I was solving some problems on diagonalization of matrixes, and I came across a particular question which seemed a bit odd.
$$\left[\begin{array}{rrr}
-8 & -6 & 2\\
-6 & 7 & -4\\
2 & -4 & 3
\end{array}\right]$$
I found the eigen values to be $\lambda_1 = -10.13$, $\lambda_2 = 11.44$, and $\lambda_3 = 0.69$
All the values are rounded off to 2 decimal places.
The problem comes when I try to find the eigenvectors, I tried to substitute the value of lambda and find the null space, But I get a zero vector. I know that for a eigen value there should be a eigen vector, Then why am I not getting it.
Have I done something wrong? Or is my approach Wrong?
Best Answer
One rough and ready way to find an (approximate) eigenvector, given an (approximate) simple eigenvalue $\lambda$ of $n \times n$ matrix $A$, is the following. Let $B$ be the top left $n-1 \times n-1$ block of $A-\lambda$, and $C$ the top $n-1$ entries of the last column of $A-\lambda$. $\lambda$ being a simple eigenvalue, $A-\lambda I$ should have rank $n-1$. Usually, $B$ will have rank $n-1$: if not, you'll have to adapt this method to use another row and/or column instead of the last one. Then the approximate eigenvector is $\pmatrix{u\cr 1\cr}$, where $B u = -C$.
In the given example, with $\lambda = −10.130887$, we have $B = \pmatrix{2.130887 & -6\cr -6 & 17.130887}$, $C = \pmatrix{2\cr -4\cr}$, so $u = \pmatrix{-20.361293 \cr -6.897936}$. And indeed $\pmatrix{-20.361293 \cr -6.897936 \cr 1\cr}$ is a good approximation to the eigenvector.