Prove that if $M$ is rotation matrix, then so is $TMT^{-1}$ Prove that if $M$ is rotation matrix, then so is $TMT^{-1}$
I already proved that the determinant and eigenvalues are the same for $M$ and $TMT^{-1}$, but I'm not sure how to prove they are both rotations.  Do i need to write out the matrix and its components, or is their a rotation equation/relationship I can use to show they are both rotations.
T is any invertible matrix
 A: The claim is false in general. If one takes $$M = \pmatrix{0&-1\\1&0}$$ (anticlockwise rotation by $90^{\circ}$) and $$T = \pmatrix{1&0\\0&\lambda}$$ (dilation by $\lambda$ in the $y$-direction). Then, $$TMT^{-1} \pmatrix{1\\0} = \pmatrix{0\\ \lambda} .$$
In particular, unless $\lambda = \pm 1$, we see that multiplication by $TMT^{-1}$ does not preserve lengths and so is not a rotation matrix.
A: Check that $\begin{bmatrix}
1 & 1 \\
0 & 1 
\end{bmatrix} \begin{bmatrix}
0 & -1 \\
1 & 0 
\end{bmatrix} \begin{bmatrix}
1 & 1 \\
0 & 1 
\end{bmatrix}^{-1} =  \begin{bmatrix}
1 & -2 \\
1 & -1 
\end{bmatrix}$.
So it seems not to be true for some $T $ invertible. 
And determinant of this result matrix $= 1$ and eigenvalues $+i,-i$ as for the middle rotation matrix (rotation by $ \pi /2 $) but the result matrix is not a rotation matrix.
A: M is rotation matrix so ,$T^{-1}$exist $$|M|=1,\\M_{\alpha}M_{\beta}=M_{\alpha+\beta} ,\\M_{\alpha}^n=M_{n\alpha}\\M_{\alpha}^{-1}=M_{-\alpha}$$
check these properties for $TMT^{-1}$
$$|TMT^{-1}|=|T|.|M|.|T^{-1}|=|T|.1.\dfrac{1}{|T|}=1\\$$$$
TM_{\alpha}T^{-1}.TM_{\beta}T^{-1}=TM_{\alpha}M_{\beta}T^{-1}=TM_{\alpha+\beta}T^{-1}\\$$$$
(TM_{\alpha}T^{-1})^n=TM_{\alpha}T^{-1}.TM_{\alpha}T^{-1}.TM_{\alpha}T^{-1}...TM_{\alpha}T^{-1}=TM_{\alpha+\alpha+\alpha+...}T^{-1}=TM_{n\alpha}T^{-1}\\$$$$
(TM_{\alpha}T^{-1})^{-1}=(T^{-1})^{-1}M_{\alpha}^
{-1}T^{-1}=TM_{-\alpha}T^{-1}$$
