I am trying to prove the following theorem:

A Bilinear form $ B$ on a finite-dimensional vector space $ V$ is symmetric if it is diagonalizable.

Here is my attempt at a proof.

Suppose $ B$ is diagonalizable. Since $ B$ is a bilinear form on a finite-dimensional vector space for any basis $ S=\{s_1,\dots,s_n\}$ of $ V$ we can find a matrix $ A = (c_{ij})$ such that for all $ v,w\in V$ :

$ $ B(v,w) = [v]_S^T A [w]_S $ $

Since $ B$ is diagonalizable choose $ S’ = \{s_1′,\dots,s_n’\}$ to be a basis for $ V$ that diagonalizes $ A$ . We can write $ [v]_S’ = \left(\begin{array}{c} a_1 \ \vdots \ a_n\end{array}\right)$ , and $ [w]_s’ = \left(\begin{array}{c} b_1 \ \vdots \ b_n\end{array}\right)$ , and $ A=(\alpha_{ij})$ where $ \alpha_{ij} = \left\{ \begin{array}{cc} c_{ij} & i=j \ 0 & \text{otherwise} \end{array}\right.$ .

Then if we compute $ B(v,w)$ we obtain:

$ $ \begin{align*} B(v,w) &= [v]_{S’}^TA[w]_{S’} \ &= \left( \begin{array}{ccc} a_1 & \cdots & a_2\end{array}\right)A\left( \begin{array}{c} b_1 \ \vdots \ b_n\end{array}\right) \ &= \left( \begin{array}{ccc} c_{11}a_1 & \cdots & c_{nn}a_n\end{array}\right)\left( \begin{array}{c} b_1 \ \vdots \ b_n\end{array}\right) \ &= \sum_{i=1}^n c_{ii}a_ib_i \ &= \sum_{i=1}^n c_{ii}b_ia_i \ &= \left( \begin{array}{ccc} c_{11}b_1 & \cdots & c_{nn}b_n\end{array}\right)\left( \begin{array}{c} a_1 \ \vdots \ a_n\end{array}\right) \ &= \left(\begin{array}{ccc} b_1 & \cdots b_n \end{array}\right)A\left( \begin{array}{c} a_1 \ \vdots \ a_n\end{array}\right) = B(w,v) \end{align*} $ $

$ \blacksquare$

In class we proved it by induction on the dimension of $ V$ , I am guessing because you cant always assume your field is commutative and so the fifth line in the computation would be invalid. However, over a commutative field would this proof work?