# A matrix and its transpose have the same set of eigenvalues

Let $\sigma(A)$ be the set of all eigenvalues of $A$. Show that $\sigma(A) = \sigma(A^T)$ where $A^T$ is the transposed matrix of $A$.

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This is a bit more advanced than what you need, but: an interesting article. –  Ｊ. Ｍ. Mar 24 '12 at 13:10
I guess your work in an algebraically closed field. In this case, use the fact that $r$ is an eigenvalue of $A$ if and only if $r$ is an eigenvalue of $A^T$. In fact, it can be shown that $A$ and $A^T$ are similar. –  Davide Giraudo Mar 24 '12 at 13:11
Here's one possible simpler problem that will get you started on the right path. If $A$ is an n by n singular matrix, can you show that $A^T$ is also singular? –  Sam Lisi Mar 24 '12 at 13:25
Please don't post your questions in the imperative; please tells us what your thoughts are about the question, so that people don't tell you things you already know; please tell us the context in which you encountered the question, so that people can write their answers at an appropriate level. –  Arturo Magidin Mar 24 '12 at 21:03

The matrix $(A - \lambda I)^{T}$ is the same as the matrix $(A^{T} - \lambda I)$, since the identity matric is symmetric.

Thus:

$$\det(A^{T} - \lambda I) = \det((A - \lambda I)^{T}) = \det (A - \lambda I)$$

From this it is obvious that the eigenvalues are the same for both $A$ and $A^{T}$.

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@PeterTamaroff Obviously! It's so obvious! I mean, how obvious can it get? You're right, it's truly so obvious. Spot on. –  Trancot May 20 '13 at 21:11
@Trancot What are you talking about? –  Pedro Tamaroff May 20 '13 at 21:12
I just don't like the word obvious. It's a personal thing. Just ignore me. –  Trancot May 20 '13 at 21:12
@Trancot OK. ${}{}{}{}$ –  Pedro Tamaroff May 20 '13 at 21:14
The question is asked by someone wishing to solve a problem about eigenvalues...therefore I expect the person reading/grading the answer to understand eigenvalues. I shouldn't need to write an answer explaining it for anyone else! –  fretty May 25 '13 at 17:58
$$\operatorname{det}(A-tI) = \operatorname{det}(A-tI)^T = \operatorname{det}(A^T-tI)$$ A matrix and its transpose have the same determinant. If you apply properties of transposition, you get that both $A$ and its transpose have the same characteristic polynomial.