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- 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.
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: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.
I really admire your comment here. could you please give a good understanding of hilbert nullspace theorem?My rifle would be Hilbert's Nullstellensatz but I assume that the canonical Jordan normal form will do.
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 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.
ok, seems I can't understand it fully(crying face), but thanks! i didnt learn well on abstract algebra(crying)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.
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.