# finding the jordan canonical with one eigenvalue

Let $$A= \left[ \begin{array}{ c c } -2 & -3 & 1 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{array} \right]$$ Determine the Jordan canonical form of A.

The only eigenvalue I found was $\lambda=-2$

So finding the eigenvector of this is $$\left[ \begin{array}{ c c } 0 & -3 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]. \left[ \begin{array}{ c c } x \\ y \\ z \end{array} \right]=0\implies -3x+z=0\iff z=3x$$ So let $x$ and $y$ be free variables so $x=a$ and $y=b$ with $a,b\in\mathbb{F}$. So $$\left[ \begin{array}{ c c } a \\ b \\ 3b \end{array} \right]=a\left[ \begin{array}{ c c } 1 \\ 0 \\ 0 \end{array} \right]+b\left[ \begin{array}{ c c } 0 \\ 1 \\ 3 \end{array} \right]$$ So we have two eigenvectors that are LI: $$v_1=\left[ \begin{array}{ c c } 1 \\ 0 \\ 0 \end{array} \right], v_2=\left[ \begin{array}{ c c } 0 \\ 1 \\ 3 \end{array} \right]$$

But what would $v_3$ be? Since correct me if I am wrong but our Jordan canonical matrix $J=P^{-1}AP$ where $P$ is the matrix made up of $v_1,v_2,v_3$.

• Your last sentence is only correct if $A$ has a basis of eigenvectors, which this matrix doesn't. Its JCF has a $2\times 2$ block and a $1\times 1$ block. – vadim123 Dec 8 '14 at 22:44
• so what type of block does this one have? – snowman Dec 8 '14 at 22:45
• $J=\left(\begin{smallmatrix}-2&1&0\\0&-2&0\\0&0&-2\end{smallmatrix}\right)$. – vadim123 Dec 8 '14 at 23:36

Hint: You can find a third generalized and linearly independent eigenvector by solving (use Gaussian Elimination):

$$[A-\lambda I]v_3 = [A + 2I]v_3 = v_1$$

You should get:

$$v_3 = \left(0, -\dfrac 13, 0 \right)$$

You can then use the eigenvectors to find (this uses your current eigenvectors) and $v_3$ as:

$$J = P^{-1} A P = \begin{bmatrix}-2 &0&0 \\0 & -2 & 1 \\ 0 & 0 & -2\end{bmatrix}$$

Where $P$ is made up of your column eigenvectors.

• Why is your first equation equal to $v_1$ and not $v_2$? – snowman Dec 9 '14 at 11:36
• that is a question I'd also like to be answered. – etothepitimesi Dec 14 '14 at 3:50
• I understand it would return $v_3=(0,0,0)$. What I'd like to know is if trial and error is the only way to proceed. More generally, I'd like to know if there's a way to find a priori the number of elements of the chain of generalized eigenvectors to which each eigenvector "belongs" :) – etothepitimesi Dec 14 '14 at 4:44
• thanks for the nice reference (and for your time). so, if I understood correctly, the answer to the question "Is there a way to find a priori the number of elements of the chain of generalized eigenvectors to which each eigenvector belongs?" is simply "No", right? – etothepitimesi Dec 14 '14 at 5:23