If the matrix A has eigenvalues λ1, λ2, . . . λn then prove that A2 has eigenvalues λ21
, λ22
, . . . λ2
n. Also find
the eigenvalues of A−1 if A is non-singular.

### Part 1: Proving that \( A^2 \) has eigenvalues \( \lambda_1^2, \lambda_2^2, \dots, \lambda_n^2 \)

Let \( A \) be an \( n \times n \) matrix with eigenvalues \( \lambda_1, \lambda_2, \dots, \lambda_n \). This means that for each eigenvalue \( \lambda_i \), there exists a non-zero eigenvector \( \mathbf{v}_i \) such that:
\[
A \mathbf{v}_i = \lambda_i \mathbf{v}_i
\]
Now, consider the matrix \( A^2 \). We want to prove that the eigenvalues of \( A^2 \) are \( \lambda_1^2, \lambda_2^2, \dots, \lambda_n^2 \).

To do this, apply the matrix \( A^2 \) to the eigenvector \( \mathbf{v}_i \):
\[
A^2 \mathbf{v}_i = A(A \mathbf{v}_i) = A(\lambda_i \mathbf{v}_i) = \lambda_i A \mathbf{v}_i = \lambda_i (\lambda_i \mathbf{v}_i) = \lambda_i^2 \mathbf{v}_i
\]
This shows that \( \mathbf{v}_i \) is also an eigenvector of \( A^2 \), but with the eigenvalue \( \lambda_i^2 \).

Therefore, the eigenvalues of \( A^2 \) are \( \lambda_1^2, \lambda_2^2, \dots, \lambda_n^2 \).

---

### Part 2: Finding the eigenvalues of \( A^{-1} \) if \( A \) is non-singular

If \( A \) is non-singular, it means that \( A \) is invertible, and none of its eigenvalues are zero.

Let \( \lambda_i \) be an eigenvalue of \( A \), and let \( \mathbf{v}_i \) be the corresponding eigenvector. Then:
\[
A \mathbf{v}_i = \lambda_i \mathbf{v}_i
\]
Multiplying both sides by \( A^{-1} \), we get:
\[
A^{-1} A \mathbf{v}_i = A^{-1} \lambda_i \mathbf{v}_i
\]
Since \( A^{-1} A = I \), the identity matrix, this simplifies to:
\[
\mathbf{v}_i = \lambda_i A^{-1} \mathbf{v}_i
\]
Multiplying both sides by \( \frac{1}{\lambda_i} \), we obtain:
\[
A^{-1} \mathbf{v}_i = \frac{1}{\lambda_i} \mathbf{v}_i
\]
This shows that \( \mathbf{v}_i \) is an eigenvector of \( A^{-1} \) with eigenvalue \( \frac{1}{\lambda_i} \).

Thus, the eigenvalues of \( A^{-1} \) are \( \frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \dots, \frac{1}{\lambda_n} \).

---

### Summary:
- The eigenvalues of \( A^2 \) are \( \lambda_1^2, \lambda_2^2, \dots, \lambda_n^2 \).
- If \( A \) is non-singular, the eigenvalues of \( A^{-1} \) are \( \frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \dots, \frac{1}{\lambda_n} \).

---

Would you like more details on any of these steps, or have any questions?

Here are some related questions for deeper understanding:
1. What are the conditions under which a matrix is non-singular?
2. How do eigenvalues relate to the determinant and trace of a matrix?
3. Can a matrix with a zero eigenvalue be invertible? Why or why not?
4. How does the eigenvalue-eigenvector relationship change for a diagonalizable matrix?
5. How do eigenvalues behave under matrix exponentiation?

**Tip:** For any matrix transformation, eigenvalues provide critical insights into how that matrix scales vectors along its eigenspaces.

If the matrix A has eigenvalues λ1, λ2, . . . λn then prove that A^2 has eigenvalues λ1^2, λ2^2, . . . λn^2. Also find the eigenvalues of A−1 if A is non-singular.

This problem explores the proof that a matrix A with eigenvalues λ1, λ2, ... λn has eigenvalues λ1^2, λ2^2, ... λn^2 for A^2, and also finds the eigenvalues of A^−1 if A is non-singular. Learn key linear algebra concepts like matrix powers, matrix inverses, and eigenvalue properties in this detailed solution.

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