uncool actual school on linear algebra

uncool · 05-10-2016, 07:16 AM

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.

Heinz Hershold · 05-10-2016, 03:06 PM

OK. That was easy. I have a feeling *easy* will not last much longer.

Bring it on!

Testycalibrated · 05-10-2016, 04:43 PM

Wow. Nice job.

Pingu · 05-10-2016, 08:29 PM

cool uncool

MikeB · 05-14-2016, 03:34 AM

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 = 4-2b

Then substituting this for a into the other equation:

2(4-2b) +b =5 yields b=1

And so forth, same results...

ETA: Oh yeah, "Bite me Heinz" you so-called superior member! Heh.

uncool · 05-14-2016, 07:49 AM

:

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 = 4-2b

Then substituting this for a into the other equation:

2(4-2b) +b =5 yields b=1

And so forth, same results...

ETA: Oh yeah, "Bite me Heinz" you so-called superior member! Heh.

Yup, both good ways to do it - and both are widely taught. Linear algebra does it a slightly different way - in fact, it standardizes it, and that's what the next post will be.

Pandora · 05-17-2016, 08:25 PM

:

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 = 4-2b

Then substituting this for a into the other equation:

2(4-2b) +b =5 yields b=1

And so forth, same results...

ETA: Oh yeah, "Bite me Heinz" you so-called superior member! Heh.

My brain is insisting that your approach is the standard algebraic way to go about it - but it involves moving things from one side of the equation to the other. It involves rearranging things. Linear algebra doesn't let you rearrange things - you manipulate the entire formula, and you work across formulas rather than within.

Traditional algebra is intra-formula solving; Linear algebra is inter-formula solving

Pandora · 05-17-2016, 08:26 PM

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.

Brother Daniel · 05-18-2016, 12:49 PM

I don't want to derail uncool's thread when it has barely gotten started, so I'll just note that this

:

Linear algebra doesn't let you rearrange things

...is not true, and I'll bite my tongue regarding the rest of my objections.

Pandora · 05-18-2016, 04:48 PM

:

I don't want to derail uncool's thread when it has barely gotten started, so I'll just note that this

:

Linear algebra doesn't let you rearrange things

...is not true, and I'll bite my tongue regarding the rest of my objections.

We could get all sorts of techincal with that, but generally, teh shape of the formulas stay the same. If you start with ax+by=cz, you don't really shift to by = cz-ax. Sometimes, but the whole point of a matrix is kind of that you're not moving from one side of the equality to the other, you're manipulating the formulas as a whole entity in and of themselves.

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!

Brother Daniel · 05-18-2016, 04:54 PM

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.

Pandora · 05-18-2016, 05:01 PM

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?

Brother Daniel · 05-18-2016, 05:53 PM

"Regular algebra" isn't a term I use.

Linear Algebra is about what you can do with vector spaces.

Pingu · 05-18-2016, 06:10 PM

Why "linear"?

Pandora · 05-18-2016, 06:49 PM

:

"Regular algebra" isn't a term I use.

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.

:

Linear Algebra is about what you can do with vector spaces.

And that means what? How are they different?

Pingu · 05-18-2016, 06:55 PM

Well, the variables in linear algebra can be vectors and matrices. So you can solve y=bx, where y and x are matrices.

amirite?

Mike PSS · 05-18-2016, 07:07 PM

Brother Daniel · 05-18-2016, 08:28 PM

:

Why "linear"?

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.

Brother Daniel · 05-18-2016, 08:33 PM

:

And that means what?

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.

Pingu · 05-18-2016, 08:46 PM

:

Why "linear"?

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.

OK, thanks

uncool · 05-18-2016, 09:59 PM

:

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?

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.

Pandora · 05-20-2016, 06:53 PM

:

And that means what?

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.

Fair enough

el jefe · 05-20-2016, 11:34 PM

:

Why "linear"?

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.

OK, thanks

one might add that linearity turns out to be an enormously useful property for simplifying and facilitating calculations and problem-solving in all kinds of contexts. furthermore, linear approximations are a standard and indispensible tool for studying an even broader class of problems. that's why so much studying was done on it in the first place, and hence why there is so much to learn about linear functions/equations/etc. people have developed a lot of mathematical machinery to take full advantage of linearity, everywhere it comes up.

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.

Pingu · 05-21-2016, 07:19 AM

Well solving linear (or linearized) equations is mostly what I use it for

05-10-2016, 07:16 AM	#2647837 / #1
uncool Senior Member : Jul 2008 : 15,106	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.

05-10-2016, 03:06 PM	#2647954 / #2
Heinz Hershold Superior Member : Sep 2011 : 2,727	OK. That was easy. I have a feeling easy will not last much longer. Bring it on! __________________ 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.

05-10-2016, 08:29 PM	#2648202 / #4
Pingu I did. F. Poste. GLaDOS : Mar 2008 : UK : 60,846	cool uncool __________________ Join me at The Skeptical Zone

05-14-2016, 03:34 AM	#2649933 / #5
MikeB Senior Member : Apr 2010 : So. Cal. USA : 322	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 = 4-2b Then substituting this for a into the other equation: 2(4-2b) +b =5 yields b=1 And so forth, same results... ETA: Oh yeah, "Bite me Heinz" you so-called superior member! Heh. Last edited by MikeB; 05-14-2016 at 03:38 AM.

05-17-2016, 08:26 PM	#2650806 / #8
Pandora Betty Beep Boop : Oct 2008 : 14,480	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. __________________ *This is going to hurt me a lot more than it hurts you.*

05-18-2016, 04:54 PM	#2651183 / #11
Brother Daniel predisposition to antagonism Kodos the Executioner Mod: HASH, Philo, Math, THC : Jun 2008 : on the shore of the North Atlantic : 15,670	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. __________________ Requiem: Everything we humans do is fully deserving of ridicule and mockery. Without a God to laugh, it falls to us enlightened fools to make up the difference.

05-18-2016, 05:01 PM	#2651189 / #12
Pandora Betty Beep Boop : Oct 2008 : 14,480	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? __________________ *This is going to hurt me a lot more than it hurts you.*

05-18-2016, 05:53 PM	#2651214 / #13
Brother Daniel predisposition to antagonism Kodos the Executioner Mod: HASH, Philo, Math, THC : Jun 2008 : on the shore of the North Atlantic : 15,670	"Regular algebra" isn't a term I use. Linear Algebra is about what you can do with vector spaces. __________________ Requiem: Everything we humans do is fully deserving of ridicule and mockery. Without a God to laugh, it falls to us enlightened fools to make up the difference.

05-18-2016, 06:10 PM	#2651221 / #14
Pingu I did. F. Poste. GLaDOS : Mar 2008 : UK : 60,846	Why "linear"? __________________ Join me at The Skeptical Zone

05-18-2016, 06:55 PM	#2651239 / #16
Pingu I did. F. Poste. GLaDOS : Mar 2008 : UK : 60,846	Well, the variables in linear algebra can be vectors and matrices. So you can solve y=bx, where y and x are matrices. amirite? __________________ Join me at The Skeptical Zone

05-18-2016, 07:07 PM	#2651244 / #17
Mike PSS Senior Member : Mar 2008 : Florida : 11,478

05-21-2016, 07:19 AM	#2652677 / #24
Pingu I did. F. Poste. GLaDOS : Mar 2008 : UK : 60,846	Well solving linear (or linearized) equations is mostly what I use it for __________________ Join me at The Skeptical Zone

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