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I already know that, for a given complex square matrix, the right eigenvalues are also left eigenvalues, with same algebric multiplicity.

But what about geometric multiplicity?

I'm searching for a proof that, if the right eigenvectors form a basis, the same goes for left eigenvectors, but up till now I've found nothing.

ygh
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  • In a finite-dimensional setting, the right eigenvectors are a basis iff the matrix vanishes a nonzero polynomial which is split and with simple roots. That condition is stable under transposition (which corresponds to considering left, instead of right, eigenvectors). – Aphelli Oct 28 '20 at 18:47
  • The Jordan normal form theorem can be used to show that a matrix is similar to its transpose, so the geometric multiplicities also match. This might be overkill, I don’t know. – Qiaochu Yuan Oct 28 '20 at 19:08

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