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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

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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

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  • $\begingroup$ Sir I could not understand how to algebra of delta is working? it is very new for me.Please can you tell me ? $\endgroup$ – MathLover Nov 25 '18 at 15:42
  • $\begingroup$ @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 $i = j$, $$\sum_i a_i \delta^i_j = a_j$$ $\endgroup$ – caverac Nov 25 '18 at 15:45

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