Jordan Decomposition (I tried to find the null space for (A-xI)^3.)

Steven G

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I was given a matrix A. I found the Jordan matrix for A
Let x = lamda. The characters equation for |A-xI| = (x-1)3
So I tried to find the null space for (A-xI)3. However, that matrix was the zero matrix. So I am at a loss as how to find P as I only have two Linear Independent vectors for the null space of (A-xI)

A = row 1 = (-1 2 -3), row 2 = (1 2 -2), row 3 = (1 -1 2)
 
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I get a different characteristic equation for 'A' with three different roots.
 
So I tried to find the null space for (A-xI)3. However, that matrix was the zero matrix. So I am at a loss as how to find P as I only have two Linear Independent vectors for the null space of (A-xI)
According to this section in Wikipedia, you first find null space, a.k.a. kernel, of AIA-I, which in your case has one vector, then find another vector in ker((AI)2)ker(AI)\ker((A-I)^2) \setminus \ker(A-I), then the third vector from ker((AI)3)ker((AI)2)\ker((A-I)^3) \setminus \ker((A-I)^2). These three vectors should produce, maybe after normalization, your matrix PP.
Does this make sense?
 
Yes, it makes sense. The problem is that (A−I)3=0
Not really a problem as long as AI0A-I \neq 0 and (AI)20(A-I)^2 \neq 0. I am pretty sure I've figured out how to build PP such that P1APP^{-1}AP is in Jordan normal form after seeing, without really understanding, another section in Wikipedia.
 
Now that one week has lapsed, here is my solution:

If $\lambda$ is a multiplicity 3 eigenvalue of a 3x3 matrix A\mathbf A, then we can bring it to Jordan normal form by forming matrix P\mathbf P with column vectors $\mathbf B^2 \mathbf v, \mathbf B\mathbf v, \mathbf v$, where B=AI\mathbf B=\mathbf A-\mathbf I, and v\mathbf v is any (column) vector for which B2v0\mathbf B^2 \mathbf v \neq \mathbf 0:
P=(B2v,Bv,v)\mathbf P = \left( \mathbf B^2 \mathbf v , \mathbf B \mathbf v , \mathbf v \right)In the basis (B2v,Bv,v)(\mathbf B^2 \mathbf v, \mathbf B\mathbf v, \mathbf v) matrix B\mathbf B is represented by:
B=(010001000)\mathbf B^\prime = \left( \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right)i.e.
B=P1BP\mathbf B^\prime = \mathbf P^{-1}\mathbf B \mathbf PBut then
A=    P1AP=    P1(B+λI)P=    P1BP+P1λIP=    P+λI=    \mathbf A^\prime =\;\; \mathbf P^{-1}\mathbf A \mathbf P =\;\; \mathbf P^{-1} (\mathbf B+\lambda \mathbf I) \mathbf P =\;\; \mathbf P^{-1}\mathbf B \mathbf P + \mathbf P^{-1}\mathbf \lambda \mathbf I \mathbf P =\;\; \mathbf P^\prime + \lambda \mathbf I =\;\;=(λ100λ100λ)=\left( \begin{array}{ccc} \lambda & 1 & 0\\ 0 & \lambda & 1\\ 0 & 0 & \lambda \end{array}\right)Note: I actually don't know how to prove that columns of P\mathbf P form a basis, i.e., are linearly independent, but it might not be too difficult in the case of a specific A\mathbf A.
 
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