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Rational canonical form

I'm quite confused on how they deduced that there's a factor of f^k(x) in the minimal polynomial with k>=2 and m>=1.

m_A(x) means minimal polynomial of A, c_A(x) means characteristic poly of A, C(f) means companion matrix with characteristic poly f. The question itself is asking you to find the maximal N such that for n>N the statement is false.

https://imgur.com/a/Y994LCz
(edited 2 months ago)
Reply 1
This is a bit of a ramble - I've only read a bit about this, not studied it, so I'm somewhat winging it - I didn't find it obvious.

It's also a bit close to a full solution perhaps, so I've put spoilers. Do try to just look at one bit at a time and see where it gets you, rather than rushing through to the end.

To start: if every irreducible factor only divides mAm_A once, what can you say about the block structure of the RCFs of A & B? (Remember that they share the same minimal/characteristic polys, so they share the same factors).

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So one of A and B contains a 3-block decomp of the form suggested.
(edited 2 months ago)
Reply 2
Original post by DFranklin
This is a bit of a ramble - I've only read a bit about this, not studied it, so I'm somewhat winging it - I didn't find it obvious.

It's also a bit close to a full solution perhaps, so I've put spoilers. Do try to just look at one bit at a time and see where it gets you, rather than rushing through to the end.

To start: if every irreducible factor only divides mAm_A once, what can you say about the block structure of the RCFs of A & B? (Remember that they share the same minimal/characteristic polys, so they share the same factors).

Spoiler



Spoiler



Spoiler



Spoiler




Thank you for the response, for your first statement is it because if the irreducible polys divide minimal poly A once then it implies the RCF would just have 1 block for each irreducible factor thus making the RCF of A and B equal contradicting the non similar assumption?

For the second part, isn't the companion matrix of the highest power of the irreducible factor in the minimal poly always a block in the RCF thus wouldn't k_a=k_b.

I'm a bit lost from the 3rd and 4th hints that you give. You wrote that if neither contains C(f^k_A) or C(f^k_B) for A,B respectively then A decomposes as two blocks with one of them being C(f^k_A)
(edited 2 months ago)
Reply 3
Original post by Student 999
Thank you for the response, for your first statement is it because if the irreducible polys divide minimal poly A once then it implies the RCF would just have 1 block for each irreducible factor thus making the RCF of A and B equal contradicting the non similar assumption?
No. E.g. if I is the 2x2 identity matrix then it's minimal polynomial is (x-1) (and so (x-1) only divides it once) but it has 2 blocks in its decomposition.

For the second part, isn't the companion matrix of the highest power of the irreducible factor in the minimal poly always a block in the RCF thus wouldn't k_a=k_b.
You might be right. I wasn't sure and I didn't need to assume it, so I didn't.
I'm a bit lost from the 3rd and 4th hints that you give. You wrote that if neither contains C(f^k_A) or C(f^k_B) for A,B respectively then A decomposes as two blocks with one of them being C(f^k_A)

No, I said "of the form described" - i.e. the 3-block decomposition they were claiming exists. The point being the only way it doesn't exist is if there are only 2 blocks corresponding to powers of f.
(edited 2 months ago)
Reply 4
@Student 999 - having thought about it, I think you're right - both A and B must have one block corresponding to the highest power of f dividing the minimal polynomial. So that will make it a bit easier when you do the next part.
Reply 5
Original post by DFranklin
@Student 999 - having thought about it, I think you're right - both A and B must have one block corresponding to the highest power of f dividing the minimal polynomial. So that will make it a bit easier when you do the next part.

I’m still stuck on the first part you wrote, would you be able to hint further or tell me what the reasoning is?

Also in your last spoiler, you wrote there’s a relationship between the sum of block sizes and the multiplication that f divides m_A. I know for JCF there’s the relationship that the sum of block sizes of the same evalue is equal to it’s algebraic multiplicity but I haven’t come across anything similar related to RCF
Reply 6
Original post by Student 999
I’m still stuck on the first part you wrote, would you be able to hint further or tell me what the reasoning is?

Also in your last spoiler, you wrote there’s a relationship between the sum of block sizes and the multiplication that f divides m_A. I know for JCF there’s the relationship that the sum of block sizes of the same evalue is equal to it’s algebraic multiplicity but I haven’t come across anything similar related to RCF

No I didn't. I said c(A) not m(A).

But the sum of block sizes corresponding to the (algebraic multiplicity of f) * (Deg f) should be obvious if you think about what those blocks actually look like in the RCF. [The Deg f here is because I'm saying a simple block C(f) has size Deg f; possibly you'd just say it has size 1 and lose the Deg f).
Reply 7
As I said, I wasn't that familiar with this topic (and it might well be I'm getting things wrong).

Anyhow, I found this video (and successors) helpful: https://youtu.be/PEuf8E_OSB4?si=k1Gg1rFOSYVMExq4

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