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Math Help - Matrix Ring

  1. #1
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    Matrix Ring

    My professor gave us this query at the end of class, it contained two parts.
    1. Show a ring is idempotent
    2. Consider the degree one polynomial f(x) is an element of M2(R)[x] given by
    f(x) = [0 1
    ______0 0]x + B
    (so f(x) = the matrix []x + B).
    For which B is an element of M2(R), if any, is f(x) idempotent?

    I proved part 1:
    If a2 = a, then a2 - a = a(a-1) = 0. If a does not equal 0, then a-1 exists in R and we have a-1 = (a-1a)(a-1) = a-1[a(a-1)] = a-10 = 0, so a-1 = 0 and a = 1. Thus 0 and 1 are hte only two idempotent elements in a division ring.

    Part 2 I simply have no idea on though. How do I do this with a matrix?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Well, call the matrix M. Then you want that (Mx+B)^2=Mx+B in M_2(\mathbb{R})[x]. So just perform the calculation and see what you get,

    (Mx+B)^2=MxMx + MxB + BMx+B^2 = M^2x^2+(MB + BM)x + B^2 (the x-terms act like this (commutatively) by definition).

    So, what is M^2? What must B be for M^2x^2+(MB + BM)x + B^2 = Mx+B? (Put B=\left(\begin{array}{cc}a&b\\c&d\end{array}\right  ) and see what happens)
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  3. #3
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    So for this to be true, M^2 would have to equal M/x, and B would have to be it's own multiplicative inverse right?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by DanielThrice View Post
    So for this to be true, M^2 would have to equal M/x, and B would have to be it's own multiplicative inverse right?
    No. You know what M^2 is, you can just work it out (as you know what M is). And if B was its own multiplicative inverse than B^2=I, but you need B^2=B.

    Basically, you have \left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)^2+\left(\left(\begin{arra  y}{cc}0 & 1\\0&0\end{array}\right)B + B\left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)\right)x + B^2 = \left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)x+B.

    Now, if B=\left(\begin{array}{cc}a & b\\c&d\end{array}\right) then you basically have to solve the equation,

    \left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)^2+\left(\left(\begin{arra  y}{cc}0 & 1\\0&0\end{array}\right)\left(\begin{array}{cc}a & b\\c&d\end{array}\right) + \left(\begin{array}{cc}a & b\\c&d\end{array}\right)\left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)\right)x + \left(\begin{array}{cc}a & b\\c&d\end{array}\right)^2  = \left(\begin{array}{cc}0 & 1\\0&0\end{array}\right)x+\left(\begin{array}{cc}a & b\\c&d\end{array}\right) for a, b, c and d. So do this.
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