I am given the following matrix:
$$A = P\left(\begin{matrix} -5 & -6 & 3\\3 & 4 & -3\\0 & 0 & -2\end{matrix}\right)$$
After finding the following eigenvalues by finding the characteristic polynomial I get:
$\lambda_1 = \lambda_2 = -2$
and $\lambda_3 = 1$
After finding the I now need to find the eigenvectors for $\lambda_1$ and $\lambda_2$. After putting matrix into reduced-row echelon form for $\lambda_1$:
$$\left(\begin{matrix} 1 & 2 & -1\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right)$$
I now want to find the eigenvector from this, but am I bit puzzled how to find it an then find the basis for the eigenspace (I know this involves putting it into vector form, but for some reason I found the steps to translating-to-vector-form really confusing and still do).
A step-by-step explanation on this point would be very helpful for a linear algebra newbie.
Best Answer
Hint: If $A v_i = \lambda_i v_i$ then $(A - \lambda_i I) v_i = 0.$ You can solve fox $v_i$ by Gaussian elimination on the augmented matrix $$\left( A-\lambda_i I \quad {\bf 0} \right).$$ For $\lambda = -2,$ we have $$ A - (-2) I = \begin{pmatrix} -3 & -6 & 3\\ 3 & 6 & -3 \\ 0 & 0 & -1 \end{pmatrix} $$ And the reduced echelon form of $(A-\lambda_i I )v = 0$ is $$ \begin{pmatrix} 1 & 2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\y \\z \end{pmatrix} = 0. \tag{1} $$ This gives you a solution $z = 0, y = t, x = -2.$ i.e., the set of vectors satisfying $Av = -2v$ are of the form $t-$multiple of $$ \begin{pmatrix} -2 \\ 1 \\0 \end{pmatrix} t. $$ The's the basis.