Conversation started Feb 24, 2023 at 21:01.
D.C. the III
D.C. the III
Feb 24, 2023 21:01
I just did a first pass on reading on Jordan Canonical Forms......seems OK, but I need clarification on a concept. Anybody available?
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We have this matrix and after doing all the necessary work we get to a basis of the solution space for $K_{\lambda_2}$. In the book they chose the vector $(-1,2,0)$ to begin their cycle. I tried using the vector $(1,-3,-1)$ initially. I was expecting to be able to get the vector $(-1,2,0)$ after I multiplied by the correct matrix, but I didn't.
Isn't it the case that I should be able to get every vector in that solution space by cycling through it?
When I did it the other way and started with $(-1,2,0)$ I got the second vector, but it didn't work the other way around
Astyx
Astyx
Feb 24, 2023 21:54
One thing to notice is that solutions to $(A-2I)v$ are solutions to $(A-2I)^2v$
i.e. you have $E_{\lambda_2}\subset K_{\lambda_2}$
If you have a basis of the former you can complete it into a basis of the latter
Since $\dim E_{\lambda_2} =1$, it is sufficient to find one nonzero vector in this space to get a basis
To do this, one thing you may do is pick a vector of $K_{\lambda_2}$ and apply $A-2I$ to it – if you're lucky you will find a nonzero vector, which has to be in $E_{\lambda_2}$
D.C. the III
D.C. the III
So even though they didn't show the computation, then $(-1,2,0)$ was an eigenvector?
and that is what allowed for the creation of the "cycle" per se?
Astyx
Astyx
If you do this with $(1,-3,-1)$ it works. If you do it with $(0,-1,-1)$ it works. If you do it with $(-1,2,0)$ it does not work.
Here you can take any vector in $K_{\lambda_2}$ that is not colinear to $(-1, 2, 0)$ and it will work (can you prove that?)
Right, they did not actually do the computation, they just claimed it works (and it does)
D.C. the III
D.C. the III
Yea. As I finished the section they said the next section will show the ways to compute these things because I one o fthe things I was wracking my head about was how they were getting some of these results
So when they solved for the basis of the solution space and got the vectors $(1,-3,-1), (-1,2,0)$, they "found" the acceptable vector to be $(-1,2,0)$. I take it won't always be the case that a basis of such a homogenous system will provide me with a cycle of generalized eigenvectors?
Astyx
Astyx
Feb 24, 2023 22:18
It depends what you mean
if your generalized eigenspaces staired well, you will always be able to find such a thing
more precisely, here $\dim E_{\lambda_2} = \dim K_{\lambda_2} - \dim E_{\lambda_2} = 1$
if there were three independent vectors in K and only one in E, then you should be able to convince yourself that there is never a cycle which yields a basis
however if it is straired (I don't know the actual word) nicely, then you alway will be, by taking one in the upmost space that is not in the one below it
D.C. the III
D.C. the III
Yea, they showed that idea so it makes sense to me.
Hmmm. I'll pause the rest of my inquiries until a little later when I've read the next section but I do have a bit of better understanding. Thanks for the help
Astyx
Astyx
Glad to help
 
Conversation ended Feb 24, 2023 at 22:22.