# Process of finding orthonormal basis of given bilinear form

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

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
• @MathLover $$\delta^i_j = \begin{cases}1 &,& i = j \\ 0 &,& {\rm otherwise}\end{cases}$$ That means that in a sum of the form $\sum_i a_i \delta^i_j$ all terms vanish, except the one for which $i = j$, $$\sum_i a_i \delta^i_j = a_j$$ – caverac Nov 25 '18 at 15:45