• Support PF! Buy your school textbooks, materials and every day products Here!

Eigenvalues/Vectors with bizarre 3x3 matrix

  • Thread starter david118
  • Start date
  • #1
4
0

Homework Statement



Let B :=
2 1 5
0 2 3
0 0 2


. [Hint: Write B as a diag-matrix
plus a nilpotent matrix.]

Then B^2005 = ?



Homework Equations





The Attempt at a Solution



so i found the eigenvalue to be 2, with a multiplicity of 3. When plugging the eigenvalue back into B, the original matrix, im left with all zeros except for a 1 in top row 2nd column, a 5 next to it on right, and a 3 below the 5. Thus i figured 3x3 must = 0, thus x3=0. and this x2 =0, and no eigenvectors. So IDK what to do..

i took the hint, and made it the diag matrix with 2s for each entry of hte 3x3 matrix. the nilpotent thus is the remaining terms, which goes to zero for the matrix cubed.



nevermind i got it. thanks anyway yall
thanks
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,847
1,644
No probs - well done:
BTW: wot he said (below)
 
Last edited:
  • #3
lurflurf
Homework Helper
2,426
126
B=d+n
b^2005=(d+n)^2005
=b^2005+2005 d^2004 n+2009010d^2003 n^2+1341349010d^2002 n^3+...
 
  • #4
382
4
Yeah remember B and d commute.
 

Related Threads on Eigenvalues/Vectors with bizarre 3x3 matrix

  • Last Post
Replies
1
Views
1K
Replies
8
Views
1K
  • Last Post
Replies
3
Views
9K
Replies
1
Views
2K
  • Last Post
Replies
5
Views
11K
  • Last Post
Replies
12
Views
1K
Replies
11
Views
2K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
Top