generalized eigenvector example

Question

$$A=\left(\begin{matrix} 2 & 1 & 0\\ 0 &2 &0\\ 0 &-1 &2 \end{matrix}\right) $$ Want to find the eigenvalues, eigenvectors, and generalized eigenvectors of $A$ and put $A$ in Jordan normal form.

Here is my attempt.

So, the characteristic polynomial is $(2-\lambda)^3=0$. So the eigenvalue is $2$ with multiplicity $3$. Solving $(A-2I)v=0$ to obtain span$\left\{v_1=\left(\begin{matrix} 1\\ 0 \\ 0 \end{matrix}\right), w_1=\left(\begin{matrix} 0\\ 0 \\ 1 \end{matrix}\right) \right\}$ is the eigenspace of eigenvalue $\lambda=2$. So there should be one more generalized eigenvector. Using Jordan chain, I need to solve $(A-2I)v_2=v_1$ and $(A-2I)w_2=w_1$. Note that
$$(A-2I)\left(\begin{matrix} a\\ b \\ c \end{matrix}\right)=\left(\begin{matrix} 0 & 1 & 0\\ 0 &0 &0\\ 0 &-1 &0 \end{matrix}\right) \left(\begin{matrix} a\\ b \\ c \end{matrix}\right) =\left(\begin{matrix} b\\ 0 \\ -b \end{matrix}\right), $$ which can't be equal to $v_1$ or $w_1$. Not sure what went wrong....

Answer 1

You know that if you have a vector $$u = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$ then you have $$(A- 2I)u = \begin{pmatrix} b \\ 0 \\ -b \end{pmatrix} = v$$

You also know that $v$ is an eigenvector of $A$, with eigenvalue 2, since it is in your calculated eigenspace (it can be written as $v = bv_1 - bw_1$).

Hence $u$ will be a generalized eigenvector as long as $b \neq 0$.

If you want to find a transformation matrix that takes your matrix to its Jordan normal form, you can first pick a generalized eigenvector, say $$u = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$$ and then pick $$u_1 = (A-2I)u = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$$ as your first eigenvector, and then another vector in the eigenspace, linearly independent of $u_1$, e.g. $$u_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$ and then you can form the matrix $$T = \begin{pmatrix} | & | & | \\ u & u_1 & u_2 \\ | & | & | \end{pmatrix}$$ so that $T^{-1}AT = J$ where $J$ is the Jordan normal form of $A$.

Answer 2

You find the "true" or first order eigenspace by solving $\det({\bf A}-\lambda{\bf I}) = 0$, then second order eigenvector $\det(({\bf A}-\lambda{\bf I})^2)=0$ which is true for the full space but not in the first space. This leaves the only one left $(0,1,0)^T$. ${\bf A}(0,1,0)^T = (1,2,-1)^T$ which gives us a hint that it is actually $(1,0,-1)^T$ it belongs to (by projection). So we choose that vector from the ordinary eigenspace to make an eigenvector ( Because that gives the simplest image of our new found generalized eigenvector ) Now the second ordinary eigenvector (which is left) is chosen to fill out the space but contribute as little as possible to the previous ones. This would be $(1,0,1)^T$ as it is orthogonal to the projection found.

So our Jordan form $\bf A = SJS^{-1}$ becomes:

$${\bf S} = \left[\begin{array}{rrr} 1&0&1\\ 0&1&0\\ -1&0&1 \end{array}\right], {\bf J} = \left[\begin{array}{rrr}2&1&0\\0&2&0\\0&0&2\end{array}\right]$$