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c0uchm0nster

Help with linear algebra

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So, I'm going to school for the first time in 4 years, and my math is a bit rusty. I'm in a linear algebra/pre-calc class, and im struggling quite a bit (although so far i manage to understand a topic just enough to get by). My class is supposed to have tutors, but doesn't, and the book we're using doesn't explain stuff nearly enough, and doesn't provide hardly any exercise problems. I need to figure out some way to get a handle on this stuff and quickly. Does anyone have any suggestions?

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So show some problems, and where you were stuck. If you'd use source tags for large amounts of math, it surely will not harm this forum.

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Hello fellow vermonter. My linear algebra and calculus textbooks had great examples. Last year I used Lee Johnson and Dean Riess textbook for linear algebra which you can get for about $1 online. It showed lots of good examples and worked me through everything and kept me at the top of the class despite using an out of date book (math doesnt really go out of date). As for Calculus, I can't say anything bad about 4th ed James Stewart's Calculus book, which served me well through Calc 1 in high school, Calc II and Calc III in undergrad.

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If you know some programming, you can probably figure out Maple or Matlab easily enough. You can then program out a lot of stuff from Linear Algebra and do examples rather quickly until you have a full grasp of the material.

Also, David Lay's text on Linear Algebra is great. I still reference it in my Linear Algebra classes in graduate school.

mathworld.wolfram.com/topics/LinearAlgebra.html may help you out, but you'll have to wade through some advanced theory.

Finally, as Raghar mentioned, you can always just post a problem and I'm sure people will help you out.

Or PM me, and I'll see what I can do.

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Thanks for the advice guys. The reason I held off from posting problems is back when I used to use these forums a lot more for cpp help I thought I saw a few threads that said that helping with school homework wasn't allowed.

Here's an example of one of my current problems:

Given the vectors v and w, find the orthogonal projection of u of w onto v. decompose w into components u and u^perp.

The books example of an orthogonal projection is as follows:

u = (||w||cos(theta))(v/||v||) = ((v*w)/||v||^2)v




I know that cos(theta) can become (v*w)/(||v||*||w||)... so I think ||w||cos(theta) turns into (v*w)/||v||... and then (v*w)/||v|| * v/||v||
My regular old arithmatic gets a little hazy here, but I think this becomes ((v*w)/(||v||^2))v because of the homogeneous properties? If that's not correct could someone explain this step out a bit?

And then from there I guess it's trivial to find u^perp = w - u

So, I think I get the theory/forumla behind it, and I try one with actual values for the vectors, and I'm stuck at pretty much square one:

v=(1,-1) w=(3,2)
u= ([1,-1]*[3,2])/2 * [1,-1] = [0.5, -0.5] <-- how the hell?

thanks to anyone brave enough to try and help

[update] of course right after posting i think i've figured it out...

([1,-1]*[3,2])/2 becomes (1*3 + -1*2) / 2 becomes 1/2... then .5 * [1,-1] = [.5,-.5]
correct?
[/update]

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That's absolutely correct. A way to check your answer is to verify that u^perp * v = 0, which is indeed the case.

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Let me show you the way I usually think about this problem, in case it helps you.

You want to find a vector proportional to v, which we call u. This means we can write u = t*v, where t is some real number. The vector u is the one that gets as close as possible to w, in the sense that it makes u_perp's length minimum, where u_perp is defined as w-u.

So let's see how long is u_perp. In order to do that, we can look at the square of its length, which is <u_perp,u_perp>. I am using the notation <X,Y> to mean "the dot product of X and Y."

<u_perp,u_perp> = <w-u,w-u>
= <w,w>+<u,u>-2<w,u>
= <w,w> + <t*v,t*v> - 2<w,t*v>
= <w,w> + t^2 <v,v> - 2t <w,v>

In order to maximize that second-degree polynomial in t, we need to pick t such that the derivative is zero. The derivative is
2t <v,v> - 2 <w,v> = 0 => t=<w,v>/<v,v>

So u = t*v = (<w,v>/<v,v>)*v

With w=(3,2) and v=(1,-1), you get
u = (<(3,2),(1,-1)>/<(1,-1),(1,-1)>)*(1,-1) = 1/2*(1,-1) = (.5,-.5)

Instead of minimizing the function by finding the point where the derivative is zero, you can also use the theorem that says that u and u_perp are orthogonal. Doing it that way is easier if you are projecting onto a space of higher dimension.

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If you like Gilbert Strang's video lectures, you should check out his textbook, Introduction to Linear Algebra (ISBN 0-9614088-9-8). It's definitely one of the better (if not best) linear algebra books I've read. It starts from the basics and reads very much like a novel rather than a textbook, and provides an excellent foundation for thinking about problems in linear algebra.

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haha it just so happens i have this


http://en.wikipedia.org/wiki/Linear_algebra


haha im not quite old enouph to understand it,but you might

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