# Eigenvalues/Vectors with bizarre 3x3 matrix

## 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 = ?

## 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:

Simon Bridge
Homework Helper
No probs - well done:
BTW: wot he said (below)

Last edited:
lurflurf
Homework Helper
B=d+n
b^2005=(d+n)^2005
=b^2005+2005 d^2004 n+2009010d^2003 n^2+1341349010d^2002 n^3+...

Yeah remember B and d commute.