I cannot seem to get the right eigenvalues when using the secular equation (specific case) I have a matrix A: \begin{bmatrix}7/3&-2/3&-2/3\\-2/3&7/3&-2/3\\-2/3&-2/3&7/3\end{bmatrix}
which I solve using the characteristic equation like so: https://imgur.com/gallery/TGgxpEO
The problem is, the right eigenvalues are 1 and 3. What am I doing wrong?
 A: You went wrong right off the bat when you expanded $\det(A-\lambda I)$. The main diagonal elements of $A-\lambda I$ are all $\frac73-\lambda$, so when you pulled out a factor of $\frac13$ to simplify your calculations, they should’ve become $7-3\lambda$. What you’ve written down is instead the determinant of $A-\frac\lambda3I$, so all of your eigenvalues are off by a factor of three. To put it a different way, what you’ve done is to compute the eigenvalues of $3A$. It’s OK to scale a matrix in this way to simplify things since if $Av=\lambda v$ then $(cA)v=(c\lambda)v$, but you then have to remember to scale the resulting eigenvalues by $1/c$ to get the eigenvalues of the original matrix.  
Note that for this matrix, it’s easy to find eigenvectors and eigenvalues directly, without solving the characteristic equation. It’s of the form $aI+bJ$, where $J$ is the $n\times n$ matrix of all ones and $b\ne0$. $bJ$ is a rank-one matrix with equal row sums, so $0$ is an eigenvalue with multiplicity $n-1$ and the vector of all $1$s is an eigenvector with eigenvalue $nb$. The eigenvalues of the original matrix are therefore $a$ and $a+nb$, respectively. In this case, $a=3$ and $b=-2/3$.
A: the eigenvalues of the matrix with all elements equal to $1$ (and 3 by 3) are
(0,0,3).
the eigenvalues of the matrix with all elements equal to $-2/3$ (and 3 by 3) are
(0,0,-2).
The eigenvalue of that matrix added to $3I,$ which is your actual matrix, are
(3,3,1).
