Find bases of Jordan Form for a $3\times 3$ matrix

Question

Here is my matrix: $A=\begin{bmatrix} 2 &2 &3 \\ 1 &3 &3 \\ -1 &-2 &-2 \end{bmatrix}.$

And I know that the Jordan form is $J=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &1 \\ 0 &0 &1 \end{bmatrix}.$

I have problems when I try to find the invertible matrix $P$ such that $P^{-1}AP=J.$

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Here is my trial:

Consider $AP=PJ$:

$A\begin{bmatrix} p_1 &p_2 &p_3 \end{bmatrix}=\begin{bmatrix} p_1 &p_2 &p_3 \end{bmatrix}\begin{bmatrix} 1 &0 &0 \\ 0 &1 &1 \\ 0 &0 &1 \end{bmatrix}=\begin{bmatrix} p_1 &p_2 &p_2+p_3 \end{bmatrix}.$

Then, we have $(A-I)p_1=0, (A-I)p_2=0$ and $(A-I)p_2=p_3$.

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Then, I try to consider $$A-I=\begin{bmatrix} 1 &2 &3 \\ 0 & 0& 0 \\ 0& 0 &0 \end{bmatrix}.$$

I know that $(-2,1,0)^T$ and $(-3,0,1)^T$ are two linearly independent vectors corresponding to the eigenvalue 1. (I do now know I should do this step or not...)

I have stuck for the whole days to compute different bases for different Jordan canonical form of matrices but I do not get a standard way to finish them. I am frustrated and could anyone give me a hand to do a step by step calculation for this example? Thank you so much.

Answer 1

For the matrix $A$, you have found that we only have a single eigenvalue, $\lambda = 1$, with a multiplicity of $m = 3$.

You set up and found the null space of $[A-\lambda I]v_i = 0$ with the two linearly independent eigenvectors:

$$v_1 = (-2, 1, 0), v_2 = (3,0,-1)$$

Unfortunately, we cannot find a third linearly independent eigenvector by setting up either:

$$[A - I]v_3 = v_1 ~ \mbox{or}~ [A-I]v_3 = v_2.$$

In both cases, what do you call this?

One approach to find three linearly independent eigenvectors is to find a chain of generalized eigenvectors. We calculate:

$$[A-I]^2 = \begin{bmatrix}1 & 2 & 3 \\ 1 & 2 & 3 \\ -1 & -2 & -3\end{bmatrix}^2 = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

This already tells us we should be able to find chains of length $2$, but we also calculate (we already know it is all zeros):

$$[A-I]^3 = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

Since $[A-I]^3 = 0$ is the zero matrix, then every three dimensional vector will be a solution to the equation $[A-I]^3 v = 0$. There are no restrictions, so we are free to choose the three linearly independent vectors:

$$v = (1,0,0), (0,1,0), (0,0,1)$$

These are three solutions to $[A-I]^3 v = 0$, but are not a chain of generalized eigenvectors!

Now, we pick a vector from the above list, say the third one, and keep multiplying the matrix $[A-I]$ with the resultant vector and generate the following chain:

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

This gives us a chain of length $2$.

Note: Repeat this with the two other vectors from above and you will find that each generates a chain of length $2$. Ideally, we would have loved to find a chain of length $3$. Sometimes, we will only get a chain of length $1$.

So, we now have the two linearly independent and generalized eigenvectors:

$$v_1 = (0, 0, 1), v_2 = (3,3,-3)$$

We also have the two other eigenvectors we found earlier, so are free to choose:

$$v_3 = (-2, 1, 0) ~~ \mbox{or}~~ v_3 = (3,0,-1)$$

We can now form $P$ as a linear combination as:

$$P = \begin{bmatrix} 3 & 0 & -2 \\ 3 & 0 & 1 \\ -3 & 1 & 0\end{bmatrix} ~~ \mbox{or}~~ P = \begin{bmatrix} 3 & 3 & 0 \\ 0 & 3 & 0 \\ -1 & -3 & 1\end{bmatrix}$$

Note, repeat this with those other two vectors and see how many choices you have to form $J = P^{-1}AP$. Also note that you could have used a different generalized eigenvector as the third eigenvector, so you have a lot of choices. Also note the difference in $J$ for the different choices of that initial vector!

I would strongly recommend searching the web for more examples and variations so you can practice this approach and I already pointed to one example. Some others include:

• Generalized Eigenvectors
• Wiki Generalized Eigenvectors
• Repeated Eigenvalues
• Generalized Eigenvectors and Repeated Eigenvalues

Update

Note that the order we choose for the vectors in $P$ impacts the location of the Jordan block. Since you want:

$$J=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &1 \\ 0 &0 &1 \end{bmatrix}$$

You can choose (just swap column $1$ and $3$ of the first $P$ above, for example) so we have:

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

Now:

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

You can do the same with the others.

Update 2

Since you want:

$$J=\begin{bmatrix} 1 &1 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$$

You can choose:

$$P = \begin{bmatrix} 3 & 0 & -2 \\ 3 & 0 & 1 \\ -3 & 1 & 0 \end{bmatrix} ~~\mbox{or}~~ \begin{bmatrix} 3 & 0 & 3 \\ 3 & 0 & 0 \\ -3 & 1 & -1 \end{bmatrix} $$

Now:

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

You might like to review the properties of Jordan Normal Form that tell you how many possible JNF's there are per eigenvalue and algebraic multiplicity. Also, it is possible to figure out the possible JNF forms without doing all of this work. See JNF Uniqueness.

Try this with the other two sets of vectors that were originally found!

Answer 2

First, it should be $(A-I)p_3=p_2$. That is, with $p_2$ and $p_3$ the other way around.

Second, it is not true that: $$A-I=\begin{bmatrix} 1 &2 &3 \\ 0 & 0& 0 \\ 0& 0 &0 \end{bmatrix}$$

It should be: $$(A-I)v = 0 \Rightarrow \begin{bmatrix} 1 &2 &3 \\ 0 & 0& 0 \\ 0& 0 &0 \end{bmatrix} v = 0$$

In other words, you have found that the 2 eigenvectors are perpendicular to $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.

So let's try: $p_3 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} $, since it has to be independent of $p_1$ and $p_2$: $$(A-I)\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 14 \\ 14 \\ -14 \end{bmatrix} = 14\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$$ This vector is perpendicular to $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$, so we have found $p_2$...

Answer 3

Very thanks to two of you and I nearly spent two hours more to compute again and I still get stuck :(

I outline what I did here:

(1) We have the system to solve: $$\left\{\begin{matrix} (A-I)v_1=0\\ (A-I)v_2=0\\ (A-I)v_3=v_2 \end{matrix}\right.$$

(2) I start with $(A-I)v = 0$ which gives $$\begin{bmatrix} 1 &2 &3 \\ 0 & 0& 0 \\ 0& 0 &0 \end{bmatrix} v = 0$$

(3) This system gives two eigenvectors $$\begin{bmatrix} -2 &1 &0 \end{bmatrix}^T, \begin{bmatrix} -3 &0 &1 \end{bmatrix}^T. $$

(4) However, no matter I put $\begin{bmatrix} -2 &1 &0 \end{bmatrix}^T$ or $\begin{bmatrix} -3 &0 &1 \end{bmatrix}^T $ as $v_2$ and consider $(A-I)v_3=v_2$ in the system of the step (1) to solve, e.g., $$ \left( \begin{array}{ccc|c} 1 & 2 & 3 & -2 \\ 1 & 2 & 3 & 1 \\ -1 & -2 & -3 & 0 \\ \end{array} \right) $$ It gives rref: $$ \left( \begin{array}{ccc|c} 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$

which is clearly unsolvable.

(4) I have read carefully Serena's comments but I am too stupid and I do not know what my mistakes are:( I have also read the notes Amzoti provides but it seems I do not know how to apply to this case in which the eigenvalues repeat itself 3 times. I appreciate further help very much because I nearly go mad with it:((