Help with linear algebra

Durandal · Post by **Durandal** » 2003-03-27 10:28pm

All right, I've revised the assignment. Just click the link again.

Kuroneko · Post by **Kuroneko** » 2003-03-27 10:51pm

Durandal wrote:The vector(s) removed from the set A U B form the basis for the intersection space V^W.

That's still wrong. Again, consider the analogous three-dimensional case

v1 = [1 0 0]; v2 = [0 1 0]; A = {v1, v2}; V = span(A)
u1 = [0 0 1]; u2 = [0 1 1]; B = {u1, u2}; W = span(B)

So, for example, you remove u2 from AUB since it is linearly dependent on the rest. However, u2 cannot span the intersection V^W, because it is completely independent of V!

Please consider using the my definition of z, on the assumption that u6 is dependent on the rest (this assumption can be made without loss of generality, since we can always rename things):

Code: Select all

u6 = c1u1 + c2u2 + c3u3 + c4u4 + c5u5; c's fixed scalar constants
z = c1u1 + c2u2 + c3u3

Feel free to find out why z is non-zero and in both V and W yourself, or just look at at one of my previous post where this is derived.

Durandal · Post by **Durandal** » 2003-03-27 11:52pm

Phew! Okay, I got it (again). You can check the link again, if you're still here. But thank you very much for your help.

Kuroneko · Post by **Kuroneko** » 2003-03-28 12:07am

Durandal wrote:Phew! Okay, I got it (again). You can check the link again, if you're still here. But thank you very much for your help.

No problem. I need to keep in practice anyway, so I'll be glad to help whenever I can. (Best to keep it in this thread, though.)

There is, however, one little issue:

Durandal wrote:However, since C contains six vectors but spans R5, ...

change "spans R5" to "spans a subspace of R5". There are degenerate cases like V = W, after all, in which case span(C) = V = W.

This one is more of a minor nitpick than an issue, though just for the sake of completeness:

Durandal wrote:where k1, k2, k3 is a set of scalars, not all zero.

Either "{k1,k2,k3} is a set of scalars..." or "k1, k2, k3 are scalars, ...".

Kuroneko · Post by **Kuroneko** » 2003-03-28 12:13am

By the way, what are you using to typeset? LaTeX?

Durandal · Post by **Durandal** » 2003-03-28 11:12am

Kuroneko wrote:No problem. I need to keep in practice anyway, so I'll be glad to help whenever I can. (Best to keep it in this thread, though.)

Sounds okay. What are you studying? Math?

There is, however, one little issue:
Durandal wrote:However, since C contains six vectors but spans R5, ...
change "spans R5" to "spans a subspace of R5". There are degenerate cases like V = W, after all, in which case span(C) = V = W.

Too late. I handed it in.

This one is more of a minor nitpick than an issue, though just for the sake of completeness:
Durandal wrote:where k1, k2, k3 is a set of scalars, not all zero.
Either "{k1,k2,k3} is a set of scalars..." or "k1, k2, k3 are scalars, ...".

Yeah, I noticed a few small errors right before I turned it in this morning, but nothing huge.

One of my classmates had an easier way of doing it, though. Basically, he started with my approach by combining the two sets and then stating that the reduced row echelon form would have at least one free variable in the parametric representation of its solution. I think that's what I kept trying to say, originally, but it was late, and I'd been working on the thing all day. My brain was slightly dead.

And yes, I use LaTeX for typesetting, specifically pdflatex. It's not required (considering that I'm an undergraduate), but I figured that I'd have to learn it eventually, so I might as well keep in practice with it.

RedImperator · Post by **RedImperator** » 2003-03-28 01:54pm

Seven. The answer is seven.

Queeb Salaron · Post by **Queeb Salaron** » 2003-03-28 04:28pm

RedImperator wrote:Seven. The answer is seven.

Psst! It's a proof! There's no numerical answer!

(At least I got that much right. Beyond that, I'm lost.)

Durandal · Post by **Durandal** » 2003-03-28 04:38pm

I wrote:Basically, he started with my approach by combining the two sets and then stating that the reduced row echelon form would have at least one free variable in the parametric representation of its solution.

I just looked at that again.

Jesus Christ I sound like a fucking geek.

phongn · Post by **phongn** » 2003-03-28 05:19pm

Gack, linear algebra. The course I took was more based on MATLAB, so they more or less had us writing code to solve linear algebra in addition to teaching us the pure math

Post by **Darth Wong** » 2003-03-28 05:20pm

I'm an engineer. We like calculus. We no like this linear algebra shit

Durandal · Post by **Durandal** » 2003-03-28 05:47pm

Darth Wong wrote:I'm an engineer. We like calculus. We no like this linear algebra shit

Economists love it.

Phong: Sounds like my multivariable calculus course. We did Maple labs every week.

Arthur_Tuxedo · Post by **Arthur_Tuxedo** » 2003-03-28 06:00pm

Durandal wrote: Economists love it.

Phong: Sounds like my multivariable calculus course. We did Maple labs every week.

I'm an economist, and I hate that shit

Kuroneko · Post by **Kuroneko** » 2003-03-28 07:43pm

Durandal wrote:Sounds okay. What are you studying? Math?

Yes, mathematics is my main area of interest.

Durandal wrote:Basically, he started with my approach by combining the two sets and then stating that the reduced row echelon form would have at least one free variable in the parametric representation of its solution.

In that case, it would probably be better to start with the parametric forms of the subspaces instead of their bases. I hope he was successful, as it would indeed be nice way to do it.