So far I've found the characteristic equation.
[tex] (t-1)^2(t+2) [/tex]
So the eigenvalues, are
And I know the jordan matrix is along these lines(below)
But I literally have NO clue on where the extra 1's should be.
I know a singular eigenvector towards T for sure, but I'm confused on -2, as the answer is totally different in my mark scheme :S!
so.....
For the last one i'm trying...
[tex](A-I)^2\begin{bmatrix}a&b&c\end{bmatrix}=\begin{bmatr ix}1&1&1\end{bmatrix}[/tex]
I also know
I know this is a bit scrappy but could anybody help me on the bits i'm struggling with..
1) Finding a second eigenvector for t=1, making sure my t=-2 eigenvector is correct.
2) Explaining how you decide how many, and where to put the 1's in a jordan matrix.
Thank you
If null(A-I) = 2, does that have 2 1's above the diagonal??
And yes let's move onto the T's.
Looking at the solutions, I get the correct answer to t = 1, with (1,1,1). However when I put in -2, I get (2,1,1), whereas in the solutions it says (1,2,2).
For the third t, where t=1 again, I have written the method I have been trying
Is this correct I do it but still get a different answer to the solutions again..
Since you have determined that 1 is a double root of the characteristic equation (has "algbraic multiplicity" 2) you know that the "Jordan Normal Form" is either
(if eigenvalue 1 has "geometric multiplicity 2) (two independent eigenvectors corresponding to eigenalue 1)
or
(if eigenvalue 1 has "geometric multiplicity" 1) (only one independent eigenvectors corresponding to eigenvalue 1.
The "exra ones" only go above the multiple eigenvalue so there is only one place it could be.
The point is this: every matrix satisfies its own characteristic equation so for this matix, A, it must b true that where v is any vector. Obviously if (A- 2I)v= 0 (which is the same as saying that Av= 2v) this is true. Also if (A- I)v= 0 so that Av= v, it is true. That, is if v is an eigenvector of A with eigenvalue 2 or 1, that is true. If A has two independent eigenvectors with eigenvalue 1 (1 has geometric multiplicity 2) then, since an eigenvector corresponding to eigenvalue 2 would have to be independent of them, there are 3independent eigenvectors and the matrix T having those eigevectors as columns makes diagonal.
If there are NOT two independent eigenvectors for eigenvalue 1 (it has geometric multiplicity 1) it still must be true that for every vector. What that means is there must be an vector, independent of the eigenvector for eigenvalue 1, such that ( but such that . We can write that as so that if v is not an eigenvector, must be an eigenvector!
Since you have found that one eigenvector for eigenvalue 1 is , you want to find such that
Such a vector will be the second column you need in T.
(I once made a total hash of a problem like this and Plato corrected me so I am redeeming myself.)
Thank you for this response HallsofIvy, I'd nearly resigned myself to being doomed come tommorow morning, but this is a great explanation. (Except the latex errors =P)
I now understand this problem much better than I did before.
Just as a question, would I proceed to make the Matrix rREF, to make the equations easier to solve? and Also does nullity(A-I) = geometric multiplicity?
I totally know what you mean by making a hash of these problems! The numbers are so easy to mix up with your minus signs etc. I'm just going to have to take my sweet time.
I have corrected the LaTex (had to go fix dinner just as I was typing that!). You can solve the equations , , and however you like- row reducing the augmented matrix or solving the three equations simultaneously.
Yes, the "geometric multiplicity" of an eigenvalue, [tex]\lambda[/itex, is the number of independent vectors, v, that satisfy which is, of course, the same as and so is the "nullity" of .