Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been trying to complete a proof for a while now, but I can't. It would be great if someone could finish it for me so that I can at least learn from the solution.

The problem is asked like this:

Show that $\lambda$ is an eigenvalue of A iff $\lambda$ is eigenvalue of A transpose.

There's a hint and it says:

For any $\lambda$, $(A-\lambda I )^T = A^T-\lambda I$. By a theorem (which one?), $A^T - \lambda I$ is invertible iff $A-\lambda I$ is invertible.

I know that there is a theorem, in this book called the Invertible Matrix Theorem, which says that if A is invertible then so is its transpose. But I don't see what that has to do with anything? If A has an eigenvalue, then $A-\lambda I$ is linearly dependent and is not even invertible! Is that what the hint is all about? Well, I still don't see what that says about the solution set. In order to complete this proof, I think, we need to show that the solution set of the homogenous equation $A-\lambda I = 0$ is the same as that for its transpose.

Many thanks.

share|cite|improve this question
"If A has an eigenvalue, then A is linearly dependent and is not even invertible!" - you meant "...then $\mathbf A-\lambda\mathbf I$ is linearly dependent..." I suppose. – J. M. Apr 28 '11 at 16:57
Yes that's what I meant. Thanks for pointing it out. – Pickett Apr 28 '11 at 16:59
"I still have the problem with =>" - Better now? – J. M. Apr 28 '11 at 17:03
Yes, you fixed it, thank you! – Pickett Apr 28 '11 at 17:05
I do not know what $A$ is linearly dependent nor what $A-\lambda I$ is linearly dependent means. – Did Apr 28 '11 at 17:07
up vote 8 down vote accepted

Use the definition of being an eigenvalue and the invertible matrix theorem mentioned in your post: $\lambda$ is an eigenvalue of $A$ if and only if $A-\lambda I$ is not invertible if and only if $A^T-\lambda I$ is not invertible if and only if $\lambda$ is an eigenvalue of $A^T$.

share|cite|improve this answer
I get that if $A^T - \lambda I$ is not invertible, there must be an eigenvalue. Because if there would be no solution to the homogenous equation $A^T - \lambda I = 0$, the set of rows would make up an independent set and the matrix would in fact be invertible (please tell me if I'm getting this wrong). But I don't know why the fact that $A^T$ HAS an eigenvalue neccesarily means that it is the SAME eigenvalue as for A. How can I know this? – Pickett Apr 28 '11 at 17:55
@Calle: The determinant is invariant under transpose. In particular, $\det(A - \lambda I) = \det( (A-\lambda I)^T) = \det( A^T - (\lambda I)^T) = \det(A^T - \lambda I^T) = \det(A^T - \lambda I)$. Say you know that $\lambda_0$ is an eigenvalue for $A^T$, so it's a solution for $\det (A^T - \lambda I) = 0$. Can that tell you anything about the solutions for $\det(A-\lambda I) = 0$? – Michael Chen Apr 28 '11 at 23:26
Alright it will suffice. Thank you everybody. – Pickett Apr 29 '11 at 8:51

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.