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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.

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

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  • $\begingroup$ 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? $\endgroup$ Dec 27 '20 at 14:17
  • $\begingroup$ @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. $\endgroup$ Dec 27 '20 at 14:22

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