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

Question

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?

score 28 · Accepted Answer · answered Aug 05 '15 at 05:03

28

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*}

answered Aug 05 '15 at 05:03

Sheldon Axler

6,226
1
18
25

14

Classic Axler move! – user217285 Aug 05 '15 at 05:03
8

Welcome to MSE, Professor Axler. So glad to see you here! And thanks for the book! Cheers! – Robert Lewis Aug 05 '15 at 05:23
1

This answer just made my day! – Brian Fitzpatrick Aug 05 '15 at 06:39

score 4 · Answer 2 · answered Aug 05 '15 at 04:39

4

Step 1: Prove that $\det(A^T) = \det A$ for all square matrices $A$.

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

answered Aug 05 '15 at 04:39

Alex G.

8,848

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

2 Answers2