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

[Steven G]

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)

 

[blamocur]

I get a different characteristic equation for 'A' with three different roots.

 

[blamocur]

A = row 1 = (-1 2 -3), row 2 = (1 2 -2), row 3 = (1 -1 2)

I wonder whether you meant

A = row 1 = (-1 2 -3), row 2 = (-1 2 -2), row 3 = (1 -1 2)

 

[blamocur]

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 [imath]A-I[/imath], which in your case has one vector, then find another vector in [imath]\ker((A-I)^2) \setminus \ker(A-I)[/imath], then the third vector from [imath]\ker((A-I)^3) \setminus \ker((A-I)^2)[/imath]. These three vectors should produce, maybe after normalization, your matrix [imath]P[/imath].
Does this make sense?

 

[Steven G]

Yes, it makes sense. The problem is that (A−I)³=0

 

[blamocur]

Yes, it makes sense. The problem is that (A−I)³=0

Not really a problem as long as [imath]A-I \neq 0[/imath] and [imath](A-I)^2 \neq 0[/imath]. I am pretty sure I've figured out how to build [imath]P[/imath] such that [imath]P^{-1}AP[/imath] is in Jordan normal form after seeing, without really understanding, another section in Wikipedia.

 

[blamocur]

Now that one week has lapsed, here is my solution:

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