Eigenvalues of a $A^T A$

Hint: Suppose that $v$ is an $n \times 1$ vector orthogonal to $A$ (or more precisely, $w$ is a $1 \times n$ vector orthogonal to $A$, and $v = w^t$). What do you get when you compute $A^{*}Av$?

$$A^tA = \begin{pmatrix} a_1a_1 & a_1a_2 & a_1a_3 & \dots & a_1a_n\\ a_2a_1 & a_2 a_2 & a_2a_3 & \dots & a_2a_n \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_na_1 & a_na_2 & a_na_3 & \dots & a_na_n \end{pmatrix}$$

All the columns can be obtained by mutiplying by a scalar the first column, so $rank(A^tA)=1$. So $0$ is an eigen value with multiplicity $n-1$. The sum of eigen values is equal to the trace of the matrix, thus you can easily find the last eigenvalue.