# Orthogonal basis for the Bilinear form given by $H(A,B) = \operatorname{tr}(AB)$

I've been thinking about this for a while and I've found that for the 2x2 case the matrices are just $\{e_{11}, e_{12} + e_{21}, e_{21}, e_{22}\}$ where $e_{ij}$ is the matrix with $1$ on the $ij$ entry, $0$'s everywhere else. However I can't seem to see any patterns with higher orders. I need to find the basis for the $n \times n$ case.

Any help is appreciated.

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Typeset everything in $\LaTeX$. Let me know if I messed anything up. – Dylan Moreland Dec 2 '11 at 6:02
You really want to define it as $\mbox{tr}(A B^T).$ If you want the identity matrix to give $H(I,I) = 1$ you take $\frac{1}{n} \; \mbox{tr}(A B^T).$ – Will Jagy Dec 2 '11 at 6:13
Well no, the bilinear form I'm given is defined as H(A,B) = tr(AB), not tr(AB^t) unless I misunderstood what you meant. – The eigenvalue Dec 2 '11 at 6:23
Note that $H(e_{21},e_{21})=0.$ – Will Jagy Dec 2 '11 at 7:22

You really want to take the definition as $$H(A,B) = \mbox{tr} \; A B^T$$ in which case your collection of all $e_{ij}$ does indeed give an orthonormal basis.

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That would definitely make life easy but that's not the bilinear form I'm working with. – The eigenvalue Dec 2 '11 at 6:27

By Sylvester's Law of Inertia, no orthonormal basis is possible, including the 2 by 2 case. In particular, your $H(e_{21},e_{21}) = 0.$

What is possible is to give a collection of matrices as below. There are $n$ matrices of type (A). There are $\frac{n^2 - n}{2}$ matrices of type (B).There are also $\frac{n^2 - n}{2}$ matrices of type (C).

Any two distinct matrices from the total collection of $n^2$ matrices are orthogonal. Given any matrix $\alpha$ of type (A), $H(\alpha, \alpha) = 1.$ Given any matrix $\beta$ of type (B), $H(\beta, \beta) = 1.$ However, given any matrix $\gamma$ of type (C), $H(\gamma, \gamma) = -1.$

(A) all your $e_{ii}$

(B) with all $j < k,$ take $$\frac{1}{\sqrt 2} (e_{jk} + e_{kj})$$

(C) with all $j < k,$ also take $$\frac{1}{\sqrt 2} (e_{jk} - e_{kj})$$

See SYLVESTER

EDIT, FRIDAY: this was a clever assignment, and not something I knew about. We have a quadratic form defined on square matices with real entries, given by $$q(M) = \mbox{tr}(M^2).$$ This is positive definite on the linear subspace given by the symmetric matrices, as then $q(M^2) = \mbox{tr}(M M^T)$ which is the sum of the squares of all $n^2$ entries. It is negative definite on the skew-symmetric matrices, as then $q(M^2) = - \mbox{tr}(M M^T).$ By adding the dimensions, we find that the "corank" of the quadratic form $q$ is $0.$ Finally, any symmetric matrix $M$ and any skew symmetric matrix $N$ are orthogonal, as $$\mbox{tr}(M N) = \mbox{tr}(N^T M^T) = - \mbox{tr}(N M) = - \mbox{tr}(M N),$$ so that $\mbox{tr}(M N) =0$ here.

A symmetric matrix A can always be transformed in this way into a diagonal matrix D which has only entries 0, +1 and −1 along the diagonal. Sylvester's law of inertia states that the number of diagonal entries of each kind is an invariant of A, i.e. it does not depend on the matrix S used.

The number of +1s, denoted n+, is called the positive index of inertia of A, and the number of −1s, denoted n−, is called the negative index of inertia. The number of 0s, denoted n0, is the dimension of the kernel of A, and also the corank of A. These numbers satisfy an obvious relation

$$n_0+n_{+}+n_{-}=n.$$


The clever part is that we do not really need to talk about the "matrix" of $q,$ which would be an $n^2$ by $n^2$ matrix. To repeat, see QUADRATIC

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Thanks, I actually figured this out before reading this but glad to have it confirmed :) – The eigenvalue Dec 2 '11 at 8:17