Power of orthogonal matrix

Question

Suppose $U$ is an orthogonal matrix, and $D$ is a diagonal matrix. Let $I$ denote the identity matrix. Let $k$ be a positive integer.

I think the following holds:

$$(I - UDU^T)^k = U(I - D)^kU^T$$

But I got a little lost while writing out the steps \begin{align*} (I - UDU^T)^k &= (UIU^T - UDU^T)^k\\ &= (U(I - D)U^T)^k\\ &= ? \end{align*}

What exactly are $(U^T)^k$ and $U^k$?

Answer 1

I am not sure why you are asking about $(U^T)^k$ and $U^k$, but recall that it does not generally hold that $(ABC)^k = A^kB^kC^k$; we would need to have more information about $A,B,C$ (for instance, that they commute).

We can prove the result inductively. Note that $$ \begin{align} (U(I - D)U^T)^k &= (U(I - D)U^T)^{k-1}U(I - D)U^T \\ &= [(U(I - D)U^T)^{k-1}]U(I - D)U^T \\ & = \color{red}{[U(I - D)^{k-1}U^T]}U(I - D)U^T \\ & = U(I - D)^{k-1}\color{red}{[U^TU]}(I - D)U^T \\ & = U\color{red}{(I - D)^{k-1}(I - D)}U^T = U(I - D)^kU^T. \end{align} $$

Answer 2

It would help if you write $U^{-1}$ instead of $U^T$.

Then you should be able to see the more familiar identity: $(UXU^{-1})^k = UX^kU^{-1}$.

Putting $X = I - D$ gives what you want to show.

Answer 3

$$(UMU^T)^k=(UMU^T)(UMU^T)(UMU^T)\cdots (UMU^T) \\=UM(U^TU)M(U^TU)M(U^T\cdots U)MU^T\\=UM^kU^T$$

and

$$(I-D)^k=\begin{pmatrix} (1-d_1)^k&0&0&\cdots&0 \\0&(1-d_2)^k&0&\cdots&0 \\0&0&(1-d_3)^k&\cdots&0 \\&&\cdots \\0&0&0&\cdots&(1-d_n)^k \end{pmatrix}$$