0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Great! Thank you for your help. Just to be sure I'm understanding correctly, $v_1$ can be any combination of the ordinary eigenvectors that I solved for in the first step given that it is linearly independent with the two other generalized eigenvectors? $\endgroup$ – Jake May 2 '18 at 0:37
  • $\begingroup$ @Jay yes. If you mistakenly made $v_1$ a multiple $v2,$ it would turn out that the matrix I have been calling $P$ would not be invertible $\endgroup$ – Will Jagy May 2 '18 at 0:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.