# Prove - determinant

1. May 22, 2009

### lukaszh

Hello,
I don't want how to prove: Matrix A is nilpotent, so A^k=0. Prove that det(A+I)=0.
Thank you so much :-)

2. May 22, 2009

### matt grime

What you state is trivially false - take A=0, that is nilpotent, and det(0+I) is not zero.

3. May 22, 2009

### jbunniii

In fact, det(A^k) = det(A)^k, so det(A) must be zero. Alternatively, A^k = 0, so A cannot be injective, so det(A) = 0.

By the way, this shows that 0 is an eigenvalue of A. Something stronger is true: 0 is the ONLY eigenvalue of A.

I think lukaszh's problem statement should be: prove that det(A+I) = 1.

4. May 26, 2009

### lukaszh

I'm so sorry, because I've made a mistake. Prove that det(A+I)=1. Thank you again for your reactions.

5. May 26, 2009

### jbunniii

First, find all the eigenvalues of A+I. Then, use what you know about how to calculate the determinant of a linear map given its eigenvalues.

6. May 26, 2009

### lukaszh

Hello,
so, I know that
$$A=S\Lambda S^{-1}$$
Eigenvalues of A are $$\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$$ for $$n\times n$$ matrix. If I add to both sides identity, then
$$A+I=S\Lambda S^{-1}+I=S\Lambda S^{-1}+SS^{-1}=S(\Lambda+I)S^{-1}$$
Its determinant is
$$\mathrm{det}(A+I)=\mathrm{det}S(\Lambda+I)S^{-1}=\mathrm{det}(\Lambda+I)=\prod_{j=1}^{n}(\lambda_j+1)$$
I don't know, how to utilize fact $$A^k=0$$ :-(

7. May 26, 2009

### jbunniii

$$\lambda$$ is an eigenvalue of A + I if and only if

$$(A + I)x = \lambda x$$

for some nonzero vector x. This is true if and only if

$$Ax = (\lambda - 1)x$$

which is true if and only if $$\lambda - 1$$ is an eigenvalue of A.

I claim that $$A^k = 0$$ implies that all of the eigenvalues of A are zero. If you can prove this claim then it implies the result you want.

8. May 26, 2009

### lukaszh

Beautiful, thank you.