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Homework Help: A multiplication map as a matrix in Q(sqrt2)

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the field [itex]Q(\sqrt{2})[/itex], viewed as a vector space of dimension 2
    over [itex]Q[/itex]. Let [itex] r + s\sqrt{2} \in Q(\sqrt{2})[/itex], and defi ne the multiplication map
    [itex]M_{r+s\sqrt{2}}: Q(\sqrt{2}) → Q(\sqrt{2}) [/itex] by [itex]
    M_{r+s\sqrt{2}}(\alpha)= (r+s\sqrt{2})*\alpha [/itex]
    In other words, [itex]M_{r+s\sqrt{2}}[/itex] is multiplication by [itex] r+s\sqrt{2}[/itex]

    There's a few parts, but my linear algebra is really rusty, so I can't even figure out the first step:

    (i) In terms of the basis [itex] \{1, \sqrt{2}\}[/itex] express the Q-linear map [itex] M_{r+s\sqrt{2}} [/itex] as a 2x2 matrix with entries in Q.

    (ii) If [itex]A[/itex] is a 2x2 matrix with entries in Q, what is the condition that [itex] A=M_{r+s\sqrt{2}}[/itex] with respect to the matrix?

    (iii) What is the determinant? Argue without computation that it's not zero.

    (iv) Calculate the inverse matrix and show that it is of the form [itex] M_{t+u\sqrt{2}}, t,u \in Q. [/itex] Use this to give an explicit formula for [itex] (r+s\sqrt{2})^{-1} [/itex]
    2. Relevant equations

    3. The attempt at a solution
    I honestly have no idea how to express the linear map as a matrix, and the rest of the problem basically hinges on that concept. The only part I think I understand is that because the vector space has dimension =2, it's going to have 2 elements in it's basis? And thus a 2x2 matrix? But I'm not even exactly sure I have that right...

    So just trying to figure out what this is talking about...
    I guess α is some element of the form α= a+b√2, for some a and b in Q, right? So if I just multiply out the function...
    [itex] M_{r+s\sqrt{2}}(\alpha)= (r+s\sqrt{2})*\alpha= (r+s\sqrt{2})*(a+b\sqrt{2})= (ar+2bs)+\sqrt{2}(as+br)[/itex]
    So the matrix should produce the same result if I do [itex] \alpha*A [/itex] right? So... I suppose that means my matrix would be (r s,2s r)?
    (sorry I don't remember how to format matrices)
  2. jcsd
  3. Mar 24, 2013 #2


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    It depends a bit on whether you want to picture the vector representing ##\alpha## as a column vector and let A act on the right, so the result is ##A \alpha## or have ##\alpha## be a row vector and act on the left, so it's ##\alpha A##. You've got the right matrix for the last case. If it's the first case you just need to take the transpose.
  4. Mar 24, 2013 #3
    Ah thank goodness I figured something out! Would ##A \alpha## be the more proper way of writing it out then?

    Since I have that, for the rest:
    (ii) if [itex] A= \left( \begin{array}{ccc}
    a & b \\
    c & d \end{array} \right)[/itex] Then based on my matrix from part (i), a=d and 2b=c would work? Or does a also have to be a multiple of r, and b a multiple of s?

    (iii) I'm a little confused about this... I guess since the determinant would be r2-2s2 and r,s are rational by definition, r2-2s2=0 would imply r2=2s2 → √2 is rational if both s,r≠0. Which is equivalent to saying r+s√2=0. I'm not sure if what I did counts as calculation.

    (iv) The inverse is found in the traditional way for a 2x2, but i'm not really sure how it gives me (r+s√2)-1
  5. Mar 24, 2013 #4


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    ##A \alpha## is probably more common. In which case a matrix [itex] A= \left( \begin{array}{ccc}
    r & 2s \\
    s & r \end{array} \right)[/itex] will do the job. For iii) you've got it. If r and s are rational then the only noninvertible matrix is zero because the square root of 2 is irrational. For (iv) if you find the inverse matrix style and then convert back to r and s you should get the same thing as the matrix of (r+s√2)^(-1), shouldn't you?
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