# Can a matrix be orthogonal with respect to a positive definite $\phi$?

I've a doubt on othogonal matrices.

I know that an orthogonal matrix is a matrix $O$ such that $O^{T}O=O^{-1}O=I$ and also that $O$ has on the columns and on the rows the coordinates of the vectors of an orthonormal basis. Therefore the change of basis matrix between two orthonormal basis is indeed an orthogonal matrix.

But the columns and the rows are coordinates of the vectors of an orthonormal basis with respect to which scalar product?

Usually of course the standard scalar product is used so all the matrices that I saw were orthogonal "with respect to the standard scalar product" (if I can say that).

But is it possible to build an orthogonal matrix which has on the rows the coordinates of the vectors of an orthonormal basis with respect to a generic positive definite symmetric bilinear form $\phi$ ? Is in this case the matrix "orthogonal with respect to $\phi$"?

This confuses me also because from the definition $O^{T}O=I$ and that $O^{T}O$ looks much like a standard scalar product, even if it should be $trace(O^{T}O)$.

Does this make sense? Could anyone make clear for me if a matrix that is "orthogonal with respect to $\phi$" can exist and if it should be consider orthogonal?

Thanks a lot for your help

• you need that the bilinear form $\phi$ be symmetric, positive-definite and non-degenerated, then, the formula $x^{\top}\Phi x$, where $\Phi$ is the matrix $[\phi(b_i,b_j)]$ with $\{b_i\}$ a basis and $x=\sum_s x^sb_s$ , as the recipe of the very same standard scalar product. – janmarqz Dec 9 '15 at 14:34
• Thanks for the answer, I'm sorry but I can't actually understand your point, are you saying that $\phi$ necessarily represents the standard scalar product? – Gianolepo Dec 9 '15 at 16:39
• No. I'm saying is that any other bilinear map which is symmetric, positive-definite and non-degenerated gives you another scalar product, thru selecting a square matrix $Q$ to a pair of vectors via $x^{\top}Qy$ and to form a quadratic form $x^{\top}Qx$. The std scalar product corresponds to the identity matrix $1\!\!1$=diag(1,...,1). – janmarqz Dec 9 '15 at 17:39
• So, with $Q$ having these qualities, we could say that pair of vectors are $\phi$-orthogonal if $\phi(x,y)=x^{\top}Qy=0$ and $\phi$-orthonormal with $x^{\top}Qx=1$ – janmarqz Dec 9 '15 at 17:56
• Ok that's clear now, but can we also say that a matrix is $\phi$-orthogonal? – Gianolepo Dec 9 '15 at 17:59