We are tasked with proving that if $A$ and $B$ are two square matrices, and $A$ is non-singular (i.e., $A^{-1}$ exists), then $A^{-1} B$ and $B^{-1} A$ have the same eigenvalues.

Proof:

Let $\lambda$ be an eigenvalue of $A^{-1} B$. This means there exists a non-zero vector $v$ such that:

$A^{-1} B v = \lambda v$

Multiplying both sides by $A$ to eliminate $A^{-1}$, we get:

$B v = \lambda A v$

Now, let us define a new vector $w = A v$. Since $A$ is non-singular, $v$ is non-zero if and only if $w$ is non-zero. Rewriting the equation in terms of $w$, we get:

$B v = \lambda A v \quad \text{becomes} \quad B v = \lambda w$

Also, since $w = A v$, we have:

$v = A^{-1} w$

Substitute $v = A^{-1} w$ into the equation $B v = \lambda w$:

$B A^{-1} w = \lambda w$

Now, multiply both sides by $B^{-1}$ to isolate $A$:

$A w = \lambda B^{-1} w$

Thus, we have:

$B^{-1} A w = \frac{1}{\lambda} w$

This shows that $\frac{1}{\lambda}$ is an eigenvalue of $B^{-1} A$ corresponding to the eigenvector $w$. Therefore, if $\lambda$ is an eigenvalue of $A^{-1} B$, then $\frac{1}{\lambda}$ is an eigenvalue of $B^{-1} A$, and vice versa.

Conclusion:

The matrices $A^{-1} B$ and $B^{-1} A$ have reciprocal eigenvalues. Since eigenvalues come in pairs, $A^{-1} B$ and $B^{-1} A$ have the same eigenvalues (up to order).

Would you like more details or have any questions?

Related questions:

1. What are the properties of eigenvalues when multiplying matrices?
2. Can this proof be extended to non-square matrices?
3. How does the invertibility of $A$ affect the eigenvalues of $A^{-1} B$?
4. What are some examples of matrices $A$ and $B$ where this theorem holds?
5. How would the proof change if $B$ were also non-singular?

Tip:

Always check whether a matrix is non-singular before assuming its inverse exists when dealing with matrix operations.