The image presents a problem with the following instructions:

• Given: $A = S \Lambda S^{-1}$
• Task: Take determinants to prove that $\text{det}(A) = \text{det}(\Lambda) = \lambda_1 \lambda_2 \cdots \lambda_n$.
• Additional Note: This quick proof only works when $A$ can be ______.

Solution:

1. Determinant of A: Since $A = S \Lambda S^{-1}$, take the determinant on both sides: $\text{det}(A) = \text{det}(S \Lambda S^{-1})$ By the property of determinants, the determinant of a product of matrices equals the product of their determinants: $\text{det}(A) = \text{det}(S) \cdot \text{det}(\Lambda) \cdot \text{det}(S^{-1})$

2. Determinant of the Inverse: The determinant of the inverse of a matrix $S$ is the reciprocal of the determinant of $S$: $\text{det}(S^{-1}) = \frac{1}{\text{det}(S)}$ Substituting this into the equation: $\text{det}(A) = \text{det}(S) \cdot \text{det}(\Lambda) \cdot \frac{1}{\text{det}(S)}$ The $\text{det}(S)$ terms cancel out: $\text{det}(A) = \text{det}(\Lambda)$

3. Determinant of Lambda: $\Lambda$ is a diagonal matrix with eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$ on the diagonal. The determinant of a diagonal matrix is the product of its diagonal elements: $\text{det}(\Lambda) = \lambda_1 \lambda_2 \cdots \lambda_n$

4. Conclusion: Therefore: $\text{det}(A) = \lambda_1 \lambda_2 \cdots \lambda_n$

5. Condition for Quick Proof: The quick proof only works when $A$ is diagonalizable. This is because $A = S \Lambda S^{-1}$ assumes that $A$ can be decomposed into a diagonalizable form.

Final Answer:

The quick proof only works when $A$ can be diagonalized.

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Would you like more details on this solution or have any further questions?

Related Questions:

1. What conditions must a matrix satisfy to be diagonalizable?
2. How is the determinant of a product of matrices related to the determinants of the individual matrices?
3. What is the significance of eigenvalues in determining the determinant of a matrix?
4. Can non-diagonalizable matrices be decomposed similarly? If so, how?
5. What are the implications of a matrix being non-diagonalizable?

Tip:

Always verify if a matrix is diagonalizable by checking if it has a full set of linearly independent eigenvectors.