Solve B^3 = A^2 Matrix 2x2 on C

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In summary: No, the Borel functional calculus is not the best way to go. The reason is that the set of solutions to the equations ##B^3-A^2=0## is not a Borel set.
  • #1
maria papadakh
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TL;DR Summary
if A matrix 2x2 on C show that there is a 2x2 matrix B on C that B^3=A^2
i know that there is the Cayley -Hamilton theorem but i don't know if i can use it and how.Do you have any ideas about it?Please give me any help.
 
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  • #2
My rifle would be Hilbert's Nullstellensatz but I assume that the canonical Jordan normal form will do.
 
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  • #3
i will try with the jordan normal form.thank you!
 
  • #4
could it be finished? btw, jordan normal form contains the hamilton-caley theorem. if you regard jordan normal as correct, then it's obvious that hamilton-caley holds.
 
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  • #5
Spectral decomposition might work. Put ##A^2 = T^{-1}XT##, where ##T## is the matrix with column vectors as eigenvectors and ##X## is a diagonal matrix with eigenvalues on the diagonal. Pick ##Y^3=X##, then ##B := T^{-1}YT## should do the trick.
 
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  • #6
nuuskur said:
Spectral decomposition might work. Put ##A^2 = T^{-1}XT##, where ##T## is the matrix with column vectors as eigenvectors and ##X## is a diagonal matrix with eigenvalues on the diagonal. Pick ##Y^3=X##, then ##B := T^{-1}YT## should do the trick.
NO. I have finish this... is simply a calculation question after using jordan normal form(maybe the "scary"(at least I am scared)hilbert nullspace theorem could work out easier ). And you said that X is diag, which is not ture because Jordan normal form include and must include sth like this:
1 1 0 0 0
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1
0 0 0 0 1
 
  • #7
sry I haven't come to PF for a long time. And sry for I have worked out it many weeks ago but I forget to reply to you guys. my 'general math' way:
if A got two different eigenvalue, of course A's jordan normal form is a diagnal matrix, then B choose to be its ^1/3 (also diagnal) could work. but sometimes A could got two same eigenvalue and only one eigenvector, in that case A could be regarded as :
lambda 1
0 lambda
though it seems tough, but in fact you can sinply asume B is in a form like:
x y
0 x
then B^3=(x*Id+y*J)^3, where the Id refers to identity matrix,J refers to
0 1
0 0
J^2, viewed as linear functions could easily seems to be zero.(sth more should be said about J but I don't have the time to type it, sry)
then B^3 immediately could be calculated:
x^3 3x^2y
0 x^3
so you could easily choose x,y to make B^3=A^2(A= its jordan normal form)
 
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  • #8
fresh_42 said:
My rifle would be Hilbert's Nullstellensatz but I assume that the canonical Jordan normal form will do.
I really admire your comment here. could you please give a good understanding of hilbert nullspace theorem?
I thought in matrix maybe we have over any algebraic closed field F, any polynomial k[x] where the coefficient is in F, x choose from M_n[F], has a zero point? like a general version of the foundamental theorem of algebra.
 
  • #9
graphking said:
I really admire your comment here. could you please give a good understanding of hilbert nullspace theorem?
I thought in matrix maybe we have over any algebraic closed field F, any polynomial k[x] where the coefficient is in F, x choose from M_n[F], has a zero point? like a general version of the foundamental theorem of algebra.
Yes, that's what Hilbert's Nullstellensatz says. ##B^3-A^2=0## are four polynomial equations over ##\mathbb{C}[a_{ij},b_{kl}]## which define a proper ideal, hence there is a common zero.
 
  • #10
fresh_42 said:
Yes, that's what Hilbert's Nullstellensatz says. ##B^3-A^2=0## are four polynomial equations over ##\mathbb{C}[a_{ij},b_{kl}]## which define a proper ideal, hence there is a common zero.
ok, seems I can't understand it fully(crying face), but thanks! i didnt learn well on abstract algebra(crying)
 
  • #11
fresh_42 said:
Yes, that's what Hilbert's Nullstellensatz says. ##B^3-A^2=0## are four polynomial equations over ##\mathbb{C}[a_{ij},b_{kl}]## which define a proper ideal, hence there is a common zero.

I don't think this is a good argument. The matrix ##A## is fixed so it is really just a set of equations over the ##b_{kl}.## But anyway, how do you know that the ideal is proper? Showing this looks to be just as hard as the original problem. For example, it's not true that every matrix has a square root but how would you know that the ideal defined by ##B^3-A^2## is always proper but the ideal defined by ##B^2-A## might not be?

I think considering the Jordan type of ##A## is best way to go.
 
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  • #12
Maybe the Borel ( or otherwise) functional calculus?
 

1. How do you solve B^3 = A^2 Matrix 2x2 on C?

To solve this equation, you can use the diagonalization method. First, find the eigenvalues of matrix A and B. Then, use the diagonalization formula to find the diagonal matrices D and P. Finally, solve for B by taking the cube root of D and multiplying it by P.

2. What is the purpose of solving this equation?

Solving this equation can help in understanding the relationship between two matrices and their powers. It can also be used in various applications such as in physics, engineering, and computer science.

3. Can this equation be solved using other methods?

Yes, there are other methods such as using the Jordan canonical form or the Cayley-Hamilton theorem. However, the diagonalization method is the most commonly used and straightforward approach for solving this type of equation.

4. Are there any restrictions on the values of A and B in this equation?

Yes, both matrices A and B must be square matrices of the same size (2x2 in this case) and have complex entries. Additionally, A must be invertible and B must have a unique cube root.

5. Can this equation be solved for higher dimensions?

Yes, this equation can be solved for higher dimensions as long as the matrices are square and have complex entries. However, the diagonalization method becomes more complex and other methods may be more efficient.

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