TITLE: Tricks for quickly reading off the eigenvalues of a matrix
QUESTION [16 upvotes]: I noticed that some mathematicians have an uncanny ability to identify the eigenvalues of matrices without doing much in the way of computation.
For instance, one might notice that all the rows have the same sum, from which it follows that the sum must be an eigenvalue.
There can be no simple way to find eigenvalues of arbitrary matrices, but what special cases might we look for where finding the eigenvalues are easy? What tricks do you know?

REPLY [4 votes]: Suppose we are given a matrix $A \in \mathbb{R}^{n \times n}$.
If $A = O_n$, the one eigenvalue of $A$ is $0$, with multiplicity $n$. 
If the first $n-1$ rows or columns of $A$ are either zero or a multiple of the $n$-th row or column, then we have a rank-$1$ matrix that can be written in the form
$$A = \mathrm{u} \mathrm{v}^T$$
where $\mathrm{u}, \mathrm{v} \in \mathbb{R}^n$. As the null space of $\mathrm{v}^T$ is $(n-1)$-dimensional, the null space of $A$ is at least $(n-1)$-dimensional. Hence, $0$ is an eigenvalue of $A$ with multiplicity at least $n-1$. Since the trace is the sum of the eigenvalues, the other eigenvalue is
$$\lambda = \operatorname{tr} (A) = \operatorname{tr} (\mathrm{u} \mathrm{v}^T) = \operatorname{tr} (\mathrm{v}^T \mathrm{u}) = \mathrm{v}^T \mathrm{u} = \langle \mathrm{u}, \mathrm{v} \rangle$$

For example, suppose we are given
$$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\  1 & 1 & 1\end{bmatrix}$$
As the 2nd and 3rd rows are equal to the 1st one, we conclude that $\operatorname{rank} (A) = 1$. Hence, one of the eigenvalues of $A$ is $0$, with multiplicity $2$. The other eigenvalue is $\operatorname{tr} (A) = 3$, with multiplicity $1$.