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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.

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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?

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  • $\begingroup$ 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? $\endgroup$ – Max Dec 14 '16 at 5:08
  • $\begingroup$ There are actually infinitely many choices for generalized eigenvecotrs. In short, it doesnt matter $\endgroup$ – Daniel Lautzenheiser Dec 14 '16 at 5:09
  • $\begingroup$ 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? $\endgroup$ – Max Dec 14 '16 at 5:10
  • $\begingroup$ If you want to brute force find a generalized eigenvector, simply look at $Null(A-3I)^2$ $\endgroup$ – Daniel Lautzenheiser Dec 14 '16 at 5:11
  • $\begingroup$ Yes. Try it for yourself. You could also pick (0,0,-1) as a starting 3-cycle $\endgroup$ – Daniel Lautzenheiser Dec 14 '16 at 5:13

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