# Jordan Forms, Algebraic and Geometric Multiplicity

1. Nov 4, 2011

### shaon0

1. The problem statement, all variables and given/known data
A 20 × 20 matrix C has characteristic polynomial (λ^2 − 4)^10. It is given that ker(C−2I), ker (C − 2I)^2, ker (C −2I)^3 and ker (C −2I)^4 have dimensions 3,6,8,10 respectively. It is given that ker (C + 2I), ker (C +2I)^2, ker (C +2I)^3 and ker (C +2I)^4 have di-
mensions 3,5,7,8 respectively. What can be said about the Jordan form of C?

3. The attempt at a solution
I know the eigenvalues of C are +-2 each w/ a multiplicity of 10. So, the Jordan Forms will be; Jn(2)\...\Jn(-2) for each n=1,2,3,... where n is the algebraic multiplicities.
Any help with Jordan Forms, Algebraic and Geometric Multiplicity will be appreciated.

2. Nov 4, 2011

### micromass

Staff Emeritus
OK,

1) What is the algebraic and geometric multiplicity of your eigenvalues??

3) How would you choose your basis. Theoretically. I know you don't have a matrix here, but what is the general process for choosing the Jordan basis??

4) Can you say anything about $Ker(C-2I)^5$ or $Ker(C+2I)^5$?? And what about $Ker(C-2I)^6$ or $Ker(C+2I)^6$??

3. Nov 5, 2011

### shaon0

1) Alg. Multiplicity for each eigenvalue is 10 where eigenvalues are +-2. Not sure how to find the geometric multiplicity. Probably my biggest issue with this topic.

2) Geometric multiplicity is the no. of jordan blocks for each eigenvalue w/ 1 on top of each diagonal entry in the matrix C.

3) Gram-Schmidt, Householder's. To be honest, i'm not entirely sure.

4) Does Ker(C-2I)^6=Union {n=1 to 5} Ker(C-2I)^n. Same goes for Ker(C+2I)^6. I'm also not sure about this.

4. Nov 5, 2011

### micromass

Staff Emeritus
The geometric multiplicity is just $Ker(C-\lambda I)$ where $\lambda$ is your eigenvalue. So you are given your geometric multiplicities.

Good. And now that you know the geometric multiplicity, what is the number of Jordan blocks?

It's quite easy, you need to construct "chains". For example, if

$$Ker(C-\lambda I)=3, ~Ker(C-\lambda I)^2=4, ~Ker(C-\lambda I)^3=5$$

then you start by taking a v in $Ker(C-\lambda I)^3$ and you construct the chain

$$\{v,(C-\lambda I)v, (C-\lambda I)^2v\}$$

If none of these vectors is zero, then you found three elements of your basis. These three elements correspond to a 3x3 Jordan block.

Try to use this to figure out the Jordan blocks in your matrix.

No, all you know that for large enough n: $Ker(C-2I)^n=10$ (= the algebraic multiplicity).

5. Nov 5, 2011

### shaon0

Oh ok, thank you. I've kind of got it, will need to do a couple of more questions though to solidify the knowledge. Thanks again, really appreciate it :)