Maths and Theoretical Physics Course Thread TR031 TR035

Tears in Rain wrote: »

I'm not convinced anyone knows how to do the Jordan Normal questions :S

I do! :P

And posting a method for JNF is actually a very good idea, which I might do later on today.

Since JNF is one of the things most people find difficult and the supplemential exams are next week I thought some people might find this useful..

Suppose we have a linear operator A:V->V where A is the matrix of the operator relative to some basis, presumable the standard orthonormal basis of R^3.

The purpose of finding the Jordan Normal Form is because it makes the matrix appear simpler in a different basis so we can more easily understand what the operator does.

Right, so take the matrix A. Firstly, find its characteristic equation found by
det(A-tI)=0, where I is the identity matrix.

The solutions to the equation are the eigenvalues of our matrix.
Define a new matrix B=A-tI. There will usually only be one or two eigenvalues, I haven't seen a question of Vlad's yet where there was 3 distinct eigenvalues.

For each eigenvalue find rk(B), rk(B^2), ..., rk(B^k-1), rk(B^k)
until rk(B^k-1) = rk(B^k).
Obviously then dimKer(B^k-1) = dimKer(B^k) because of the formula
rk(A) + dimKer(A) = dim(V)

NOTE: CASE 1 AND 2 ASSUME TWO EIGENVALUES

CASE 1: rk(B) = rk(B^2)
Find a basis of ker(B) given by the vector v, Bv=0.
v is our basis vector. Since the kernels of powers stabilise instantly we get a thread of length one, and we have our one basis vector, and we are done.

CASE 2: rk(B) =/= rk(B^2) = rk(B^3) =/= 0
Reduce the vectors which span Ker(B^2) using the vector which spans Ker(B) to get a relative basis vector f, which is the non-zero vector obtained after reducing. Since the kernels of powers don't stabilize for two steps, find Bf and reverse the order to get the basis vectors Bf, f.

CASE 3: If B is nilpotent, that is to say (B^k)=0 for some k. There are certain intricacies to this case.
Firstly, assume two distinct eigenvalues.
Find the vectors that span ker(B^k-1), and our first basis vector is the vector which "makes up for the missing leading 1". If k=2, find Bf and we are done.
Reverse the order to have a basis; Bf, f.

Now assume only one eigenvalue. If k=3, do the same as above, but find f, Bf and (B^2)f. Reverse the order for the basis. (B^2)f, Bf, f.

****** Now suppose only one eigenvalue and k=2. Find f, and Bf, and then reduce ker(B) using Bf to get a vector g.
Basis given by g,Bf,f.

Right, you have your jordan basis. More than likely you will be using two of the cases above, one for each eigenvalue.
So you get a transition matrix C, which transforms A into it's Jordan Normal Form, J.

J = (C^-1)AC

In general, J is given be each of your eigenvalues along the diagonal.
The multiplicity of each factor of eigenvalue in your characteristic equation determines the size of your jordan block. Best way to illustrate is with an example. Suppose the characteristic equation is t(t-3)^2
Eigenvalues are 0 and 3, and 3 has a multiplicity of 2, so the jordan normal form is given by

.... 0 0 0
J = 0 3 1
.... 0 0 3

A one is placed over the intersection of two same eigenvalues, so if your characteristic equation was (t-3)^3, J would be

.... 3 1 0
J = 0 3 1
.... 0 0 3

This is true in all cases except the one marked with ******. In this case you essentially have two jordan blocks of different size, but each has the same eigenvalue, so you'd get

.... 3 0 0
J = 0 3 1
.... 0 0 3

This is a result of the Cayley-Hamilton theorem and minimal polynomials etc.

One last thing, you put the basis vectors in corresponding order of the eigenvalues of your JNF. Example, if your eigenvalues are 2 and 3, and for 2 you get a basis of f, and for 3 you get a basis of g and Bg, and you arrange your J as

.... 2 0 0
J = 0 3 1
.... 0 0 3

then the columns of your transition matrix C is f, Bg, g


So there you have it, Jordan Normal form explained.

What's the story like with the maths schols?
Are the papers generally the same standard as final exam papers or way more hardcore including stuff you've never done in class?

fh041205 wrote: »

Didn't Stalker do GR last year? Why wouldn't he do it again this year? Actually has anyone left the maths dept since last year lecturer-wise?

Yeah Vlad left this year. As in he was here for all of this year but wont be here from September on. Pete is taking over Linear Algebra but apparently they don't have someone to do analysis yet. I'd imagine Stalker could do it.