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 ${B}^{-1}AB$. $[{B}^{-1}AB{]}^{T}=I$.

This suggests that ${B}^{T}B=I$ which means it is orthogonal, but that is a contradiction.

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 ${B}^{-1}AB$. $[{B}^{-1}AB{]}^{T}=I$.

This suggests that ${B}^{T}B=I$ which means it is orthogonal, but that is a contradiction.

hildiadau0o

Beginner2022-06-06Added 21 answers

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.

${B}_{1}=\{{e}_{1},\dots ,{e}_{n}\}$, ${B}_{2}=\{{f}_{1},\dots ,{f}_{n}\}$ 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.

$\mathcal{A}({e}_{i})=\text{the}i\text{th colume of}{A}_{1};\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathcal{A}({f}_{i})=\text{the}i\text{th colume of}{A}_{2}$

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

${B}_{1}=\{{e}_{1},\dots ,{e}_{n}\}$, ${B}_{2}=\{{f}_{1},\dots ,{f}_{n}\}$ 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.

$\mathcal{A}({e}_{i})=\text{the}i\text{th colume of}{A}_{1};\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathcal{A}({f}_{i})=\text{the}i\text{th colume of}{A}_{2}$

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

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