Number of possible zero entries in orthogonal matrices

Question

It's easy to check that in an orthogonal matrix $Q$ dimension $2 \times 2$ if there is entry $0$ in the matrix then necessary one additional zero must be present and the total number of zeros is $2$.

In an orthogonal matrix dim. $3 \times 3$ number of zeros can be (if they are present) , I suppose from observations, only $4$ or $6$ - once again we obtain an even number of possible zeros.

Examples: $ \begin{bmatrix} 0.6 & -0.8 & 0 \\ 0.8 & 0.6 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \ \ $ , $ \ \ \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix}$

• Can this observation be extended for other orthogonal matrices of greater dimensions? The number of zeros is always even? How to prove this?

• Maybe, it is known the explicit formula for the number of possible zeros in orthogonal matrices of any dimension?

Answer 1

An interesting observation, but it doesn't pan out unfortunately! It already fails in dimension 3. Wikipedia has the following counterexample of a rotoinversion: $$ \begin{bmatrix} 0 & -0.8 & - 0.6 \\ 0.8 & -0.36 & 0.48 \\ 0.6&0.48&-0.64 \end{bmatrix} $$

(WolframAlpha agrees that this is indeed orthogonal, the example is from here)

Answer 2

Rotations are orthogonal. Most examples of rotation matrices in any dimension will provide a counter example to this. In the $2 \times 2$ case consider $$\begin{pmatrix} Cos(x)&-Sin(x)\\ Sin(x)&Cos(x) \end{pmatrix}$$ This is orthogonal for any $x$. There are many cases where there are no zeros.