How do I prove that the sum (or difference) of two orthogonal matrices are not orthogonal?

Question

I am currently struggling with this problem:

Let $U$ and $V$ be two $n\times n$ orthogonal matrices. Are

$U+V\quad \text{and}\quad UV-VU$

also orthogonal? Explain your answer.

I know that products of orthogonal matrices are also orthogonal, so $UV$ and $VU$ are orthogonal. With that out of the way I approached the problem like this:

$(U+V)(U+V)^T=(U+V)(U^T+V^T)=UU^T+UV^T+VU^T+VV^T\\\hspace{3.36cm}=2I+UV^T+VU^T\neq I.$

How do I prove that the final inequality is true or not? I know that the products of $U$ and $V$ are orthogonal but can I prove that $UV^T+VU^T\neq -I$? I can't see how. Is there some other approach to this problem I am not seeing?

Similarly, if I use the same approach for the difference $UV-VU$, after some playing around with the expression I end up with

$(UV-VU)(UV-VU)^T=2I-UVU^TV^T-VUV^TU^T\neq I.$

So once again the same problem, can I somehow prove that the left-hand side is indeed not equal to $I$?

Thanks for any answers.

Answer 1

Since $U+V=U\left(I-(-U^TV)\right)$ and $UV-VU=UV(I-V^TU^TVU)$, the two cases are the same in spirit. You are essentially asking when $I-Q$ is orthogonal, given that $Q$ is orthogonal.

Since all eigenvalues of an orthogonal matrix lie on the unit circle, if $I-Q$ is orthogonal and $z$ is an eigenvalue of $Q$, we must have $|z|=|1-z|=1$. Thus $z=e^{\pm i\pi/3}$. (As non-real eigenvalues of a real square matrix must occur in conjugate pairs, this occurs only when $n$ is even.) Conversely, if all eigenvalues of $Q$ are equal to $e^{\pm i\pi/3}$, then $Q^2-Q+I=0$ and hence $I-Q=-Q^2$ is orthogonal.

In short, when $U$ and $V$ are orthogonal:

• $U+V$ is orthogonal precisely when all eigenvalues of $Q=-U^TV$ over $\mathbb C$ are equal to $e^{\pm i\pi/3}$, e.g. when $U=I_2,\,V=-\pmatrix{\frac12&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac12}$ and $U+V=\pmatrix{\frac12&\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}&\frac12}$;
• $UV-VU$ is orthogonal precisely when all eigenvalues of $Q=V^TU^TVU$ are equal to $e^{\pm i\pi/3}$, e.g. when $U=\pmatrix{1&0\\ 0&-1},\,V=\pmatrix{\frac{\sqrt{3}}{2}&-\frac12\\ \frac12&\frac{\sqrt{3}}{2}}$ and $UV-VU=\pmatrix{0&-1\\ -1&0}$.

Answer 2

Hint for part 1: What if $U=-V$?

Use that idea to come up with a similar argument for part 2.