# Eigenvalue Properties

When I took the math subject GRE I learned that in order to solve common linear algebra questions on the exam quickly, you should keep in mind a few facts about eigenvalues.

• The product of the eigenvalues of a matrix equals the determinant of the matrix
• The sum of the eigenvalues of a matrix equals the trace of the matrix
• If a matrix is invertible, the inverse matrix has reciprocal eigenvalues

I’m going to prove each of these facts.

## Product of eigenvalues equals determinant

Thinking back to your linear algebra, remember that we compute the characteristic polynomial by expanding $$\det(A - \lambda I)$$. Since the eigenvalues, $$\lambda_1, …, \lambda_n$$, are the roots of the characteristic polynomial, we can factor it as follows

$\det(A-\lambda I) = (\lambda_1 - \lambda) ... (\lambda_n - \lambda).$

If we set $$\lambda = 0$$ we get

$\det(A) = \lambda_1 \cdot ... \cdot \lambda_n.$

Which shows that the product of the eigenvalues of a matrix equals its determinant (and this is the constant term for the characteristic polynomial).

## Sum of eigenvalues equals trace

Again, we can find the characteristic polynomial

$\det(A-\lambda I) = (\lambda_1 - \lambda) ... (\lambda_n - \lambda)$ $= (-1)^n (\lambda^n + (\lambda_1 + ... + \lambda_n) \lambda^{n-1} + ... + (-1)^n \det(A)).$

Notice that the coefficient of $$\lambda^{n-1}$$ is the sum of the eigenvalues, but this term is also the trace of $$A$$ (look at the characteristic polynomial of a $$2 \times 2$$ matrix for an example). This is more of a proof sketch, but it’s essentially the same idea of manipulating the characteristic polynomial and making an observation about it.

## Inverse has reciprocal eigenvalues

This is pretty simple. Remember that an eigenvalue satisfies

$A x = \lambda x$

multiplying by $$A^{-1}$$ yields

$x = \lambda A^{-1} x$

and rearranging gives

$A^{-1} x = \frac{1}{\lambda} x.$

This is symmetric, so for each eigenvalue of $$A$$ (or $$A^{-1}$$) there is a reciprocal eigenvalue for the inverse. Notice that if the matrix is invertible it doesn’t have 0 as an eigenvalue since the determinant is non-zero.

Published on March 27, 2019