I'm doing an exercise from an old exam that asks for calculating the Singular Value Decomposition (SVD) of the following matrix, for any $p\in\mathbb{R}$:
$$A_p=\begin{bmatrix} \ \ 1+p &\ \ 1-p & -1+p & -1-p\\ \ \ 1-p&\ \ 1+p&-1-p&-1+p\\ -1+p&-1-p&\ \ 1+p&\ \ 1-p\\ -1-p&-1+p&\ \ 1-p&\ \ 1+p \end{bmatrix}$$
Now I can see that $A_p={A_p}^T$ and thus by the spectral theorem there exists an $X\in O_4(\mathbb{R})$ such that $X^{-1}A_pX=\Lambda:=\mathrm{diag}(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$, and thus the singular values equal the absolute values of the eigenvalues $\lambda_1,\lambda_2,\lambda_3,\lambda_4$. Also, it is easy to see that with $r_i$ for the $i$'th row of $A_p$, $r_1=-r_4$ and $r_2=-r_3$ and thus $\dim\ker(A_p)=2$, and thus $A_p$ has eigenvalue $0$ with multiplicity $2$ for any $p\in\mathbb{R}$.
Now from here everything I tried and expanded didn't lead me anywhere. Mathematica confirms my work this far and gives the remaining two eigenvalues which are $4$ and $4p$. However, I want to know how to calculate these eigenvalues myself. I'm quite sure there is some trick that I am not seeing since calculating $\det(A_p-\lambda I)$ is tedious and this old exam question does not yield that much points.
Just a note: A follow up question also asks for the best $\mathrm{rank}\ 1$ approximation so I would also need the eigenvectors corresponding to the largest eigenvalue, which can be either $4$ or $4p$.
Any help is much appreciated.
I just found out that $$A_p=4\begin{bmatrix}\ \ \frac12\\ \ \ \frac12\\-\frac12\\-\frac12 \end{bmatrix}\begin{bmatrix}\frac12& \frac12&-\frac12&-\frac12 \end{bmatrix}+4p\begin{bmatrix}\ \ \frac12\\-\frac12\\\ \ \frac12\\-\frac12 \end{bmatrix}\begin{bmatrix}\frac12& -\frac12&\frac12&-\frac12 \end{bmatrix},$$ so this fully solves my question and directly yields the eigenvectors I was looking for.