# An orthogonal matrix that satisfies a property must be a permutation matrix

Let $$A$$ be an $$n\times n$$ orthogonal matrix such that $$\sum_{k=1}^na_{ik}^3a_{jk}=\sum_{k=1}^na_{jk}^3a_{ik}$$ for every $$1\le i,j\le n$$.

Original question which is solved by a counterexample given (For the new question see Edit 2): I want to show that $$A$$ is either a permutation matrix (i.e., all but one entry in each row and column is zero, and that non-zero entry is $$1$$ or $$-1$$), or all entries of $$A$$ have absolute value $$1/\sqrt n$$.

This question happens in my calculation on some functions from the tangent space of $$A$$.

I will provide a proof of $$n=2$$ without using the $$\sin$$ & $$\cos$$ representation of $$A$$:

$$a_{11}^3a_{21}+a_{12}^3a_{22}=a_{11}a_{21}^3+a_{12}a_{22}^3$$, so $$a_{11}a_{21}(a_{11}^2-a_{21}^2)=a_{12}a_{22}(a_{22}^2-a_{12}^2)$$. Since they are unit vectors, $$a_{11}^2-a_{21}^2=a_{22}^2-a_{12}^2$$, so it's easy to deduce the desired result. However, I find it hard to generalize this method.

Any suggestions will be appreciated.

Edit: Ok let's add the condition that $$\sum_{k=1}^na_{ki}^3a_{kj}=\sum_{k=1}^na_{kj}^3a_{ki}$$ for every $$1\le i,j\le n$$. I have proved that this must be true from the assumptions.

Edit 2: the comment gives a counterexample of this. As Ilya Bogdanov suggested, it is true that if $$A$$ is irreducible, then each non-zero entry of $$A$$ has the same absolute value?

• @LSpice What I meant is from the assumption of $\sum_{k=1}^na_{ik}^3a_{jk}=\sum_{k=1}^na_{jk}^3a_{ik}$. – ryanriess May 10 '20 at 18:39
• What if you take the direct sum of a (normalised) 2x2 Hadamard and a 2x2 identity matrix? – Padraig Ó Catháin May 10 '20 at 18:44
• @PadraigÓCatháin Well you find a counterexample. Thanks! – ryanriess May 10 '20 at 18:53
• So, @Padraig's comment siggests that any direct sum of normalized Hadamard matrices, with rows and columns being permuted arbitrarily, works. Perhaps, it is better to ask about irreducible matrices... – Ilya Bogdanov May 10 '20 at 18:57
• Check out weighing matrices -- they are nxn orthogonal matrices with k non-zero entries in each row and column. There should be also lots of irreducible examples of these. – Padraig Ó Catháin May 10 '20 at 19:14

It seems that here is an irreducible counterexample: $$\frac12\begin{bmatrix} 1& 1& -1& 1& 0& 0& 0& \cdots& 0& 0& 0& 0& 0\\ 1& 1& 1& -1& 0& 0& 0& \cdots& 0& 0& 0& 0& 0\\ 0& 0& 1& 1& -1& 1& 0& \cdots& 0& 0& 0& 0& 0\\ 0& 0& 1& 1& 1& -1& 0& \cdots& 0& 0& 0& 0& 0\\ \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \vdots& \ddots& \vdots& \vdots& \vdots& \vdots& \vdots\\ 0& 0& 0& 0& 0& 0& 0& \cdots& 0& 1& 1& 1& -1\\ -1& 1& 0& 0& 0& 0& 0& \cdots& 0& 0& 0& 1& 1\\ 1& -1& 0& 0& 0& 0& 0& \cdots& 0& 0& 0& 1& 1\\ \end{bmatrix}$$ Surely, one can find many more similar examples.