# A multiplication map as a matrix in Q(sqrt2)

1. Mar 24, 2013

### Gale

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

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 $\{1, \sqrt{2}\}$ express the Q-linear map $M_{r+s\sqrt{2}}$ as a 2x2 matrix with entries in Q.

(ii) If $A$ is a 2x2 matrix with entries in Q, what is the condition that $A=M_{r+s\sqrt{2}}$ 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 $M_{t+u\sqrt{2}}, t,u \in Q.$ Use this to give an explicit formula for $(r+s\sqrt{2})^{-1}$
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...
$M_{r+s\sqrt{2}}(\alpha)= (r+s\sqrt{2})*\alpha= (r+s\sqrt{2})*(a+b\sqrt{2})= (ar+2bs)+\sqrt{2}(as+br)$
So the matrix should produce the same result if I do $\alpha*A$ right? So... I suppose that means my matrix would be (r s,2s r)?
(sorry I don't remember how to format matrices)

2. Mar 24, 2013

### Dick

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.

3. Mar 24, 2013

### Gale

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 $A= \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right)$ 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

4. Mar 24, 2013

### Dick

$A \alpha$ is probably more common. In which case a matrix $A= \left( \begin{array}{ccc} r & 2s \\ s & r \end{array} \right)$ 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?