Given a square matrix with orthogonal, non-zero rows (that are not orthonormal); must the rows each have exactly and only one non-zero element in order for the columns of the matrix to be orthogonal as well?
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No, for example $\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} $ or more generally $\begin{pmatrix} x & y \\ -y & x \end{pmatrix} $ for $x,y \neq 0$.
Ryan Vitale
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Erk. Just realized I had downvoted this instead of upvoting... (and can't change that unless the question gets edited.) – Semiclassical Jun 12 '16 at 02:11
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I'll edit my answer by adding a "." at the end. – Ryan Vitale Jun 12 '16 at 21:36
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Rotation matrices $$\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ are a counterexample.
angryavian
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Some complements:
As you may have seen, there are a lot of counterexample. If you want one with the norm of the rows not all equal, just take $$ \begin{pmatrix} 1 & 1 & 0\\ 1 & -1 & 0\\ 0 & 0 & 1 \end{pmatrix} $$
Your claim is true only if the norm of the rows are all distinct. In fact, in that case, you have (since $A$ is invertible) $$ AA^T =D \quad A^TA=D' \implies A^{-1}DA = D' $$ where $D$ and $D'$ are diagonal matrices, so $A$ is forced to be almost a permutation matrix.
Exodd
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