anyone good at linear algebra?

[sanantoniosooner]

I may need a little assistance.

[yermom]

maybe

:les: Let 'Er Rip

[Ike]

ummm...its easier than normal algebra.

[Stanley1]

Maybe.

[yermom]

it may be easier than Higher Algebra...

i'll say Linear Algebra is deceptively named

[crawfish]

How hard can it be? It's a line.

[yermom]

exactly

[IronSooner]

Here's a hint:

y = mx + b

Just move some of those letters around as needed. YWIA.

[yermom]

this was fun:

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

i barely remember it now, but we used it in Physical Math too

[Vaevictis]

What do you need to know wrt linear algebra?

[Ike]

this was fun:

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

i barely remember it now, but we used it in Physical Math too


thats like, 90% of quantum mechanics right there.

[Al Gore]

Just guess the answer like I do.........It's ok as long as it's close.......

[sanantoniosooner]

sample test (http://www.math.utsa.edu/~feng/courses/linear_algebra/e2-sample.pdf)

I've got a lot of cramming to do the next few days.

[TUSooner]

'Zat like numbers AND letters together?
Cain't hep ya.

[Penguin]

Diagonalizable? Hell, I can't even pronounce diagonalizable.



Can't help you there, bub.

[Howzit]

Isn't linear algebra solving matrices?

Oh, yeah, it is. You're about 20 years too late for me to help...

[Al Gore]

Pay someone to take your test for you.....

[OzarkSooner]

nope

[slickdawg]

Slickwife can tear the *** outta this stuff, she's a smart gal.
She's asleep right now too.

[Vaevictis]

I can remember how to do all the stuff with linear algebra that's really useful for solving linear systems (as it's useful in my field), but all the stuff on basis math left my head as soon as I completed the final exam.

[sanantoniosooner]

Slickwife can tear the *** outta this stuff, she's a smart gal.
She's asleep right now too.
show it to her tomorrow

[Al Gore]

I thought your business involved bushes, flowers and stuff.....Why the algebra???

[sanantoniosooner]

I thought your business involved bushes, flowers and stuff.....Why the algebra???
I'm switching careers.

I should have never tried to cut down the forrest with a herring.

[slickdawg]

show it to her tomorrow


will do. She's got a physical oceanography class in the am, then she'll be out
in the afternoon. I'll ask then.

[MiccoMacey]

Pretty simple stuff, actually.

The premier basis of linear algebra is Chinese synch mathematics.

It was developed by some Buddhist monks in the 12th century.

If you can visualize, all numbers and powers of all finite and infinite vectors can be, at some point, both systematically as well as geometrically joined by their opposite counterpoint on a "linear" line graph.

To wit: (and this juxtaposition is where it makes it's money):

Any representation of a valid and joined system can and will be a dual host for both members in any equal or bilateral subset of either vector on a geographical plane.

[sanantoniosooner]

Pretty simple stuff, actually.

The premier basis of linear algebra is Chinese synch mathematics.

It was developed by some Buddhist monks in the 12th century.

If you can visualize, all numbers and powers of all finite and infinite vectors can be, at some point, both systematically as well as geometrically joined by their opposite counterpoint on a "linear" line graph.

To wit: (and this juxtaposition is where it makes it's money):

Any representation of a valid and joined system can and will be a dual host for both members in any equal or bilateral subset of either vector on a geographical plane.
put yer money where yer mouth is (http://www.math.utsa.edu/~feng/courses/linear_algebra/e2-sample.pdf)

[GottaHavePride]

sample test (http://www.math.utsa.edu/%7Efeng/courses/linear_algebra/e2-sample.pdf)

I've got a lot of cramming to do the next few days.

I could have done all of that really fast about 3 years ago. Not so much now. Don't ask me about my "Applications of Modern Algebra" class - I flunked that one.

[proud gonzo]

i hate matrices.

[proud gonzo]

I could have done all of that really fast about 3 years ago. Not so much now. Don't ask me about my "Applications of Modern Algebra" class - I flunked that one.
did they teach you how to use all the games on your calculator? :confused:

[MiccoMacey]

Dude, what is that... a test?

He can either accept my help or not.

[sanantoniosooner]

Dude, what is that... a test?

He can either accept my help or not.
it's a sample.

[GottaHavePride]

did they teach you how to use all the games on your calculator? :confused:

Judging from my grade, they didn't teach me a damn thing. ;)

[Penguin]

i hate matrices.

I hate mattresses, too. I'd rather sleep on the floor.

[MiccoMacey]

Forgot who the author of the thread was..thought it was someone else.

Check your spek...just finished #1.

[MiccoMacey]

The problem with linear algebra, as with most non-rational and organized charts, is that eventually there will be a formula of any division that cannot regroup itself in the same pattern as its original offset.

[sanantoniosooner]

If you don't stop googling I'll kick your butt :D

[proud gonzo]

hey, NO HABLAN "NERDO"

[Sooner Born Sooner Bred]

nerds. ;)

[Beano's Fourth Chin]

Wait until you get to spherical algebra.

[MiccoMacey]

MiccoMacey = Rubik's cube champion circa 1984.

I see patterns really well. Which is why linear algebra is kind of easy for me.

"I just see it".

Look, basically there are two perpendicular blocs that can be paralleled on any given finite plane. If these two intercede at any rate, the resulting projection has the same rate as its original AND counteropposing vector on its inverse plane.

[BajaOklahoma]

SAS, how soon do you need this?
My younger son may know how to do this, but he is in the middle of finals this week. I think he is done with them on Friday.

[sanantoniosooner]

this thread is WAY too long for the amount of actual assistance I've received.

[sanantoniosooner]

SAS, how soon do you need this?
My younger son may know how to do this, but he is in the middle of finals this week. I think he is done with them on Friday.
that is a sample test.

the actual test is thursday.

[Vaevictis]

Been awhile on this, but on the first problem, can't you just do a piece-wise dot product (ie, dot product the last one against each of the others), set each equation to zero, and solve the linear system?

[MiccoMacey]

Been awhile on this, but on the first problem, can't you just do a piece-wise dot product (ie, dot product the last one against each of the others), set each equation to zero, and solve the linear system?

Can't.

If each product is a correlation to the inverse of its vector, a piece-wise dot product won't show any relationship on a geographical plane.

So if you set each equation to zero, you'll end up with a subset of each finite product not in its own system.

[Ike]

sample test (http://www.math.utsa.edu/~feng/courses/linear_algebra/e2-sample.pdf)

I've got a lot of cramming to do the next few days.

1) require that the determinant of the coefficent matrix is non-zero. any choice of beta that does that will make those vectors basis vectors.

2) you can answer the second part first. the dimension must be 3, because that equation is essentially a 3-dimensional "plane" that exists in a 4 dimensional space. therefore, any basis of that subspace must have dimension 3. for the other part, I'd have to look that up...
3)the exact answers aren't obvious, but the short version is this. this matrix projects a 5 dimensional space onto a 4 dimensional space. the kernel has 2 answers then, the null vector, and the nullspace of A. The nullspace of A is the solution space of the set of equations that satisfy A*vec(x)=vec(0)
The image, or range of A is the column space, or the set of vectors spanned by the column vectors of A.
4) let T be the coefficent matrix of the 3 new basis vectors (3 5 1 / 5 3 0 / 0 0 1). The transformation A expressed in this basis is Inv(T)AT.
5) there is exactly 1
6) (1*4*27)-(1*9*8)+(2*9*1)-(2*1*27)+(3*1*8)-(3*4*1)=108-72+18-54+24-12 =12
7)form the coefficent matrix of these vectors, find the determinant. if its nonzero, they are linearly independent
8)Assume A is invertable. Show that this is equivalent to det(A)!=0 (ie, det(Inv(A))=1/det(A) )
then show that this is equivalent to A*vec(x)=vec(0) having only the trivial solution.
then A*vec(x)=vec(0) can be written as x1*vec(c1)+x2*vec(c2)*....xn*vec(cn) = vec(0) if and only if the vectors c1, c2, ... , cn are linearly independent.
9) and 10)
use the equation det(cI-A)=0. If this has a solution, then the matix is diagonalizable. solve this equation for c to determine eigenvalues. to find the eigenvectors, plug each eigenvalue found into the equation (cI-A)vec(x)=vec(0), and solve for the vector x.
once the eigenvalues are found, the diagonal matrix D can be constructed by hand by placing the eigenvalues along the diagonal. the transformation matrix S can be constructed from the inner products of the eigenvectors where S_ij =<v_i|v_j>
I didnt really want to do the math for those last 2.

Anyway, I hope this helps.

[sanantoniosooner]

good stuff

[MiccoMacey]

1) require that the determinant of the coefficent matrix is non-zero. any choice of beta that does that will make those vectors basis vectors.

2) you can answer the second part first. the dimension must be 3, because that equation is essentially a 3-dimensional "plane" that exists in a 4 dimensional space. therefore, any basis of that subspace must have dimension 3. for the other part, I'd have to look that up...
3)the exact answers aren't obvious, but the short version is this. this matrix projects a 5 dimensional space onto a 4 dimensional space. the kernel has 2 answers then, the null vector, and the nullspace of A. The nullspace of A is the solution space of the set of equations that satisfy A*vec(x)=vec(0)
The image, or range of A is the column space, or the set of vectors spanned by the column vectors of A.
4) let T be the coefficent matrix of the 3 new basis vectors (3 5 1 / 5 3 0 / 0 0 1). The transformation A expressed in this basis is Inv(T)AT.
5) there is exactly 1
6) (1*4*27)-(1*9*8)+(2*9*1)-(2*1*27)+(3*1*8)-(3*4*1)=108-72+18-54+24-12 =12
7)form the coefficent matrix of these vectors, find the determinant. if its nonzero, they are linearly independent
8)Assume A is invertable. Show that this is equivalent to det(A)!=0 (ie, det(Inv(A))=1/det(A) )
then show that this is equivalent to A*vec(x)=vec(0) having only the trivial solution.
then A*vec(x)=vec(0) can be written as x1*vec(c1)+x2*vec(c2)*....xn*vec(cn) = vec(0) if and only if the vectors c1, c2, ... , cn are linearly independent.
9) and 10)
use the equation det(cI-A)=0. If this has a solution, then the matix is diagonalizable. solve this equation for c to determine eigenvalues. to find the eigenvectors, plug each eigenvalue found into the equation (cI-A)vec(x)=vec(0), and solve for the vector x.
once the eigenvalues are found, the diagonal matrix D can be constructed by hand by placing the eigenvalues along the diagonal. the transformation matrix S can be constructed from the inner products of the eigenvectors where S_ij =<v_i|v_j>
I didnt really want to do the math for those last 2.

Anyway, I hope this helps.

THAT'S what I've been sayin' all along.

Listen up people, class is in session.

Ike and Micco are gonna learn you on this here algebra.

[Ike]

THAT'S what I've been sayin' all along.

Listen up people, class is in session.

Ike and Micco are gonna learn you on this here algebra.


heh. some of these I haven't had to use in a loooong time, and had to look em up. but it came back pretty easy when I did. the last 2 though, those are things I am intimately familiar with.

except for the part about A^100. I have no idea what that means.

[MiccoMacey]

Two was the hardest for me.

But I haven't used my abacus in awhile. :D

[Ike]

Two was the hardest for me.

But I haven't used my abacus in awhile. :D

heh. thats why I skipped that one really, and only answered part b. the first part I left as an excercise to the reader :D

[MiccoMacey]

We can't do his homework for him.

How will he learn.

Give a man a linear algebra equation, and he'll pass a test for one day.

Teach him a linear algebra equation, and he'll do math for life. :D

[Ike]

lol

[skeeterou]

I can't do this stuff, but I'm really good at crosswords and soduku. BOOYAH

[skeeterou]

nerds. ;)
http://i49.photobucket.com/albums/f258/skeeterou/ogre2.jpg

[MiccoMacey]

Interestingly enough, sudoku is an inverted form of linear algebra.

If each co-efficient of each column is set at any finite power or rational number, there will be a resulting counter power or rational number in each row or square.

The trick is to find the first co-efficient (usually sudoku makers use the number 1 because they aren't willing to put in more effort). If the inverse of each corresponding effort of that number 1 is diagonalized, the resulting row will have a proportioned amount equal to its pre-ordinate subset.

Hence, it's "inverse" linear algebra.

You are wiser than you know, young padwan. :D

[skeeterou]

Interestingly enough, sudoku is an inverted form of linear algebra.

If each co-efficient of each column is set at any finite power or rational number, there will be a resulting counter power or rational number in each row or square.

The trick is to find the first co-efficient (usually sudoku makers use the number 1 because they aren't willing to put in more effort). If the inverse of each corresponding effort of that number 1 is diagonalized, the resulting row will have a proportioned amount equal to its pre-ordinate subset.

Hence, it's "inverse" linear algebra.

You are wiser than you know, young padwan. :D


wow i didn't know that's what I was doing in my head. cool, my brain is more powerful than I ever imagined. I like to use a pen to fill em out cuz it makes me think harder.

[Beano's Fourth Chin]

Interestingly enough, sudoku is an inverted form of linear algebra.

If each co-efficient of each column is set at any finite power or rational number, there will be a resulting counter power or rational number in each row or square.

The trick is to find the first co-efficient (usually sudoku makers use the number 1 because they aren't willing to put in more effort). If the inverse of each corresponding effort of that number 1 is diagonalized, the resulting row will have a proportioned amount equal to its pre-ordinate subset.

Hence, it's "inverse" linear algebra.

You are wiser than you know, young padwan. :D

pre-ordinate=ipodnat Most would write that as: "tan ipod", but that's simply a matter of preference.

[MiccoMacey]

...my brain is more powerful than I ever imagined.

"People think we only use ten percent of our brains. I think we only use ten percent of our hearts".

[Beano's Fourth Chin]

http://getyourtech.biz/images/piel%20frama%20ipod%20mini%20tan.jpg

[MiccoMacey]

pre-ordinate=ipodnat Most would write that as: "tan ipod", but that's simply a matter of preference.

Tan ipod is cool if you're using a sine or co-sine efficient. If you're using a base with more than one dimension or with more than one counter position, i.e. sudoku or any geometrically designed puzzle with an inverse correlation, you need a pre-ordinate subset to rationalize your first real power or finite number.

That's basic stuff, Sally. :D

[Beano's Fourth Chin]

I think we finally figured out who Micco is:

http://www.studio360.org/images/math/hunting1.jpg

[MiccoMacey]

My beard isn't as white.

[Vaevictis]

Can't.

Ah hah. That's right. That trick only works if you're looking for an orthonormal basis, yes?

[Ike]

Ah hah. That's right. That trick only works if you're looking for an orthonormal basis, yes?
orthogonal.

[Vaevictis]

orthogonal.

Yeah, but orthonormal basis implies orthogonal vectors, yes? :)

[Vaevictis]

My beard isn't as white.

http://img145.imageshack.us/img145/4290/rainman3id.jpg (http://imageshack.us)

Guy on the right? :D

[MiccoMacey]

On the second subset, yes. For any rational power or number to equal a resulting or inverse population, the basis for an orthonormal axis will run parallel to its pre-existant and coordinating factor. This works in this case because it is an exact co-efficient of its original power, hence an orthonormal basis will unilaterally be biased against any non-diagonalized matrices.

On the first subset, no. And for the reasons I already stated.

[MiccoMacey]

http://img145.imageshack.us/img145/4290/rainman3id.jpg (http://imageshack.us)

Guy on the right? :D

KMart sucks. :D

[MiccoMacey]

orthogonal.

You say orthogonal.
I say orthonormal.
Orthogonal.
Orthonormal.
Let's call the whole thing off. :D

[Ike]

Yeah, but orthonormal basis implies orthogonal vectors, yes? :)

yes. but orthonormal is a subset of orthogonal. so your dot product method works for all orthogonal bases. The orthonormal bases are just the normalized subset of those.

[SicEmBaylor]

Well, I'd help you if I could. However, I am absolutely horrible at math and know next to nothing of any practical use which is why I became a political scientist.

[sanantoniosooner]

except for the part about A^100. I have no idea what that means.
It has something to do with using a transitional matrix......along the lines of

B=S-1A100S (those ought to be superscript)

When you square, or raise the matrix to any power, all of the S inverse and S stuff cancels except the first and last leaving the A to whatever power you are raising.

I know that's the plan, I don't know if the execution is as easy.

[sanantoniosooner]

Someone gave my plea for help ahttp://www.soonerfans.com/forums/images/rating/rating_1.gif.

sucker.

[slickdawg]

Slickwife hasn't had a chance to look at this in detail, her physical oceanography class is very time consuming,