Finding Jordan Canonical form but equation for finding the generalized eigenvector becomes inconsistent

Question

I have been given a problem to find the Jordan Canonical form of the following $ 3 \times 3 $ matrix

$$ A= \begin{bmatrix} 3 & 0 & 1 \\ -4 & 1 & -2 \\ -4 & 0 & -1 \end{bmatrix} $$

I have been told to do this using the transformation matrix way, wherein $A$ can be written as

$$ A = TJT^{-1} $$ where $T$ is the modal matrix analogue which contains the eigenvectors and/or the generalized eigenvectors as columns and $J$ is Jordan Canonical form of A. Doing some matrix operations on the above equation, $J$ can be obtained as

$$ J = T^{-1}AT $$

Coming back to the problem, I have found the characteristic equation and got all three eigenvalues as $1$. I found eigenvectors by

$$ Ax_1 = \lambda x_1 $$

Solving this I got two linearly independent eigenvectors, since I was getting only 1 pivot after row transformation operations on $A$ the above equation must give me two solutions. I got these

$$ x_1 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix} $$

For the next step, according to my textbook, I should get a generalized eigenvector by solving for $x_2$

$$ Ax_2 = \lambda x_2 + x_1 $$

The textbook example only shows one possible value for $x_1$, so clearly I've got a problem regarding which one of the two eigenvectors I got, should be chosen for the next step.

But even without considering that, I have yet another problem

$$ Ax_2 = \lambda x_2 + x_1 \\ (A - \lambda I)x_2 = x_1 $$

If I find out the matrix $A - \lambda I = A - I$, I get

$$ \begin{bmatrix} 2 & 0 & 1\\ -4 & 0 & -2\\ -4 & 0 & -2 \end{bmatrix} $$

which forms an inconsistent set of equations to solve, because of the repeated row but different values in the rows of (either) $x_1$. So I can't proceed further to get $x_2$.

Have I done something wrong, or is there some pre-condition to be checked on $A$ that I missed?

Answer 1

You need to choose an eigenvector that is also in the column space of the matrix $A-\lambda I$. In this case, by looking at the matrix you have and at your eigenvector basis, one sees that an eigenvector that will work is $\begin{bmatrix}1\\0\\-2\end{bmatrix}-2\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}1\\-2\\-2\end{bmatrix}$.

Hence, you have two eigenvectors $ v_1=\begin{bmatrix}1\\0\\-2\end{bmatrix}$ and $v_2 = \begin{bmatrix}1\\-2\\-2\end{bmatrix}$. Define $v_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}$ and note that $(A-I)v_3 =v_2$. Now using $T=[v_1,v_2,v_3]$ we will get the Jordan form.

Note that if we just wanted the Jordan form $J$, without caring about what T is, then it is sufficient to know that A has $2$ eigenvectors. For then, as A is a $3\times3$ matrix the only possibility for the matrix $T$ is that it contains $2$ eigenvectors and one generalized eigenvector, which is mapped to one of the eigenvectors under $A-I$. Hence, we will get two Jordan blocks, one of size $1\times 1$ and one of size $2\times 2$. So up to ordering of the blocks $J=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$.

Answer 2

I like the backwards method: choose the generalized eigenvector(s) with integer elements and see what is forced. Since $A-I$ gives two genuine eigenvectors, we hold off on that... The minimal polynomial gives the size of the largest Jordan block (always!). That is, $(A- I)^2 = 0, $ so we look for any nice looking vector for which $(A-I)^2 w = 0$ but $(A-I) w \neq 0.$ I like $w=(0,0,1)^T$ Next we are forced to use $v= (A-I)w = (1,-2,-2)^T.$ A genuine eigenvector that is independent of $v$ could be $u = (0,1,0)^T$

The resulting matrix, your $T,$ is those three as columns in order $u,v,w$

This method allows us to force $T$ to have all integers, with the likelihood of some rational entries in $T^{-1}$ because $\det T$ is most likely not $\pm 1$

Alright, they set this one up with determinant $-1.$

$$ T= \left( \begin{array}{rrr} 0&1&0 \\ 1&-2&0 \\ 0&-2&1 \\ \end{array} \right) $$

$$ T^{-1}= \left( \begin{array}{rrr} 2&1&0 \\ 1&0&0 \\ 2&0&1 \\ \end{array} \right) $$