# Thread: Defect of an eigenvalue

1. ## Defect of an eigenvalue

Hi,

$A = \left( \begin{array}{ccc}
13 & 5 & 2 \\
2 & 7 & -8 \\
5 & 4 & 7 \end{array} \right)$

The characteristic polynomial is $-(9-\lambda)^3=0$, giving $\lambda_{1,2,3}=9$

So the algebraic multiplicity $M_{9}=3$. I seem to be getting an incorrect answer for the geometric multiplicity $m_{9}$.

By Gaussian Elimination: $4a_{11} + 5a_{12} + 2a_{13}=0 => a_{11} = -0.5[5a_{12} + 2a_{13}]$

It seems to me that we can obtain three linearly independent (LI) eigenvectors from this by setting $a_{11}, a_{12},$ and $a_{13}$ to zero in turn. That would give $m_{9} = 0$. My textbook however gives $m_{9}=2$ (suggesting that we can only obtain one LI vector)

Could you please clarify where I am going wrong in the LI part?

Thanks.

2. ## Re: Defect of an eigenvalue

$\dim V_9=3-\mbox{rank}(A-9I)=3-\mbox{rank}\begin{bmatrix}{4}&{\;\;5}&{\;\;2}\\{2} &{-2}&{-8}\\{5}&{\;\;4}&{-2}\end{bmatrix}=\\3-\mbox{rank}\begin{bmatrix}{4}&{5}&{2}\\{0}&{9}&{18 }\\{0}&{9}&{18}\end{bmatrix}=3-2=1$

A basis of $V_9$ is $\{(2,-2,1)\}.$

Thanks.