. If
1 2 , ,....,   n
are the latent roots of a matrix A, show that the latent roots of A3
are
3 3 3
1 2 , ,....,   n
.

To show that the latent (or eigen) roots of \( A^3 \) are \( \lambda_1^3, \lambda_2^3, \dots, \lambda_n^3 \), we will use the following properties of eigenvalues (latent roots) and eigenvectors.

### Key Concepts:
1. **Eigenvalue Definition**: If \( A \) is a square matrix, and \( \lambda \) is an eigenvalue of \( A \), there exists a non-zero vector \( v \) (called the eigenvector) such that:
   \[
   A v = \lambda v
   \]
   where \( \lambda \) is the eigenvalue (latent root) corresponding to eigenvector \( v \).
   
2. **Eigenvalue of Powers of a Matrix**: If \( \lambda \) is an eigenvalue of \( A \), then the eigenvalue of \( A^k \) is \( \lambda^k \) for any positive integer \( k \).

### Proof:

Let \( A \) be a square matrix of order \( n \), and let \( \lambda_1, \lambda_2, \dots, \lambda_n \) be the eigenvalues of \( A \).

#### Step 1: Eigenvalue relation
Suppose \( v_i \) is an eigenvector of \( A \) corresponding to the eigenvalue \( \lambda_i \). By definition of eigenvalues, we have:
\[
A v_i = \lambda_i v_i
\]
for each \( i = 1, 2, \dots, n \).

#### Step 2: Powers of the matrix
Now, consider the matrix \( A^3 \). To find the eigenvalues of \( A^3 \), we apply \( A^3 \) to the eigenvector \( v_i \):
\[
A^3 v_i = A(A(A v_i)) = A(A(\lambda_i v_i)) = A(\lambda_i A v_i)
\]
Since \( A v_i = \lambda_i v_i \), we can substitute this back into the equation:
\[
A^3 v_i = A(\lambda_i (\lambda_i v_i)) = A(\lambda_i^2 v_i) = \lambda_i^2 A v_i
\]
Again, using \( A v_i = \lambda_i v_i \), we get:
\[
A^3 v_i = \lambda_i^2 (\lambda_i v_i) = \lambda_i^3 v_i
\]

#### Step 3: Conclusion
Thus, \( v_i \) is also an eigenvector of \( A^3 \), and the corresponding eigenvalue is \( \lambda_i^3 \).

Since this holds for each eigenvalue \( \lambda_i \), the eigenvalues (latent roots) of \( A^3 \) are \( \lambda_1^3, \lambda_2^3, \dots, \lambda_n^3 \).

### Conclusion:
The eigenvalues of \( A^3 \) are the cubes of the eigenvalues of \( A \), i.e., \( \lambda_1^3, \lambda_2^3, \dots, \lambda_n^3 \).

Would you like further clarification or additional examples?

Here are some related questions to explore:
1. How do eigenvalues behave under matrix addition?
2. Can a matrix with complex eigenvalues have real eigenvalues for its powers?
3. What are the eigenvalues of the identity matrix?
4. How do eigenvalues of a matrix relate to its trace and determinant?
5. Are there any conditions under which the eigenvalues of \( A^3 \) are the same as those of \( A \)?

**Tip**: Eigenvalues are invariant under similarity transformations, which makes them a powerful tool in simplifying matrix problems.

If λ₁, λ₂, ..., λₙ are the latent roots of a matrix A, show that the latent roots of A³ are λ₁³, λ₂³, ..., λₙ³.

Explore the proof that the eigenvalues of the matrix A³ are the cubes of the eigenvalues of A. This problem involves key concepts in linear algebra, such as eigenvalues, eigenvectors, and matrix powers, suitable for undergraduate students.

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