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Prove the following results for the eigenvalues of an $n \times n$ matrix $A$:

(a) $0$ is an eigenvalue for $A$ if and only if $A$ is not invertible.

(b) $A$ and $A^T$ have the same eigenvalues.

carmichael561
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Sammy
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    http://math.stackexchange.com/questions/755780/is-a-matrix-a-with-an-eigenvalue-of-0-invertible , http://math.stackexchange.com/questions/123923/a-matrix-and-its-transpose-have-the-same-set-of-eigenvalues – Moo May 27 '16 at 04:46
  • $\det (0 \cdot I -A) = 0$ iff $A$ is singular, and $\det (sI-A) = \det (sI-A)^T$. – copper.hat May 27 '16 at 04:47
  • (b) See [link] (http://math.stackexchange.com/questions/123923/a-matrix-and-its-transpose-have-the-same-set-of-eigenvalues) – mattapow May 27 '16 at 04:47

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(a) Suppose $0$ is an eigenvalue of $A$. Then $Av=0v$ for some nonzero vector $v$. What does that mean?

(b) Look at the characteristic polynomial of the matrix $A$ and the characteristic polynomial of its transpose. What can you say about them?

parsiad
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