# How to tell if a Linear Function is Orthogonal over some Inner Product

Let $$V$$ be a finite dimensional $$K$$-Vector Space ($$K= \Bbb R$$ or $$\Bbb C$$). I'm trying to characterize all linear transformations $$f:V \rightarrow V$$ such that $$f$$ is orthogonal (or unitary) for some inner product over $$V$$. Said in another way, given a linear function $$f:V \rightarrow V$$, how can I tell if there exists an inner product over $$V$$ that makes $$f$$ and orthogonal (or unitary) transformation.

For example, a necessary condition is that $$| \det (f)|=1$$ . Another necessary condition is that if $$\lambda \in \Bbb C$$ is an eigenvalue of $$f$$ then $$|\lambda|=1$$.

A necessary and sufficient condition for such an inner product to exist is "There exists a bases $$B$$ in $$V$$ such that the columns of $$|f|_B$$ form an orthonormal basis over $$\Bbb R^n$$ (or $$\Bbb C^n$$) with the usual inner product". This is not something hard to prove. So we could change the question and ask the following

Let $$K= \Bbb R$$ or $$\Bbb C$$, we say that a matrix $$M\in K^{n \times n}$$ is orthonormal if the columns of $$M$$ form an orthonormal basis over $$K^n$$ with the cannonical inner product. Given a matrix $$A\in K^{n \times n}$$, how can we tell if there is an invertible matrix $$C\in K^{n \times n}$$ such that $$C \cdot A \cdot C^{-1}$$ is an orthonormal matrix?

Im more interested in working over $$K=\Bbb R$$ but maybe is easier to work over $$K=\Bbb C$$ so we can use the Jordan Normal Form.

• Maybe$f$ should be normal and all eigenvalues have unit length. Dec 27 '20 at 3:08
• @copper.hat I think it should be $f$ is conjugate to a normal. Normality is not preserved by arbitrary $C$. Dec 27 '20 at 8:11

A matrix $$A$$ is orthogonal over some inner product when it is diagonalizable with eigenvalues of unit length. Diagonalizability can be checked from its minimal polynomial, so this is not simply a restatement of the problem.

$$A$$ is orthogonal over some inner product $$S$$ when $$A^*SA=S$$ (as in the usual proof when $$S=I$$). For $$S$$ to be an inner product it must be positive definite, $$S=C^*C$$, with $$C$$ invertible. Hence $$A^*C^*CA=C^*C\implies (CAC^{-1})^*=(CAC^{-1})^{-1}\implies A=C^{-1}UC$$ for some unitary $$U$$. Since $$U$$ is diagonalizable this shows that $$A$$ is also diagonalizable, with the same eigenvalues as $$U$$, namely of unit length.

Conversely, if $$A=C^{-1}e^{iD}C$$ then $$A^*(C^*C)A=C^*e^{-iD}C^{-*}C^*CC^{-1}e^{iD}C=(C^*C)$$ so $$S=C^*C$$ is the required inner product.

• Thanks!!! I'm already going to put your answer as the correct one, but I'm curious to ask. This works for $K= \Bbb C$ but not for $K= \Bbb R$ (rotation Matrix are not diagonalizable over $\Bbb R$). May be we could use the complex case to solve the real one? Dec 27 '20 at 14:17
• @MarcosMartínezWagner In the case $K=\mathbb{R}$, it is still correct, where diagonalizable should mean wrt $\mathbb{C}$. So a rotation is still diagonalizable over $\mathbb{C}$ and the argument above still works for it. Dec 27 '20 at 14:22