Prove - determinant

  • Thread starter lukaszh
  • Start date
  • #1
32
0

Main Question or Discussion Point

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 :-)
 

Answers and Replies

  • #2
matt grime
Science Advisor
Homework Helper
9,395
3
What you state is trivially false - take A=0, that is nilpotent, and det(0+I) is not zero.
 
  • #3
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
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
32
0
I'm so sorry, because I've made a mistake. Prove that det(A+I)=1. Thank you again for your reactions.
 
  • #5
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
I'm so sorry, because I've made a mistake. Prove that det(A+I)=1. Thank you again for your reactions.
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
32
0
Hello,
so, I know that
[tex]A=S\Lambda S^{-1}[/tex]
Eigenvalues of A are [tex]\{\lambda_1,\lambda_2,\cdots,\lambda_n\}[/tex] for [tex]n\times n[/tex] matrix. If I add to both sides identity, then
[tex]A+I=S\Lambda S^{-1}+I=S\Lambda S^{-1}+SS^{-1}=S(\Lambda+I)S^{-1}[/tex]
Its determinant is
[tex]\mathrm{det}(A+I)=\mathrm{det}S(\Lambda+I)S^{-1}=\mathrm{det}(\Lambda+I)=\prod_{j=1}^{n}(\lambda_j+1)[/tex]
I don't know, how to utilize fact [tex]A^k=0[/tex] :-(
 
  • #7
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,394
179
Hello,
so, I know that
[tex]A=S\Lambda S^{-1}[/tex]
Eigenvalues of A are [tex]\{\lambda_1,\lambda_2,\cdots,\lambda_n\}[/tex] for [tex]n\times n[/tex] matrix. If I add to both sides identity, then
[tex]A+I=S\Lambda S^{-1}+I=S\Lambda S^{-1}+SS^{-1}=S(\Lambda+I)S^{-1}[/tex]
Its determinant is
[tex]\mathrm{det}(A+I)=\mathrm{det}S(\Lambda+I)S^{-1}=\mathrm{det}(\Lambda+I)=\prod_{j=1}^{n}(\lambda_j+1)[/tex]
I don't know, how to utilize fact [tex]A^k=0[/tex] :-(
[tex]\lambda[/tex] is an eigenvalue of A + I if and only if

[tex](A + I)x = \lambda x[/tex]

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

[tex]Ax = (\lambda - 1)x[/tex]

which is true if and only if [tex]\lambda - 1[/tex] is an eigenvalue of A.

I claim that [tex]A^k = 0[/tex] implies that all of the eigenvalues of A are zero. If you can prove this claim then it implies the result you want.
 
  • #8
32
0
Beautiful, thank you.
 

Related Threads for: Prove - determinant

Replies
12
Views
1K
Replies
4
Views
3K
Replies
2
Views
577
Top