let $\{e_1,e_2,\ldots, e_n\}$ be any basis of $V$. Let us write as $M$ the matrix of $B$ with respect to this basis. Then fpr each $i,j \in \{1,2,\ldots, n \}$, the $ij$-th coordinate of $M$ is $B(e_i,e_j)$, and the $ji$-th coordinate of $M$ is $B(e_j,e_i)$. But because $B$ is symmetric, the equation $$B(e_i,e_j)=B(e_j, e_i)$$ holds. [Indeed set $v_1=e_i$ and $v_2=e_j$, and use the fact that $B(v_1,v_2)=B(v_2,v_1)$ for each $v_1,v_2$ by def'n of $B$.] So the $ij$-th coordinate of $M$ is the same as the $ji$-th coordinate of $M$ for each such $i,j$. So $M$ satisfies the definition of a symmetric matrix.