I have $$ A = \left( \begin{array}{ccc} -4 & 9 & -4 \\ 0 & 0 & 0 \\ 6 & -13 & 6 \end{array} \right) $$ whose eigenvalues are $\{0,0,2\}$.

For $\lambda=2$, I have $(A-\lambda\,I)\,\vec{v^{(3)}} = \bf{0}$ leading to $v^{(3)}=\left( \begin{array}{c} 1 \\ 0 \\ -3/2 \end{array} \right)$.

For $\lambda_{1,2} = \{0,0\}$, I get $$ A-0*I = 0 \longrightarrow \left( \begin{array}{ccc} -4 & 9 & -4 \\ 0 & 0 & 0 \\ 6 & -13 & 6 \end{array} \right)\,\vec{v^{(1)}} = \vec{0} $$ (I know vector symbols is inappropriate in case the vector field doesn't correspond to a vector space, but will use it here.)

I can see that the column space (rank) may not span a 3D space, and its determinant is zero. As the 1st and 3rd columns are the same, and the 2nd is not equal to the other two, I come up with $\operatorname{Rank}(A-\lambda_{\left|_0 \right.}\,I) = 2$.

By the rank-nullity theorem, I see that $3= \operatorname{nullity} + 2$, implying that the geometric multiplicity is 1 (the nullity).

This 2nd eigenvector is just $\vec{v^{(1)}} = \left( \begin{array}{c} 1 \\ 0 \\ -1 \end{array} \right)$

My question then becomes, how do I find a full set of eigenvectors? Do I find the 3rd one by $$ (A-0*I)^1\;\vec{v^{(2)}} = \vec{v^{(1)}} \;? $$ I tried but get the same set of 2 equations as for the 2nd eigenvector $\vec{v^{(1)}}$.

Stepping away from the calculations, I know I have a 1D nullspace (a 1D set of vectors that get sent to zero), then I have one eigenvalue with algebraic multiplicity 1 (and therefore at most 1 geometric multiplicity) associated with the eigenvalue 2 and eigenvector $\vec{v^{(3)}}$. This leaves just the eigenvector $\vec{v^{(1)}}$. Is this the `full' set?

Didn't mean to forget this part...

Is the generalized eigenvector then to represent the scaling of all vectors in the domain that get sent to zero? What is the physical interpretation of this generalized eigenvector, as couched in the context of a symplectic manifold or physical phase space? Is there an answer to this that is always true?


Since $\dim \ker A = 1$, there is only one eigenvector corresponding to the zero eigenvalue. We see that $\ker A = \operatorname{sp} \{ (1, 0 -1)^T \}$.

Since the eigenvalue $0$ has multiplicity $2$, this means there is a generalized eigenvector in $\ker A^2$. This means that we want to solve $A^2 v = 0$, but with $Av \neq 0$. That is $Av \in \ker A$, but $v \notin \ker A$. Since we know $\ker A$, we solve $A u = (1,0,-1)^T$. This gives $u = (1,1,1)^T + x (1,0,-1)^T$ (for any $x$).

  • $\begingroup$ Thanks to all! @copper.hat, where did the last expression come from? Specifically the [1,1,1]? I get a more geometrical feel for my last questions from your answer btw. $\endgroup$ – nate Sep 15 '13 at 22:52
  • $\begingroup$ Just grunt work. In this case I solved $\begin{bmatrix} A \\ (1,0,-1) \end{bmatrix} u = (1,0,-1)^T$ to get the $(1,1,1)^T$ and then added on $\ker A$, since I know that $\ker A^2$ is two dimensional, this gives the whole space. $\endgroup$ – copper.hat Sep 15 '13 at 23:00
  • $\begingroup$ Okay - got it, Thanks! $\endgroup$ – nate Sep 15 '13 at 23:01

Yes, we have:

$$\lambda_2 = 0, [A-0I]v_2 = v_1$$

Doing the RREF, the generalized eigenvector is:

$$v_2 = (-2,-1,0)$$

I also got:

  • $v_1 = (-1,0,1), \lambda_1 = 0 $
  • $v_3 = (-2,0,3), \lambda_3 = 2$ (Note: this is basically the same answer as yours.)
  • $\begingroup$ Nice work, Amzoti! $\endgroup$ – Namaste Sep 16 '13 at 0:07
  • $\begingroup$ @amWhy: Thanks, hope you are having a more stress free weekend with all your appts done! Doing chores at the moment. $\endgroup$ – Amzoti Sep 16 '13 at 0:11

Your work is fine; taking $v_3=(1,0,-3/2)$, $v_1=(1,0,-1)$, and then solving $(A-0I)v_2=v_1$ using row reduction gives $v_2=(2-t,1,t)$ for any number $t$.


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