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I am trying to prove that a $n \times n$ matrix $A$ and $A^T$ have the same eigenvalues.

I can prove that $A$ and $A^T$ have the same entries on the diagonal, but I am not sure where to go from there.

Jorge
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2 Answers2

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$\lambda$ is an eigenvalue of the $n \times n$ matrix $A$ iff $\det(A-\lambda I)=0$.

Remember that the determinant of a matrix is equal to the determinant of its transpose. Thus if $\det(A-\lambda I)=0$ then $\det([A-\lambda I]^T)=0$. But $[A-\lambda I]^T = A^T-\lambda I^T = A^T - \lambda I$. Therefore $\det(A-\lambda I) = 0 \implies \det(A^T - \lambda I) =0$. Thus $A$ and $A^T$ have the same eigenvalues.

  • I follow you up until the last sentence. I'm not sure how you went from det(A^T - lambdaI) = 0 to them both having the same eigenvalues. – Jorge Mar 27 '15 at 02:49
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    The equation $\det(A^T- \lambda I) = 0$ implies that $\lambda$ is an eigenvalue of $A^T$. But that eigenvalue was arbitrary (as in this same argument will work for any of $A$'s eigenvalues), soall of $A$'s eigenvalues are also eigenvalues of $A^T$. Then to see that $A$ also has all of $A^T$'s eigenvalues, just realize that $(A^T)^T=A$. –  Mar 27 '15 at 02:57
  • Okay, I think I understand. If $\det(A-\lambda I) = 0$ is the same as $\det(A^T - \lambda I) = 0$ then they both have the same solution for the $\lambda$s? – Jorge Mar 27 '15 at 03:03
  • $\det(B-aI)=0$ is the definition of $a$ being an eigenvalue of the matrix $B$. So in your problem, we proved that $\det(A-\lambda I)=0 \implies \det(A^T -\lambda I)=0$, therefore we proved that $\lambda$ being an eigenvalue of $A$ implies $\lambda$ is an eigenvalue of $A^T$. And by a very similar argument, you can prove that $\lambda$ being an eigenvalue of $A^T$ implies it is an eigenvalue of $A$. So the set of all eigenvalues of $A$ and $A^T$ are exactly the same. –  Mar 27 '15 at 03:05
  • okay, I understand now. Thanks. – Jorge Mar 27 '15 at 03:18
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Hint: They will have the same eigenvalues if they have the same characteristic polynomial. (which can be shown that they do have the same easily)

PersonaA
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