matrix multiplied by rotation matrix on right side and transpose(rotation) on left side

Question

Would a matrix remain un-rotated if it is multiplied by an orthonormal rotation matrix on right side and transpose of same rotation matrix on the left side?

Answer 1

As @hardmath suggested, by doing what you say you'd get a similarity transformation: suppose that you have a matrix $A$ and an invertible matrix $P$. Then, the matrix $B = P^{-1}AP$ is said to be similar to A.

And here's a counterexample. Suppose that you have the Rotation matrix

$R = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\[0.3em] -\frac{1}{2} & \frac{\sqrt{3}}{2} \\[0.3em] \end{bmatrix}$

and its inverse (transpose)

$R^{-1} = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\[0.3em] \frac{1}{2} & \frac{\sqrt{3}}{2} \\[0.3em] \end{bmatrix}$

Then, if you consider a matrix

$A = \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}$

and you left-multiply it by $R^{-1}$ and right-multiply it by $R$, you get

$\begin{bmatrix} \frac{\sqrt{3}}{4}(-2+\sqrt{3}) & \frac{1}{4}(2+\sqrt{3}) \\[0.3em] \frac{1}{4}(-1+\sqrt{3}(1+\sqrt{3})) & \frac{1}{4}(1+ 2\sqrt{3}) \\[0.3em] \end{bmatrix}$

which is clearly a completely different matrix!