Do you get a Eigen value without the respective Eigenvector? 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?
 A: 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.
A: In this approach we will use numerical approximation by Mathematica:
Let $A=\begin{bmatrix}-8&-6&2\\-6&7&-4\\2&-4&2\end{bmatrix}\in M_{3}({\bf R})$ then the matrix is ${\bf R}$-symmetric matrix, then all the eigenvalues must be reals and also the algebraic multiplicity of each eigenvalues is the same of geometric multiplicity and therefore $A$ is ${\bf R}-$diagonalizable.
The characteristic polynomial is given by
$$f(t)=-t^{3}+2t^{2}+115t-80$$
the polynomial can be written as
$$f\left(t-\frac{2}{3}\right)=-\left(t-\frac{2}{3}\right)^{3}+\frac{349}{3}\left(t-\frac{2}{3}\right)-\frac{74}{27}$$
Solving that we can control a better numerical approximation of the roots (eigenvalues) to $f$, we get
$$t_{1}\approx-10.130886904486$$
$$t_{2}\approx 0.690226155068582$$
$$t_{3}\approx 11.4406607494200$$
This allows to make a better approximation avoiding the jump to zero as you pointed. Then each eigenspace is obtained by,
$$E_{t_{k}}={\bf ker}(A-tI_{3})$$
If you want you can do this by hand, I do not recommend it. Using the computational power of Mathematica, we can find a pretty good approximation for the eigenvectors
$$t_{1}\longrightarrow  v_{1}\approx \begin{bmatrix}-20.36\\-6.90\\1\end{bmatrix}$$
$$t_{2}\longrightarrow v_{2}\approx \begin{bmatrix}-0.13\\0.54\\1\end{bmatrix}$$
$$t_{3}\longrightarrow v_{3}\approx \begin{bmatrix}0.65\\-1.78\\1\end{bmatrix}$$
The idea is that the more decimal places you add in your approximation for the roots, the better approximation you get to perform the row reduction. In fact it is possible to calculate or control the error in this approximation, but this is all part of the computational theory related to numerical methods. Due to the nature of the roots it's hard to get better than this. However, using polynomial theory we can find a closed form of the roots. Thanks to all fellow users for your corrections in the comments, they are greatly appreciated.
