Eigenvalues of symmetric matrix 4x4 I have to calculate the eigenvalue of this symmetric matrix:
$\left[ \begin{array}{rrrr}
u & u & u & v \\
u & u & v & u \\
u & v & u & u \\
v & u & u & u \\
\end{array}\right] $
What is the best way to calculate the eigenvalue. I know that I can A - $\lambda$ and calculate det or I can reduce it and read the eigenvalues from the diagonal but is there a better way?
 A: your matrix is $\begin{bmatrix} u & u& u &u \\u & u& u &u \\u & u& u &u \\u & u& u &u \end{bmatrix}$+  $\begin{bmatrix} 0& 0& 0 &v-u \\0 & 0& v-u&0 \\0& v-u& 0&0\\v-u & 0& 0 &0\end{bmatrix}$. These matrices commute with eigenvalues being $0,0,0, 4u$ and $v-u, v-u, u-v, u-v$. And also noting that one of the eigenvalues is surely $3u+v$ since the row sum is constant i.e  $3u+v$. Hence the other three eigenvalues are the sum of the eigenvalues of the two matrices i.e $v-u, u-v, u-v$.
A: Lowbrow Approach: Here is a possible way to do this. Notice that for a matrix $A$, $\operatorname{tr} A=\sum_i \lambda_i$ where the $\lambda_i$ are the complex eigenvalues, and that $\det A=\prod_i \lambda_i$. By inspection, as mentioned in the comments, we detect that $u+u+u+v$ is an eigenvalue corresponding to eigenvector $[1,1,1,1]^T$. On the other hand, corresponding to $[1,-1,1,-1]^T$ is the eigenvector $u-v$. Because our matrix is "only" $4\times 4$ we can denote by $\lambda_1,\lambda_2,\lambda_3,\lambda_4$ our eigenvalues. Then we know 
$$ \lambda_1+\lambda_2+\lambda_3+\lambda_4=\operatorname{tr}(A)=4u$$
and that
$$ \lambda_1\lambda_2\lambda_3\lambda_4=\det(A).$$
Now, to compute $\det(A)$ you will want to row reduce $A$ into a reasonable form. Then because you have $2$ eigenvalues already, you have a system of two equations in two unknowns and you  can  solve for  the  other two. 
Caution: The algebra here does not appear to be super fun. That said, this will definitely work. A general guideline for these problems is that if you  have an $n\times n$ matrix ($n=3$ usually) and can guess $n-2$ eigenvalues, this method will return the remaining $2$. 
Systematic Approach: The good news is that this can  also solve this using some theory. This is an example of a circulant matrix and so its eigenvalues have a specific form. It is related to the  discrete Fourier transform, and you can read about that here: https://en.wikipedia.org/wiki/Circulant_matrix
Anyway, taking the results from there, we see (denoting $\omega_1=\exp(\frac{2\pi i}{4})=\exp(\frac{\pi i}{2})$) that eigenvectors corresponding to the distinct eigenvalues are 
$$ v_0=\begin{bmatrix}
1\\
1\\
1\\
1
\end{bmatrix},\:\:
v_1=\begin{bmatrix}
1\\
\omega_1\\
\omega_1^2\\
\omega_1^3
\end{bmatrix},\:\:
v_2=\begin{bmatrix}
1\\
-1\\
1\\
-1
\end{bmatrix},\:\:
v_3=\begin{bmatrix}
1\\
\omega_1^3\\
\omega_1^2\\
\omega_1
\end{bmatrix}.$$
Then, applying these to the first row, we see that the eigenvalues of $A$ are 
$$ \lambda_0=u+u+u+v=3u+v$$
$$ \lambda_1=u+\omega_1u+\omega_1^2u+\omega_1^3v$$
$$ \lambda_2=u-u+u-v=u-v=u-v$$
$$ \lambda_3=u+\omega_1^3u+\omega_1^2u+\omega_1v.$$
Anytime you have a circulant matrix (the rows are cyclic permutations of eachother) this trick will work.
A: It can sometimes be faster and easier to look for eigenvectors first. When the rows are permutations of each other, the row sums are all equal, which means that the vector with all $1$s is an eigenvector with eigenvalue equal to this row sum. In this case, that’s $3u+v$.  
Since the matrix is symmetric, it has an orthogonal eigenbasis, so looking at vectors orthogonal to $(1,1,1,1)^T$ could be fruitful. A simple thing to try is differences of pair of columns, which leads to $(1,0,0,-1)^T$ and $(0,1,-1,0)^T$, both with eigenvalue $u-v$.  
The last eigenvalue comes “for free:” the sum of the eigenvalues is equal to the trace, so the last eigenvalue is $v-u$. If we assume that $v\ne u$, this eigenvalue is distinct from the others, so we look for a vector that’s orthogonal to the three already found. One such is $(1,-1,-1,1)^T$.
