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Find all 4x4 A matrices so that $A^4=A^6$.

I think the method has to do something with eigenvalues, eigenvectors etc'...

Thanks in advance for any assistance!

err
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2 Answers2

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Hint :

You have $A^4=A^6$ i.e., $A$ satisfies polynomial $x^6-x^4$.

But $A$ is a $4\times 4$ matrix so its Minimal polynomial should divide $x^6-x^4$.

What are all the polynomials that divide $x^6-x^4$?

  • Why should it divide $x^6-x^4$? And for your question, $x^4$, $x^4-x^2$ divide it... Or is there another one I missed? – err Dec 04 '13 at 17:23
  • are you familiar with something like : any polynomial which satisfies given matrix should be divisible by its characteristic polynomial... I forgot the name of the theorem (assuming that there is some name to it) –  Dec 04 '13 at 17:27
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    The characteristic polynomial does not need to divide $x^6 - x^4$. For example, the identity matrix satisfies $I^6 = I^4$, but has characteristic polynomial $(x-1)^4$ – Ben Grossmann Dec 04 '13 at 17:28
  • Oh... yes.. may be i should make that as "minimal polynomial" instead of "characteristic polynomial".. –  Dec 04 '13 at 17:31
  • Yep! A minor point, but an important one. – Ben Grossmann Dec 04 '13 at 17:33
  • yes yes.. I should have been more careful.. Thank you! :) –  Dec 04 '13 at 17:33
  • @PraphullaKoushik Why is the minimal polynomial of "order" 4? The identity satisfies $x-1$. – Pablo Rotondo Dec 04 '13 at 17:34
  • @PabloRotondo : Another punch... I will edit :D –  Dec 04 '13 at 17:36
  • @err : I am extremely sorry that I had not answered for you completely... I have edited the answer to any polynomial which divides $x^6-x^4$.. so your list is not complete at this point –  Dec 04 '13 at 17:43
  • it would be better if you can just find all roots of $x^6-x^4$ (or at least some irreducible decomposition) may be some thing like $(x-1)(x+1)g(x)$.. –  Dec 04 '13 at 17:44
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Hint:

We can write $A^6 - A^4 = 0$, which is to say that $A$ "satisfies" the polynomial $x^6 - x^4 = 0$. We then know (by a theorem that is probably in your textbook) that the minimal polynomial of $A$ divides $x^6 - x^4 = x^4(x-1)(x+1)$.

What does this tell us about the Jordan-canonical form of $A$?

Connection between minimal polynomials and J-C form:

  • $q_A(x) = (x-\lambda_1)^{m_1}\cdots(x-\lambda_k)^{m_k}$, where $\lambda_1,\dots,\lambda_k$ are the eigenvalues of $A$, and $m_k$ is the length of the longest Jordan block associated with $\lambda_k$.
  • $m_k \geq 1$ for every eigenvalue of $A$. $A$ is diagonalizable if and only if $m_k=1$ for all $k$.
Ben Grossmann
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  • So the minimal polynomial of A is either $x^4$, or $x^4-x^2=(x^2-x)(x^2+x)=x^2(x-1)(x+1)$. So if the the minimal polynomial is $x^4$ then it is the zero matrix, if I'm not mistaken. If it's $x^2(x-1)(x+1)$ then this is also the characteristic polynomial, and from that we can say the Jordan canonical form of A is with one 2x2 block of eigenvalue=0, one 1x1 block of eigenvalue=1 and one 1x1 eigenvalue=-1. Is that correct? But if I found the Jordan canonical form of A, what can I say about the original form of A?! – err Dec 04 '13 at 17:33
  • Not quite. If the minimal polynomial is $x^4$, then $A$ is a nilpotent matrix of order $4$. The zero matrix has minimal polynomial $x$. Remember, the degree of the minimal polynomial is not generally the same as the dimension of the matrix, but $(x - \lambda)$ has to divide the minimal polynomial for every eigenvalue $\lambda$ – Ben Grossmann Dec 04 '13 at 17:38
  • But if the minimal polynomial has a degree the same as the order of the matrix, isn't it equal to the characteristic polynomial? Maybe this time I'll be right: If it's a nilpotent matrix then the eigenvalues are 0, and there is one 4x4 Jordan block of eigenvalue 0. Is that correct? And again, even if I found A's Jordan canonical form, how exactly do I find the 'original' $A$s that satisfies $A^4=A^6$? Thanks for your help! – err Dec 04 '13 at 17:44
  • That's right. So any $4\times 4$ matrix with min. polynomial $x^4$ has a canonical form consisting of $1$ $4 \times 4$ Jordan block of eigenvalue $0$. So, a matrix will have this minimal polynomial if and only if it is similar to this Jordan block matrix. Similarly, once we have all of the admissible Jordan forms, we can simply say that the matrices satisfying $A^4 = A^6$ will be any that are similar to those Jordan matrices. – Ben Grossmann Dec 04 '13 at 17:48
  • And of course, note that $x^6 - x^4 = x^4(x-1)(x+1)$ – Ben Grossmann Dec 04 '13 at 17:48