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05102016, 08:16 AM  #2647837 / #1 
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uncool actual school on linear algebra
So the last thread, I accidentally turned what was supposed to be a "school" on linear algebra into an exploration of linear algebra through the lens of category theory. Which was useful to me, but probably less useful to the people who were watching the thread. This thread is for those people.
So! Linear algebra. What is linear algebra? Linear algebra is the study of what you can do when you are restricted to two basic operations  adding things together, and multiplying everything by a number. Seems restrictive, right? Well, that's a lot of how math works  by restricting what you think you can do, you can make more situations similar  and so you can study things much more deeply. The thing is, we actually haven't restricted ourselves all too much. Or rather, we've restricted ourselves to a pair of very useful things we can do. Don't believe me? Every day, Alice buys two apples and one banana for breakfast. Every day, Bob buys one apple and two bananas for breakfast. Last month, Alice paid $5, while Bob paid $4. This month, Alice paid $5, while Bob paid $7. How did the prices change? Let the price of an apple be a dollars, and the price of a banana be b dollars. We then get the equations for last month: 2a + b = 5 a + 2b = 4 So let's solve. Double the second equation  multiply all the numbers in it by 2. That still gives us a valid equation. The cost of 2 apples and 4 bananas would be 8 dollars. 2a + b = 5 2a + 4b = 8 Now, if it cost 5 dollars to get 2 apples and a banana, and 8 dollars to get 2 apples and 4 bananas, it stands to reason (assuming no discounts) that the difference in cost comes from the difference in goods. So we can subtract the first equation from the second, to get: 2a + b = 5 3b = 3 Again assuming no discounts, if 3 bananas cost $3, then one banana would cost $1. 2a + b = 5 b = 1 Finally, we can remove bananas from the equation and divide by 2 to find that an apple cost $2 last month. 2a = 4 b = 1 a = 2 b = 1 Notice that the only things we did were listed above  we multiplied (or divided) by numbers, and then we added (or subtracted) two things. OK. Now let's do the calculation for this month. 2a + b = 5 a + 2b = 7 2a + b = 5 2a + 4b = 14 2a + b = 5 3b = 9 2a + b = 5 b = 3 2a = 2 b = 3 a = 1 b = 3 We have found that the new prices are $1 for an apple, and $3 for a banana. But one thing to notice  we in fact did the exact same calculations as we did before, just with different numbers on the right. This will be important later. Hopefully I have already convinced you that we already use linear equations commonly. Next up, the process of solving generally, and some notation. 
05102016, 04:06 PM  #2647954 / #2 
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OK. That was easy. I have a feeling *easy* will not last much longer.
Bring it on!
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RICHARD P. FEYNMAN Some remarks on science, pseudoscience, and learning how to not fool yourself. Caltech’s 1974 commencement address. If you’re doing an experiment, you should report everything that you think might make it invalid—not only what you think is right about it: other causes that could possibly explain your results. Details that could throw doubt on your interpretation must be given, if you know them. You must do the best you can—if you know anything at all wrong, or possibly wrong—to explain it. 
05102016, 09:29 PM  #2648202 / #4 
I did. F. Poste.
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cool uncool
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05142016, 04:34 AM  #2649933 / #5 
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This is a bit different from the solution path I was taught. I'm used to solving for one variable in terms of the other in one equation and substituting into the other equation.
So a + 2b = 4 resolves to a = 42b Then substituting this for a into the other equation: 2(42b) +b =5 yields b=1 And so forth, same results... ETA: Oh yeah, "Bite me Heinz" you socalled superior member! Heh. Last edited by MikeB; 05142016 at 04:38 AM. 
05142016, 08:49 AM  #2649935 / #6  
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Quote:


05172016, 09:25 PM  #2650804 / #7  
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Quote:
Traditional algebra is intraformula solving; Linear algebra is interformula solving
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05172016, 09:26 PM  #2650806 / #8 
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Are you going to explain eigenvalues so they make sense? Those never made sense to me. That was a case of me memorizing the steps and plugging in the values.
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05182016, 01:49 PM  #2650988 / #9 
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I don't want to derail uncool's thread when it has barely gotten started, so I'll just note that this
...is not true, and I'll bite my tongue regarding the rest of my objections.
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05182016, 05:48 PM  #2651172 / #10  
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Quote:
Of course, it's also been a very long time since I did anything more than a pretty straightforward matrix solution. Hell, it's been about 7 years since last had to do anything more complicated than spell determinant. So there's a pretty good chance that I've brain dumped this bit!
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05182016, 05:54 PM  #2651183 / #11 
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Generally, the scope of any given subfield of mathematics is defined by its subject matter, not by any artificial constraints on which techniques you're allowed to use.
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05182016, 06:01 PM  #2651189 / #12 
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Okay then. Whatever. Don't really care that much
ETA: Apparently I care a little bit. If you were pressed to make a general statement, how would you describe the difference between "regular" algebra and Linear Algebra?
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05182016, 06:53 PM  #2651214 / #13 
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"Regular algebra" isn't a term I use.
Linear Algebra is about what you can do with vector spaces.
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05182016, 07:10 PM  #2651221 / #14 
I did. F. Poste.
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Why "linear"?
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05182016, 07:49 PM  #2651236 / #15 
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Couldn't come up with a better term. When I took the class, it was just "Algebra". But that didn't seem sufficient to distinguish it from Linear algebra, so I winged it.
And that means what? How are they different?
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05182016, 07:55 PM  #2651239 / #16 
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Well, the variables in linear algebra can be vectors and matrices. So you can solve y=bx, where y and x are matrices.
amirite?
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05182016, 09:28 PM  #2651304 / #18 
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Good (and well timed) question.
My description was perhaps too broad, since there are things you can do with vector spaces that go outside the scope of Linear Algebra. But suppose the only functions of vectors that we consider are those having the properties f(u+v) = f(u) + f(v) for any vectors u and v; f(cv) = cf(v) for any vector v and any scalar c. (This is what we call a "linear function".) That's a rather restricted set of functions, but it turns out that there's a tremendous amount of stuff to learn about them, so they're worthy of having entire courses dedicated to them. That's Linear Algebra.
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05182016, 09:33 PM  #2651313 / #19 
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That's what this whole thread will be about. uncool has plans for how to present it, and I expect xe will do a better job than I would.
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05182016, 09:46 PM  #2651334 / #20  
I did. F. Poste.
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Quote:
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05182016, 10:59 PM  #2651405 / #21 
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I'd say that linear algebra is the study of what you can do with just the two operations named before: addition and scalar multiplication. "Regular" algebra, to me, would include linear algebra  but also allow more operations.

05212016, 12:34 AM  #2652641 / #23  
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Quote:
and to go further on why "linear", one can think of the linearity property BD defined as an abstraction of the elementary equation for a line: y = mx + b. in fact, I suspect (though don't know) that is the origin of the term. Last edited by el jefe; 05212016 at 12:37 AM. 

05212016, 08:19 AM  #2652677 / #24 
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Well solving linear (or linearized) equations is mostly what I use it for
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