glycleWogry

2022-06-05

Similarity transformation of an orthogonal matrix
A transformation $T$ represented by an orthogonal matrix $A$ , so ${A}^{T}A=I$. This transformation leaves norm unchanged.
I do a basis change using a matrix $B$ which isn't orthogonal , then the form of the transformation changes to ${B}^{-1}AB$ in the new basis( A similarity transformation).
Therefore ${B}^{-1}AB$. $\left[{B}^{-1}AB{\right]}^{T}=I$.
This suggests that ${B}^{T}B=I$ which means it is orthogonal, but that is a contradiction.

A transformation $\mathcal{A}$ is orthogonal iff its matrix representation is orthogonal with respect to an standard orthogonal basis. And the transition matrix between two standard orthogonal bases must be orthogonal.
${B}_{1}=\left\{{e}_{1},\dots ,{e}_{n}\right\}$, ${B}_{2}=\left\{{f}_{1},\dots ,{f}_{n}\right\}$ are two standard orthogonal bases. ${A}_{1}$ and ${A}_{2}$ are representations of $\mathcal{A}$ with respect to ${B}_{1},{B}_{2}$. Then ${A}_{1},{A}_{2}$ are orthogonal matrices.

If $P$ is the transition matrix between ${B}_{1}$ and ${B}_{2}$ ( that is $\left({e}_{1},\dots ,{e}_{n}\right)=\left({f}_{1},\dots ,{f}_{n}\right)P$), then $P$ is orthogonal.

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