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Homework Help: Linear Algebra question

  1. May 30, 2007 #1


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    Gold Member

    1. The problem statement, all variables and given/known data
    prove or disprove:
    1) (z is complex and bar(c) is the conjugate of c)
    If c and bar(c) are solutions to z^2 + az + b = 0 and c isn't real then a and b are real.
    2) If S and T are subsets of the space V and intersection of Sp(S) and Sp(T) is {0} then the intersection of S and T is an empty set. (Sp is the span)

    2. Relevant equations

    3. The attempt at a solution
    1) I wrote z as x+yi and got:
    x^2 + 2xyi - y^2 + ax + ayi +b = x^2 -2xyi - y^2 +ax - ayi +b
    and so:
    4xyi + 2ayi = 0 => 2x + a = 0 => a=-2x which is real. and by putting that into the equation I get that b is also real. So the answer is True.

    2) If S = T = {0} then the intersection of both Sp(S) and Sp(T) and S and T is {0}. So the answer is False.

    Is that right? Am I missing anything here? especially the last one seemed too easy.
  2. jcsd
  3. May 30, 2007 #2


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    Science Advisor
    Homework Helper

    Both your answers are right. Note in your work for the first one, you have "4xyi + 2ayi = 0 => 2x + a = 0." Be aware that this is only valid because y is non-zero, which is because c isn't real. Another way to do 1) is this:

    If c and [itex]\bar{c}[/itex] are the solutions to z2 + az + b = 0, then [itex](z-c)(z-\bar{c}) = z^2 + az + b[/itex]. Multiplying the left side out gives [itex]z^2 - (c+\bar{c})z + c\bar{c}[/itex], which is just z2 - 2Re(c)z + |c|2, and so a = - 2Re(c), and b = |c|2, both of which are clearly real quantities.
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