Characteristic polynomial of 10x10 matrix 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 $A^T A$?
 A: 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.
