1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    ##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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A multiplication map as a matrix in Q(sqrt2)
  1. Matrix Multiplication (Replies: 12)

  2. Matrix multiplication (Replies: 3)

  3. Matrix Multiplication (Replies: 3)

  4. Matrix multiplication (Replies: 5)

Loading...