Yes, the statement is correct. If a matrix A is nilpotent, then all its eigenvalues are indeed 0. Here's why:

1. Assume A is a nilpotent matrix, which means there exists a positive integer m such that A^m = 0.

2. Let λ be an eigenvalue of A, and let x be the corresponding eigenvector. This means that Ax = λx.

3. If we raise both sides of the equation to the power of m, we get (A^m)x = λ^m x.

4. But we know that A^m = 0, so 0x = λ^m x.

5. This implies that λ^m must be 0, because 0 times any vector is still 0.

6. The only way for λ^m to be 0 is if λ itself is 0, because any non-zero number raised to a positive power is still non-zero.

7. Therefore, all eigenvalues of a nilpotent matrix A are 0.

Question

Yes, the statement is correct. If a matrix A is nilpotent, then all its eigenvalues are indeed 0. Here's why:

1. Assume A is a nilpotent matrix, which means there exists a positive integer m such that A^m = 0.

2. Let λ be an eigenvalue of A, and let x be the corresponding eigenvector. This means that Ax = λx.

3. If we raise both sides of the equation to the power of m, we get (A^m)x = λ^m x.

4. But we know that A^m = 0, so 0x = λ^m x.

5. This implies that λ^m must be 0, because 0 times any vector is still 0.

6. The only way for λ^m to be 0 is if λ itself is 0, because any non-zero number raised to a positive power is still non-zero.

7. Therefore, all eigenvalues of a nilpotent matrix A are 0.

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