If $T$ is invertible and $S$ is an isometry, is $TST^{-1}$ an isometry?

Question

I've tried finding a counterexample, but it's hard for me to identify lots of examples where $S$ is an isometry (e.g. in $\mathbb{R}^2$ where $S(x,y)=(-y,x)$). So I haven't found a counterexample. On the other hand, I haven't been able to prove it either. Just looking for a hint, please.

Answer 1

Counterexample: (using your $S$) $$ S = \pmatrix{0&-1\\1&0}, \quad T = \pmatrix{2&0\\0&1} \implies\\ TST^{-1} = \pmatrix{0&-2\\1/2 & 0} $$

Answer 2

To me it is the same linear map. It's just a change of the representation base for the endomorphism. So according to the original base it is still the orthogonal or unitary endomorphism.

For the definition of an isometry by a matrix the rows/cols have to be orthonormal or more general a endomorphism is orthogonal or unitary if, and only if its matrix representation with respect to a orthonormal base is a orthogonal or unitary matrix.

Answer 3

The change of basis doesn't preserve the metric so there is no reason for $TST^{-1}$ to be an isometry. $TST^{-1}$ is an isometry for the metric $d \circ (T^{-1},T^{-1})$ where $d$ is the classical metric.

In general, $TST^{-1}$ is an isometry iff $T$ is a scalar mutiple of an isometry, or equivalently, iff $T$ preserves the equality of distances. (in the case when $T$ is linear)