# quick jordan normal formWatch

#1
the JNF of a 3x3 matrix can be deduced from its characteristic and minimal polynomials. give an example to show this is not the case for 4x4 matrices.

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i haven't been to any lectures on this, so forgive my stupidity. But isn't it the case that if the CP is (t-a)^A(t-b)^B, and the MP is (t-a)^A'(t-b)^B', then the JNF has A a's, B b's etc. along the diagonal, then A-A'-1 1s in the superdiagonal of the jordan block for a? so we've found the JNF.

confused.
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10 years ago
#2
I'm not sure I can be bothered to do the matrix multiplication, but what about something like and ? They look like they'd have the same polynomials, and yet clearly the first one has one Jordan block of size 3 and one of size 1, and the second has two of size 2.

Edit: wait, they don't have the same minimal polynomial. Oops. But I think that's the point - JNF is unique up to reordering of the blocks (as he defined them), not just up to random scattering of 1s and 0s.
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10 years ago
#3
An example is . In this case, I believe that there are two possible JNFs, and you need to use other considerations beyond characteristic/minimal polynomials in order to calculate which one is the JNF. (At least, that is what my notes say.)
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#4
(Original post by generalebriety)
I'm not sure I can be bothered to do the matrix multiplication, but what about something like and ? They look like they'd have the same polynomials, and yet clearly the first one has one Jordan block of size 3 and one of size 1, and the second has two of size 2.

Edit: wait, they don't have the same minimal polynomial. Oops. But I think that's the point - JNF is unique up to reordering of the blocks (as he defined them), not just up to random scattering of 1s and 0s.
the second one is a jordan normal form?? oh.
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10 years ago
#5
(Original post by Chewwy)
the second one is a jordan normal form?? oh.
Isn't it?
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