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Prove that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^T$.

I'm stucked here, i've approached the problem by looking at $\det(A-\lambda I)=0\iff\det(A^T-\lambda I)=0$. I tried some cases, and I can see it when the matrix is triangular since the main diagonal remains the same when $A$ is transposed, but this hasn't shown me a way to proceed. Any hints or ideas?

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Here is an easier proof that avoids determinants: First note that $(AB)^T = B^T A^T$ and use this to prove that $A$ is invertible if and only if $A^T$ is invertible. Thus \begin{align*} \lambda \text{ is an eigenvalue of } A &⟺ (A - \lambda I) \text{ is not invertible}\\ &⟺ (A - \lambda I)^T \text{ is not invertible}\\ &⟺ A^T - \lambda I \text{ is not invertible}\\ &⟺\lambda \text{ is an eigenvalue of } A^T. \end{align*}

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    $\begingroup$ Classic Axler move! $\endgroup$
    – user217285
    Aug 5 '15 at 5:03
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    $\begingroup$ Welcome to MSE, Professor Axler. So glad to see you here! And thanks for the book! Cheers! $\endgroup$ Aug 5 '15 at 5:23
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    $\begingroup$ This answer just made my day! $\endgroup$ Aug 5 '15 at 6:39
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Step 1: Prove that $\det(A^T) = \det A$ for all square matrices $A$.

Step 2: Prove that $(A-\lambda I)^T = A^T - \lambda I$.

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