How to Solve B^3 = A^2 Matrix 2x2 on C?

[maria papadakh]

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.

 

[fresh_42]

My rifle would be Hilbert's Nullstellensatz but I assume that the canonical Jordan normal form will do.

 

[maria papadakh]

i will try with the jordan normal form.thank you!

 

[graphking]

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.

 

[nuuskur]

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.

 

[graphking]

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

 

[graphking]

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)

 

[graphking]

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.

 

[fresh_42]

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.

 

[graphking]

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)

 

[Infrared]

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.

 

[WWGD]

Maybe the Borel ( or otherwise) functional calculus?