Finding a matrix $A$'s eigenvalues from its square $A^2$'s eigenvalues

Question

In the proof I read, a $n \times n$ matrix $A^2$ has the eigenvalues $k^2$ with multiplicity $1$ and $k-1$ with multiplicity $n-1$. We also know that the matrix $A$ itself is symmetric. Then the proof says

Since $A$ is symmetric and hence diagonalizable, we conclude that A has the eigenvalues $k$ (of multiplicity $1$) and $\pm \sqrt{k-1}$.

What I didn't understand with this part of the proof is why did we only consider $k$, but not $-k$ as the eigenvalue of $A$. Can you please explain?

Edit(Additional information in the proof that might be related):

The matrix $A^2$ is all $k's$ in the main diagonal and $1's$ everywhere else. Or we can write it as, $A^2=(k-1)I+J$ where $I$ is the $n \times n$ identity matrix and $J$ is the $n \times n$ matrix of all $1's$.

Also the matrix $A$ consists of only $1's$ and $0's$ with a main diagonal of all $0's$.

Finally, we have $k+n-1=k^2$

Answer 1

Let $n=2$, $k=1$, suppose $$A=\begin{bmatrix}-1 & 0 \\ 0 & 0\end{bmatrix}$$

then we have $$A^2 = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$$

As we can see, it is possible that the eigenvalue of $A$ is $-k$. Hence you are right unless there is other additional informations.

Edit:

Let the all one vector be $e$.

Then we have $$A^2e=(k+(n-1))e=k^2e$$

Hence we have $$Ae=\pm k e$$

However, notice that $A$ is a nonnegative matrix, $Ae$ sums up all the columns and hence the result must be nonnegative.

Hence $$Ae=|k|e$$

Since $A$ is symmetric, the eigenvalues are real, that is we need $\pm \sqrt{k-1}$ to be real, hence we have $k-1 \ge 0$, and $k$ is positive.

Hence the eigenvalue is $k$ and not $-k$.

Remark: The other eigenvectors for $A$ are $e_1-e_i$ where $i \in \{2, \ldots, n\}$.

Answer 2

We should also consider $-k$ as a possible eigen value of $A$. Consider for example $$A=\begin{pmatrix} -2&0\\ 0&1 \end{pmatrix}\\ \implies A^2=\begin{pmatrix} 4&0\\0&1 \end{pmatrix}$$ Here $k=2$ and $n=2$.