Tricks for quickly reading off the eigenvalues of a matrix
This is a soft question without objective answers, but I can list some cases where finding the eigenvalues (or at least one eigenvalue) is easy:
- Singular matrices with obvious kernels
- Diagonal or triangular matrices
- Block-diagonal matrices with easily-analyzed blocks
- Matrices with known spectrum that have been spectrally shifted (by adding or subtracting a mutiple of $I$)
- Powers (including the inverse) of a matrix with known spectrum
- Products involving a matrix of known spectrum and its transpose
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 $1$.