I'm not sure how to do the following exercise. Could someone help me by either suggesting the knowledge I should have to solve this, or pointing me out in the right direction?

I have tried to expand B and A by substituting X in the equations, and then trying to get the eigenvalues for each. But, for B, I got a 1x1 reduction, while for A, it's a N by N.

When I tried to calculate the eigenvalues, A presented the greatest problem and that's where I got stuck.

But I think that I'm taking the difficult approach and the proof should be simpler.

exercise 3

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    $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung - Reinstate Monica Jan 24 '16 at 12:44

You will be surprised by how simple the problem is if you approach it from the right angle. Start from the definitions of eigenvalues and eigenvectors. It only takes a few manipulations afterwards to get where you want. Both problems can be solved almost on the spot.

To illustrate, for the matrix $\mathbf{A}^{\prime}\mathbf{A}$ we have by definition

$$\mathbf{A}^{\prime} \mathbf{A} \mathbf{x} = \lambda \mathbf{x} $$

Multiply both sides by $\mathbf{A}$ to get

$$\mathbf{A} \mathbf{A}^{\prime} \mathbf{A} \mathbf{x} = \lambda \mathbf{A} \mathbf{x}$$

but then we see that $\lambda$ is still an eigenvalue for $\mathbf{A} \mathbf{A}^{\prime}$ and the corresponding eigenvector is simply $\mathbf{A} \mathbf{x}$. Your intuition was correct.

From this you can also get the relationship between the eigenvectors. Note that while we have equal eigenvalues, it would be unreasonable to expect equal eigenvectors as well. You can see the comment section below for a relevant discussion.

You can do the other case on your own.

  • $\begingroup$ Yes, that was the problem, approaching from the right angle. Just to be sure, I left multiply by X the definition of eigenvalues, then A turned into B. And given that lambda was the same for both. The conclusion is that they share the same eigenvalues. Is that correct? Thank you for your help! $\endgroup$ – hugo Jan 24 '16 at 13:14
  • $\begingroup$ @hugo See my updated answer. $\endgroup$ – JohnK Jan 24 '16 at 13:23
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    $\begingroup$ @JohnK wrote "Interestingly, there is a theorem in linear algebra that says that the eigenvectors are the same if and only if the matrices commute". This is false in both directions, Strang's book notwithstanding. 'Same eignevectors -> matrices commute' requires diagonlizability. 'Matrices commute implies same eigenvectors' requires distinct eigenvalues of both matrices. $\endgroup$ – Mark L. Stone Jan 24 '16 at 13:42
  • $\begingroup$ @MarkL.Stone Yes, this comes from Strang's book indeed. He also offers a proof. $\endgroup$ – JohnK Jan 24 '16 at 13:43
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    $\begingroup$ @MarkL.Stone I'll remove the comment because this is a good point and it looks like Strang has got it wrong, or at least half-right anwyay. Thanks for bringing that to my attention. $\endgroup$ – JohnK Jan 24 '16 at 14:23

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