Canonical forms of linear operators

• Apr 28th 2009, 12:58 PM
nihilanth100
Canonical forms of linear operators
Im supposed to find the canonical form and corresponding change of coordinates for a few linear operators, heres two:
0 1 0
0 0 1
-1 -3 -3

2 1 1 -2
1 2 1 -2
1 1 2 3
3 3 3 5

I know im supposed to find the eigenvalues and vectors by finding detA = 0 for the characteristic equation then the roots, but I just am stuck when it comes to finding the equation from the matrixes.

Following from that im unsure of what to do after I plug back in lambda into the matrix...
• Apr 28th 2009, 01:14 PM
Gamma
You looking for the Jordan Canonical Form or the Rational Canonical Form?

Either way the first step is the same, find the characteristic polynomial. subtract x from the diagonal and calculate the determinant. It should factor into linear terms (if not, you cannot find the JCF unless you pass to the algebraic closure). The roots to this polynomial are the eigenvalues.

You will want to find the minimal polynomial, it must divide the characteristic polynomial (it will have the same roots, but possibly with less multiplicity). Now you either find the elementry divisors (JCF) or the invariant factors (RCF) respectively depending on which form you are looking for.

Once you find the eigenvalues you find the eigenvectors by solving $\displaystyle (A-I\lambda_i)v_i=0$.
• Apr 28th 2009, 01:42 PM
nihilanth100
Im not sure about either, but one of the lecture notes mentions the Jordan Block in one of the lectures about mapping to an n-dimensional space, so im guessing that.

I think my problem right now is more of getting the characteristic equation. I understand subtracting from the diagonal, but what is the best way going about calculating the determinant for a 4x4 matrix?

Once I can do this and get the roots he gives this tree:
http://www.indexoffice.com/mathlectu...orms/img19.png
Im still trying to understand the middle part and how to solve it to get the answer. Sorry for my ignorance, im literally working off of a few pages of his own lecture material since the guy doesnt teach from a book. Checking back in a few hours, thanks for the help so far! (Rock)
• Apr 28th 2009, 02:08 PM
Gamma
JCF
Yeah, that is definitely the Jordan Canonical Form. As for calculating the determinant of a 4x4 matrix, pick your poison man. Subtract lambda from the diagonal and then take the determinant.

Method 1) You can row reduce to get it into an upper triangular matrix (if you do it right you can actually get it into the Smith Normal Form and that will tell you your JCF directly, but I think that is probably beyond the scope of your course) and multiply down the diagonal.

Method 2) The way I would suggest is to just do the cofactor expansion method though, it may take a while, but it is way less mistake prone. Just pick a particularly nice row or column to go down (one with as many 0s as possible) and get to it.

Neither way is particularly pleasant, but such is life. In the end you should be able to factor it completely. You may get some of the eigenvalues repeated with multiplicity. You will get n eigenvalues repeated with multiplicity, that is when you follow his tree.

Although I will be honest, I am not sure that last branch is true. It is certainly possible for you to have repeated eigenvalues and not necessarily get that Jordan block with a 1 in it, but maybe he has set the problems up ensuring that this is the case.