3
$\begingroup$

$$A=\left(\begin{matrix} 2 & 1 & 0\\ 0 &2 &0\\ 0 &-1 &2 \end{matrix}\right) $$ 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....

$\endgroup$
2
$\begingroup$

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 $A$.

$\endgroup$
0
$\begingroup$

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:

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

$\endgroup$

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.