Tricks for quickly reading off the eigenvalues of a matrix

Question

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?

Answer 1

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

Answer 2

One trick to know if the eigen values we found are right is to check if the sum of eiven values equals the trace (sum of the diagonal elements) . Also if the product of eigenvalues equals the determinant of that matrix rhen the eigenvalues you found are correct .

Answer 3

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

$$\lambda = \operatorname{tr} (A) = \operatorname{tr} (\mathrm{u} \mathrm{v}^T) = \operatorname{tr} (\mathrm{v}^T \mathrm{u}) = \mathrm{v}^T \mathrm{u} = \langle \mathrm{u}, \mathrm{v} \rangle$$

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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 $1$.