Calculating JNF & Basis Vectors: Jordan Normal Form

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To determine the basis vectors for a matrix in Jordan Normal Form (JNF), one must calculate matrix P using the relationship P = (T^-1)PA, where T is the original matrix and A is its JNF. The minimal polynomial (x+2)(x-4)=0 indicates that the eigenvalue -2 has a zero-dimensional null space, while the eigenvalue 4 has a two-dimensional null space, suggesting a JNF of diag(0,4,4) for a 3x3 matrix. Finding the Jordan basis involves using the kernels of the operator related to the minimal polynomial, specifically ker(T-a)^r and its extensions. The process of calculating P and the Jordan basis can be complex and may require detailed steps for clarity. Understanding these concepts is essential for working with JNF in linear algebra.
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1. If given a matrix in JNF, what would be its basis? How would you calculate it?
If you put the basis vectors (of JNF) as columns in matrix P than
P=(T^-1)PA, T and A are given.

where T is the original matrix and A is T in JNF. But I cannot explicitly calculate P since it is on both sides. How do I find P?

2. If the minimal polynomial is given as (x+2)(x-4)=0, and the -2 eigenvalue results in a 0 dimensional null space (i.e. (0,0,0) vector) what would the JNF look like given the null space of eigenvalue 4 is 2 dimensional. And the original matrix is 3 by 3.

Would it be
diag(0,4,4)?
 
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the question does not quite make sensfinding P is a bit of work and hard to say briefly.

i can say it briefly but it won't help you that much.
 
briefly, to find a jordan basis for a map T with minimalpolynomial (X-a)^r, find a basis for ker(T-a)^r/ker(T-a)^(r-1), extend to basis of ker(T-a)^(r-1)/ker(T-a)^(r-2),...

see what I mean?
 
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