I'm having trouble understanding defective eigenvalues and associated linear systems. I apologize in advance if my formatting is poorly executed. Given the following problem: $$ x' = \begin{bmatrix} 1 & 0 & 0\\ -2 & -2 & -3\\ 2 & 3 & 4 \end{bmatrix} x $$ I find that it has an eigenvalue of $$\lambda = 1$$ with k(multiplicity) = 3. Solving for the eigenvectors I get two independent solutions $$ v_1 = \begin{bmatrix} -3\\0\\2 \end{bmatrix}, v_2 = \begin{bmatrix} -3\\2\\0 \end{bmatrix} $$ but I'm having trouble after that. Since it has a defect of $1$, I solved the following equations. $$ (A - \lambda I)^2v_3 = 0\\ (A - \lambda I)v_3 = v_2\\ (A - \lambda I)v_2 = v_1\\ $$ Since the matrix squared was a zero matrix, I set $v_3$ to $[1\, 0\, 0]^T$ and solved for the other two vectors. Since the first row of $$(A-\lambda I)$$ is all zeros, it is impossible to end with the two independent solutions that I solved for in the first step (as the book suggests I should) using this method.
Can someone help me find the error in my approach?
Thank you.
you can keep $$ (A - \lambda I)^2v_3 = 0\\ (A - \lambda I)v_3 = v_2\\ $$ but then $v_2$ is already a genuine eigenvector, as $(A -I)v_2 = (A-I)^2 v_3 = 0 v_3 = 0.$ Then $v_1$ needs to be a genuine eigenvector but linearly independent of $v_2.$ For your choice $v_3 = (1,0,0)^t,$ we are forced to $v_2 = (0,-2,2)^T.$ We may then choose $v_1.$ Since our $v_2$ is the difference of the eigenvectors you first mentioned, we might as well take $v_1$ to be a multiple of their sum, or $v_1 = (-6,2,2)^T/2 = (-3,1,1)^T$
The new basis matrix is $$ P = \left( \begin{array}{ccc} -3&0&1 \\ 1 & -2&0 \\ 1&2&0 \end{array} \right) $$ with $$ P^{-1} = \frac{1}{4} \left( \begin{array}{ccc} 0&2&2 \\ 0 & -1&1 \\ 4&6&6 \end{array} \right) $$ You get $P^{-1}AP = J,$ where your matrix is $A$ and $J$ is the Jordan form. The direction you actually use is $P J P^{-1} = A,$ where $$ J = \left( \begin{array}{ccc} 1&0&0 \\ 0 & 1&1 \\ 0&0&1 \end{array} \right) $$ so you should check, by hand, that my claimed $P J P^{-1} = A$ really works. Note that you may pull out the factor of $\frac{1}{4}$ until you multiply with it at the very end, this way you are not carrying around denominators all the way and possibly causing errors.