If $\lambda$ is an eigenvalue of a square matrix $M$, then $\frac{1}{\lambda}$ is an eigenvalue of the inverse matrix $M^{-1}$, provided that $M$ is invertible and $\lambda \neq 0$.

Explanation:

1. By definition, if $\lambda$ is an eigenvalue of the matrix $M$, then there exists a non-zero vector $v$ (an eigenvector) such that:

$M v = \lambda v$

2. If $M$ is invertible, multiplying both sides of the equation by $M^{-1}$ gives:

$M^{-1} M v = M^{-1} \lambda v$

Simplifying this:

$v = \lambda M^{-1} v$

Dividing both sides by $\lambda$ gives:

$M^{-1} v = \frac{1}{\lambda} v$

Thus, $\frac{1}{\lambda}$ is an eigenvalue of $M^{-1}$.

Summary:

If $\lambda$ is an eigenvalue of a square matrix $M$, then $\frac{1}{\lambda}$ is an eigenvalue of the inverse matrix $M^{-1}$.

Do you want more details on eigenvalues and eigenvectors, or have any questions?

Here are 5 related questions for further exploration:

1. What happens to the eigenvalues of a matrix when it is transposed?
2. How do eigenvalues relate to the determinant of a matrix?
3. What are the eigenvalues of a diagonal matrix?
4. How do eigenvalues change under matrix multiplication?
5. How can eigenvectors be used in applications like principal component analysis (PCA)?

Tip: Eigenvalues of symmetric matrices are always real.