If A and B are two square matrices and A is non singular, prove that A^ -1 B and B^-1 A have same eigenvalues

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.

If A and B are two square matrices and A is non-singular, prove that A^-1 B and B^-1 A have the same eigenvalues.

This detailed solution explains how to prove that if A and B are square matrices and A is non-singular, the matrices A^-1 B and B^-1 A have the same eigenvalues. Explore the concepts of eigenvalues, matrix inverses, and key theorems relevant to linear algebra. Suitable for undergraduate students.

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