# 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.