Process of finding orthonormal basis of given bilinear form

Question

Let $V$ be vector space of $n\times n$ matrices .

$\langle A,B \rangle = {\rm tr}(A^T B)$

I wanted to find orthonormal basis of it.

I know that if there small say 3 dimension vector say then I would have find first matrix of bilinear form

But here dimension are $n$ so how to tackle such problem

Answer 1

There are $n\times n$ elements in this basis, let's label each element as $M^{(k,l)}$, for $k,l = 1,\cdots, n$

$$ M^{(k,l)}_{i,j} = \delta^k_i \delta^l_j $$

So the matrix $M^{(k,l)}$ is full of zeros, except for the row $k$ and column $l$, where the value is $1$. The product of any two elements of this basis is

\begin{eqnarray} \langle M^{(k,l)}, M^{(p,q)} \rangle &=& \sum_{i,j} M^{(k,l)}_{ji} M^{(p,q)}_{ji} \\ &=& \sum_{i,j} \delta^k_j\delta^l_i \delta^p_j \delta^q_i \\ &=& \delta^k_p \delta^l_q \end{eqnarray}

That is, the product is $1$ if $k = p $ and $l = q$, and is $0$ otherwise. So the set of matrices $\{M^{(k,l)} \}_{k,l = 1}^{n}$ is orthonormal