# Finding $P$ in $A = P^{-1}JP$ (Jordan Form)

I'm having a lot of trouble understanding the process of finding a basis for the Jordan canonical form (the "algorithm"). My textbook (Friedberg 4E) isn't very clear, and I can't seem to find anything online.

If we consider the example $A$ = $\begin{bmatrix}6 & -2 & -1\\3 & 1 & -1\\2 & -1 & 2\end{bmatrix}$, we get the characteristic polynomial $p(t)$ = ($3 - t$)$^3$.

I understand the process of finding the matrix $J$ in $A = P^{-1}JP$, but I just can't seem to grasp finding $P$. Any clarification/explanation on the steps to do so would be incredibly helpful. Thank you.

The matrix $$A$$ has only one true eigenvector $$\vec{z} = (1,1,1)$$ corresponding to the eigenvalue $$\lambda = 3$$. In other words, $$(A-3I)\vec{z} = \vec{0}$$. Though we can't find any other linearly independent vectors that have this same property, we can find linearly independent vectors such that $$(A-3I)^2\vec{y} = \vec{0}$$ i.e. $$(A-3I)\vec{y} = \vec{z}$$. Solving this system is just solving a 3x3 linear system. Similarly, it is possible to find a third vector $$\vec{x}$$ such that $$(A-3I)\vec{x} = \vec{y}$$. Let's say we have the eigenvector $$\vec{z}$$ above and we have two generalized eigenvectors $$\vec{x}$$ and $$\vec{y}$$ such that

$$(A-3I): \vec{x} \mapsto \vec{y} \mapsto \vec{z} \mapsto \vec{0}$$

A "3-cycle" if you will (regular eigenvectors are all "1 cycles").

Construct the following matrix as column vectors: $$\left(A\vec{z} |A\vec{y} |A\vec{x} \right) = \left(3\vec{z}|3\vec{y}+\vec{z}|3\vec{x} + \vec{y} \right)$$ $$(A)(\vec{z}|\vec{y}|\vec{x}) = (\vec{z}|\vec{y}|\vec{x}) \begin{pmatrix}3&1&0\\0&3&1\\0&0&3 \end{pmatrix}$$ Letting $$P = (\vec{z}|\vec{y}|\vec{x})$$ gives us $$AP = PJ$$.

That's the theory. Now in practice, you can usually just guess at a vector (not a multiple of $$\vec{z}$$) that will be a desired "3-cycle". Check it out: $$(A-3I)(1,0,0) = (3,3,2)$$ $$(A-3I)(3,3,2) = (1,1,1) = \vec{z}$$ $$(A-3I)(1,1,1) = (0,0,0)$$ So, your 3-cycle is $$(1,0,0) \mapsto (3,3,2) \mapsto(1,1,1) \mapsto \vec{0}$$. So, your change of basis matrix $$P$$ is then $$\begin{pmatrix} 1&3&1\\1&3&0\\1&2&0\\ \end{pmatrix}$$

Does this help?

• Yeah, it does. But, I still don't quite get what to do in 2 scenarios: 1) There are 2 or more eigenvectors to "choose from" to construct a generalized eigenspace. Which one do I know is the right one to choose? 2) If the dimension of null space is 0, i.e. there are no eigenvectors to "choose" from, what do we do then? – Max Dec 14 '16 at 5:08
• There are actually infinitely many choices for generalized eigenvecotrs. In short, it doesnt matter – Daniel Lautzenheiser Dec 14 '16 at 5:09
• Oh, that's very neat then. So, if the case were that N($A - \lambda I$) = $0$, then I can just choose any arbitrary vector (say, (1 0 0)) to construct a gen. eigenspace? – Max Dec 14 '16 at 5:10
• If you want to brute force find a generalized eigenvector, simply look at $Null(A-3I)^2$ – Daniel Lautzenheiser Dec 14 '16 at 5:11
• Yes. Try it for yourself. You could also pick (0,0,-1) as a starting 3-cycle – Daniel Lautzenheiser Dec 14 '16 at 5:13