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Consider the matrix A = $\begin{bmatrix}1&0&1&0&1&0&1&0&1&0&\\0&2&0&2&0&2&0&2&0&2&\end{bmatrix}$. What is the characteristic polynomial of the 10×10 matrix $A^T A$?

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  • $\begingroup$ What is it about this problem that you do not know how to do? $\endgroup$
    – Lee Mosher
    Commented Nov 12, 2014 at 1:48
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    $\begingroup$ Can you determine what is the rank of $A^TA$? And, if so, how will that help you answer the question? $\endgroup$
    – GEdgar
    Commented Nov 12, 2014 at 1:51
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    $\begingroup$ Don't know where to start. I guessing that the intention of the question isn't to find the determinant of a 10x10 matrix, so I'm wondering if there's a theorem that would provide a shortcut. I know the rank of the matrix will be 2 and thus there will be two non-zero eigenvalues. $\endgroup$
    – Zach
    Commented Nov 12, 2014 at 1:51
  • $\begingroup$ So my prof said that the non-zero eigenvalues of $A^T A$ are the same as those of $A A^T$, meaning the non-zero eigenvalues are 5 and 20 and the rest are zero. Why are the non-zero eigenvalues of those matrices the same, and does that imply the characteristic polynomial is λ^8*(λ-5)(λ-20)=0? $\endgroup$
    – Zach
    Commented Nov 12, 2014 at 2:00
  • $\begingroup$ I didn't check the eigenvalues of $AA^T$ but if they are $5$ and $20$, then you are correct with respect to the characteristic polynomial. Your other question, however, is not something that can/should be answered in a comment. If you wish, ask another question asking that. In my opinion it is not trivial and I would guess that that fact was given in class or something. $\endgroup$
    – Git Gud
    Commented Nov 12, 2014 at 2:08

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The point is, as you correctly pointed out that the (nonzero!) eigenvalues of $A^T A$ and $A A^T$ are the same, with the larger matrix having zero eigenvalues in addition.

If one calculates $A A^T$ one gets

$A A^T = \begin{pmatrix} 5 & 0 \\ 0 & 20 \end{pmatrix}$

so the eigenvalues are readily obtained as $5 $ and $20$ (as you already said). Since the rank of $A^T A$ is the same of $A A^T$ (which is 2 due to nonzero eigenvalues), we infer that $A^T A$ has an additional zero eigenvalue of degree 8. It needs to be degree 8, since the degree of the characteristic polynomial is 10 and we already know that the matrix has two simple nonzero eigenvalues (i.e. 5 and 20). So

$\chi_{A^TA}(\lambda) = \lambda^8 (\lambda-5)(\lambda-20)$.

and for reference

$\chi_{AA^T}(\lambda) = (\lambda-5)(\lambda-20)$.

EDIT: The key point is to realize that the nonzero eigenvalues to $A^TA$ are also eigenvalues to $A A^T$. Then everything follows, including that the rank is the same (because no assumptions on $A$ are made it hold also in the other direction by replacing $A$ with $A^T$). Since the rank is the same and the nonzero eigenvalues correspond to each other, it follows that the other eigenvalues (for the larger matrix) must be zero.

If

$A^TA w = \lambda w$

with $\lambda \neq 0$ we get

$A A^T A w = \lambda A w$

which shows nothing else than $\lambda$ is an eigenvalue to $A A^T$ with $v = A w$ the eigenvector ($A A^T v = \lambda v$). The $\lambda \neq 0$ condition is needed to ensure $v \neq 0$, i.e. that it is a true eigenvector.

EDIT2: Hmm, as Git Gud pointed out, there is the question about the multiplicities. I guess one can prove that the following way. It turns out that the minimal polynomials are the same. let be the minimal polynomial of $A^TA$ be given by

$0 = a_0 I + a_1 A^TA + \ldots a_n (A^TA)^n$

if we multiply with $A^T$ from the right and with $A$ from the left we get

$0 = a_0 A A^T + a_1 A A^T A A^T + \ldots a_n A (A^TA)^n A^T$

or

$0 = a_0 (A A^T) + a_1 (A A^T)^2 + \ldots a_n (A A^T)^{n+1}$

So one sees that the minimal polynomials for $A^TA$ and $AA^T$ are essentially the same with an additional $(x-0)$ factor. Ok, in a more rigorous fashion one needs to consider both ways and that the minimal polynomial of $AA^T$ only divides the last expression. In any case one can show that the eigenvalues are the same and the multiplicities in the minimal polynomial as well. Since we have symmetric matrices the algebraic (degree in the chracteristic polynomial) and geometric multiplicities (degree in the minimal polynomial) are also the same, so no problem arises from that.

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  • $\begingroup$ Ah, right. In a comment to the question I said it wasn't trivial, but indeed it is. What's not trivial is proving that even the multiplicities are the same for non-null eigenvalues. $\endgroup$
    – Git Gud
    Commented Nov 12, 2014 at 2:50
  • $\begingroup$ @GitGud: true, I added a proof sketch showing that the minimal polynomial are essentially the same. Not rigorous but maybe sufficient here... $\endgroup$
    – Andreas H.
    Commented Nov 12, 2014 at 3:04
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    $\begingroup$ Cool idea with the minimal polynomials. Essentially the same as what I have in my notes, but more motivated than what I got. $\endgroup$
    – Git Gud
    Commented Nov 12, 2014 at 3:12

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