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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!

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pwhodges:
Reading that (for nostalgia's sake, though it's already more set theory than I ever did), I had the thought that my maths prowess was ultimately limited by the fact that I picture mathematical relationships geometrically - I cannot abstract the ideas to the point that I can handle them without that crutch, and so I am also limited to ideas that can fit into a three- (or maybe four-) dimensional space in my head.

That may not make sense to you, but I assure you it did to me.

snalin:
I, on the other hand, really struggle with geometry. I'm pretty good at maths, but going to linear algebra lectures were often pointless, because they modelled something that's basically just elements consisting of ordered sets of numbers as points in space, which became confusing.

The same thing goes for linear programming. It's basically the use of the very powerful simplex algorithm to solve a class of programming problems. The thing is, there's an algebraic way of looking at how it works and a geometric way of looking at how it works, and the geometric one is always used in explanations. I absolutely hate it. I can always see how the maths can model the geometry, but working with the geometry is just adding a layer of complication on something that's as simple as just numbers.

ankhtahr:
So we're sitting here, working on the exercise sheet.

We have a solution for the 2a), but to me it seems overly complex, and formally wrong. I can see what the creator of the solution wanted to ge to, but I don't think you're allowed to write it that way.

(click to show/hide)
I don't know how it can be done easier though. (this is not my solution)

ankhtahr:
So there's another linear algebra sheet due Monday. But as I don't have time to go to university on monday I need to be done with it on Friday. Damn.

Here's the exercise sheet.

I'm not happy about it.

Skewbrow:
I've been gone for a couple weeks - nurturing that freshman course to a conclusion. My admiration for the level of freshman courses at your university is reinforced.

For 1A: As a homomorphism of groups f respects addition. Your task is to show that f(qv)=q f(v) for all rational numbers q and all vectors v. I would do this in steps as follows:
1) first prove it by induction whenever q is a positive integer (an example like this has probably been done).
2) a group homomorphism takes the additive inverse (=negative) to the additive inverse, so we can conclude that the claim holds whenever q is any integer.
3) next we do it in the case q=1/n for some positive integer n. Here a trick is needed. Write v=(n/n)v=n*((1/n)v). By step 1 we have f(v)=f(n*((1/n)v))=n f((1/n)v). Because W is a vector space over Q, this implies (multiply both sides by 1/n) that (1/n)f(v)=f((1/n)v), which is what we want for this step.
4) The general case q=m/n then follows by combining steps 2 and 3. Leaving this to you.

1B) Hint: Consider the Q-linear mapping f defined by f(1)=1, f(sqrt(3))=-sqrt(3). Show that it is not linear over Q(sqrt(3)).

2A) This you can do simply by using the subspace criterion. If phi1 and phi2 are two linear mappings sending all the elements of the subspace U to 0, then you can easily deduce that the same thing happens with the linear combination a*phi1+b*phi2, a and b in k.
2B) Barring suitable examples/auxiliary results in the course notes this makes your hands a bit dirty. Start with a basis of U, extend that to a basis B of V. You have probably covered the bit that in the finite dimensional case you get a basis for V* by mapping one of the elements of B, say b, to 1, and the rest to 0. Call that mapping phib. You should be able to show that the elements phib, b ranging over B, such that b is not an element of U, form a basis of U0.
2C) Many ways to do this. One is to find two linearly independent elements, v1 and v2,of V that are orthogonal to (1,1,2). Both of those give rise to an element in U0 by the recipe phii: x -> the inner product of x and vi, i=1,2. Another way (perhaps the intended way?) is to redo 2B in this specific instance: find such a basis of V that (1,1,2) is one of the basis vectors. Then describe in detail the two mappings in V* that map (1,1,2) and one of those other basis vectors to zero, and the remaining one to 1.

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