$A^TA$ eigenvalues

Question

Let $A\in M_{1,3}(\mathbb{R})$ be any $1\times 3$ matrix. Find the eigenvalues of $A^{T}A$.

I understand that this will obviously be a symmetric matrix, but I still can't see the solution.

Also, is there any way to see the eigenvalues of symmetric matrices in general?

EDIT: I tried some examples and it turns out that the first two eigenvalues are 0 and 0 and the third one is $$a_{11}^{2}+a_{12}^{2}+a_{13}^{2}.$$ Any way to show this is true in general?

Answer 1

Notice that $A^T$ is one eigenvector corresponding to the eigenvalue $AA^T = a_{11}^2 + a_{12}^2 + a_{13}^2$ \begin{equation} (A^TA)A^T = A^T(AA^T)= \underbrace{AA^T}_{\lambda}A^T \end{equation}

Also notice that we have $N-1=3-1=2$ zero eigenvalues because the matrix $I - \frac{1}{AA^T}A^TA$ is of rank $2$, \begin{equation} (A^TA)(I - \frac{1}{AA^T}A^TA) = A^TA - \frac{1}{A^TA} A^TAA^TA = A^TA - A^TA = 0 \end{equation}

Answer 2

Hint: Note that $A^TA$ has rank $1$, and therefore a nullspace of dimension $3 - 1 = 2$. That is $A^TA$ will have $0$ as an eigenvalue, with two associated linearly independent eigenvectors.

One convenient way to compute the remaining (non-zero) eigenvector is by considering the trace of $A^TA$.