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**Ackbeet** I think it likely. But it needs a proof. The eigenvectors of a matrix would not be, I think, the same as the eigenvectors of its transpose. Therefore, it's not inherently obvious, at least to me, that there would be the same number of linearly independent eigenvectors.

I asked if you knew what the size of $\displaystyle A$ was. Of course it's square, or the whole eigenvalue process would be undefined. I'm wondering if it's n x n or not. Because if it is, there might be a very nice way of relating the eigenvectors of $\displaystyle A$ to those of $\displaystyle A^{T}.$