- #1
Gale
- 684
- 2
Homework Statement
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]
Homework Equations
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)