help: linear algebra

mikau · 2007-08-18 13:15:07

i don't get the proof of this one theorem. We're talking about how you can represent a linear mapping by a matrice. Like if you have a linear mappin f between two vector spaces f: V -> W, you can describe the mapping entirely by a matrice that relates the basis elements.

Anyway, the proof relates to two other thoerems. Theorem 7.3 and 6.5

theorem 7.3.
if the mappings f and g: V -> W, and A and B are the matrices that represent f and g respectivly, then f ° g is represented by A*B.

theorem 6.5:

and here's the theorem i don't get:

the underlined sentance is what I don't get at all. how did knowing that g ° f: V -> V (surjectively) show us that f was isomorphic?

I WAS able to prove it must be so using about a 12 line proof, but the book somehow see's it as obvious, and i'd LOVE to know why!

i should also mention that i'm not entirely certain what that little 'idv' stands for. I assumed it meant g°f(v) = v where v is an element of V.

Last edited by mikau (2007-08-18 13:18:15)

Ricky · 2007-08-19 00:44:23

g ° f is injective if and only if f is injective. The identity map is always a bijection.

Edit: If you've never heard of this before, it should be said that g ° f is surjective if and only if g is surjective.

mikau · 2007-08-19 06:24:05

hmm.. so we have that f is injective, but how do we know it is surjective?

Ricky · 2007-08-19 06:59:18

Theorem 6.5 directly says that an injective function is equivalent to a bijection.

mikau · 2007-08-19 07:03:02

IF it maps two vector spaces of equal dimension. How do we know that f(V) is of dimention n?

Ricky · 2007-08-19 07:58:53

Take another read at thm 6.5. It says nothing about f(V), rather, only about W. In this case, V = W. It is not the image of f that has to be of equal dimension, it is the codomain.

mikau · 2007-08-19 08:21:33

so since f(V) maps to a subspace of W which is of dimension n, (which may or not be W entirely) and since f(V) is injective, then the theorem applies?

would you say that underlined sentance was a bit non obvious? Or am I just having a premature senior moment?

Ricky · 2007-08-19 08:30:23

The statement you made is technically correct but very misleading and quite convoluted. I think you also may be missing the "gist" of the theorem as well. To meet the requirements of the theorem, it's just that f is a linear map from V to W where dim(V) = dim(W). But the way you phrased it:

so since f(V) maps to a subspace of W which is of dimension n, (which may or not be W entirely) and since f(V) is injective, then the theorem applies?

Since f(V) is injective, f(V) = W by the theorem. So it isn't "may or may not be W entirely", it must be W entirely. You just didn't know that before the theorem was given. But the theorem puts no restrictions on f(V) other than it is a linear map.

would you say that underlined sentance was a bit non obvious? Or am I just having a premature senior moment?

Perhaps. A good way to tell is to do the mini-proof in your head and if you don't feel like writing anything down, it's obvious.

f is a linear map from V to V, it must be injective since it's composition with g is injective, and so 6.5 applies and thus f is a bijection. But I didn't really feel like writing any of that.

mikau · 2007-08-19 09:38:05

well what i mean by 'which may not be W entirely' I meant that when we say V maps to W, it does not necessarily imply a surjective mapping, (W entirely) but since W itself is of dimension n it still works.

Correct?

What they never told me was that if g°f was injective then f is injective. Its obvious but it never occured to me.

Ricky · 2007-08-19 10:42:44

Still not a big fan of the wording, but it is correct.

What they never told me was that if g°f was injective then f is injective. Its obvious but it never occured to me.

That should be covered in an intro to proofs class right after set theory. Have you not taken one?

mikau · 2007-08-19 11:50:46

i've not taken either. Not in my curriculem. x_x maybe thats why this course is beating me up so bad.

i was looking at some other books on linear algebra, and some had a chapter deticated to doing proofs which seemed nice. This book is a mere 180 pages, plus a 20 page chapter on using Maple 7. I think its meant to be brief and concise, but i've always found short books to be horrible for self study.

lili · 2007-08-19 12:10:02

mikau wrote:

i've not taken either. Not in my curriculem. x_x maybe thats why this course is beating me up so bad.
i was looking at some other books on linear algebra, and some had a chapter deticated to doing proofs which seemed nice. This book is a mere 180 pages, plus a 20 page chapter on using Maple 7. I think its meant to be brief and concise, but i've always found short books to be horrible for self study.

hi hey wen are u going to be online.
u aren such asmarty tell ricky hi send me back a message

Ricky · 2007-08-19 14:02:00

Yes, they do that in my university as well. You can take Linear Algebra without taking an intro to proofs course. However, I think it is a big mistake to do so. Much of the time in linear algebra was spent going over how to do proofs anyways. So many students in my class were clueless on the easiest of problems. They didn't seem to get that there is no method for doing proofs, that there isn't a laundry list of steps you do to "solve" a problem.

It was actually to a point where I didn't go to a single class. I would hop on a bus, do my homework while riding it, and have it done before I got off to hand it in when the class got out. But then again, I had also taken an intro to abstract algebra course, so vector spaces and mappings were extremely easy to handle.

Edited to add: And set theory is just a part of a proofs course, typically. You can have classes on advanced set theory, but that stuff is really challenging.

mikau · 2007-08-19 15:30:57

So many students in my class were clueless on the easiest of problems. They didn't seem to get that there is no method for doing proofs, that there isn't a laundry list of steps you do to "solve" a problem.

clueless on the easiest problems pretty much describes my scenario. I WAS able to figure out that there is no direct approach. I noted there is a handful of proof styles (induction, contradiction, etc) and i also noticed that proofs usuaully exploit every condition of the IF clause, (otherwise they wouldn't be needed in the condition, right?) but i think i most often get stuck on proofs becasuse i don't understand the subject matter well enough. Which, I think, is because i'm teaching myself so i can't ask a teacher questsions as he explians, etc.

so your reccomendation would be to..say...get a book on an introduction to proofs?

here's a quick question. Can a vector space contain more than one of the same element?

Ricky · 2007-08-19 16:08:35

I've been finding mostly that I prefer contradiction. You get to reach any contradiction you want, and assume much more stuff. But that may just be me. Have you heard of contrapositive? If you wish to prove If A then B, it is enough to prove If not B, then not A.

And if you are able to understand the proof in this thread, even with difficulty, you are beyond what I was talking about.

but i think i most often get stuck on proofs becasuse i don't understand the subject matter well enough.

It's just experience, it takes time. What you should find is that the more problems you do the easier they become, the more you see the actual structure behind the math.

here's a quick question. Can a vector space contain more than one of the same element?

No, in the same sense that a set can not contain duplicates. When a collection does contain duplicates, it is called a multiset.

mikau · 2007-08-22 09:49:32

stuck again.

learning about determinants and this one line got me. Since they refer to some properties of determinants as they labeled them, I'll post them here.

okay, now they're showing me the following theorem. which uses this definition

here's the theorem

the first paragraph compeletely goes over my head. What do they mean depends linearly? They don't mean linearly dependant do they? (either way, i don't get it)

(and btw, i'm only asking for help with the first paragraph. So far... )

Last edited by mikau (2007-08-22 09:57:32)

mikau · 2007-08-23 04:59:52

it seems like a contradiction. It makes sense that D(Aij) is 'independant' of the jth collumn in the sense that it is not effected by it, becasue the jth collumn is deleted. But then it says 'so aijD(Aij) depends linearly on the jth colllumn of A.' first we say we can ignore the jth collumn, and then we say we are dependant on it. Even wierder they suddenly deduce this makes the addition and scaler properties for determinants true. O_o does this make ANY sense to anyone?

Ricky · 2007-08-23 06:38:59

Yes, I agree, some horrible wording.

Remember that D(A_ij) is a scalar, an element of F. So by "is independent of the jth column" I think they are saying that it's value is not affected by the jth column. Now it says that a_ijD(A_ij) depends linearly on the j-th column. To see this, set n as some number and write the summation out. I'm going to ignore the little details because they aren't really that important:

We see that this is a linear combination of the elements in the jth column.

mikau · 2007-08-23 07:16:48

isn't that actually a linear combination of the elements of the ith row?

mikau · 2007-08-24 06:08:12

hehehe, does your silence mean "oh dang, you're right, i'll let you know when i think of something" or "ugh...no, stupid! i shouldn't even have to explain it!" ?

is it not true that is actually a linear combination of the row elements? I mean its a scaler multiple of each item from a fixed row, and the collumns change.

Ricky · 2007-08-24 12:32:00

Add an 's' on to the very end of my post.

We see that this is a linear combination of the elements in the jth columns.\

Which is of course in total, the ith row.

mikau · 2007-08-24 13:59:30

ah, you mean, for each j. So we have then that the sum is equal to a linear combination of the elements in the ith row. So how exactly does this show that D1 and D2 hold?

sorry if I seem a bit blockheaded.

Ricky · 2007-08-25 04:40:26

Make sure you go through and write out explicitly what it is you want to show. Let the ith row and jth column of A be a + b. We wish to show that:

f_i(..., a + b, ...) = f_i(..., a, ...) + f_i(..., b, ...)

So now compute f_i(..., a+b, ...):

... + (a + b)D(A_ij) + ... = (... + aD(A1_ij) + ...) + (... + bD(A2_ij) + ...)

Remember that A1 and A2 are the matricies with a in the ith row and jth column, and b in the ith row and jth column. Keeping in mind that D itself is determinantal, the above statement shouldn't be too hard to prove.

George,Y · 2007-08-26 10:02:19

Believe me, Mikau. The Linear Algebra is a Jungle of many theorems, you need at least two professional - easy to read, but covering advanced topics like long proofs - textbooks to guide you walk out of it, if you just want to understand it instead of simply memorizing it.

But it may be worth it. Linear Algebra uses so many advanced proof techniques so after learning it, your logic sense and your IQ improve.

Ricky · 2007-08-26 10:28:11

Linear algebra, like any field of math, can become highly specialized. And it is a huge field, mostly used to solve problems in other fields. But an introductory book should be giving you the basic ideas behind things. The definition of a vector space, linear independence, basis, linear mappings, the 4 vector spaces of a matrix, representing linear maps with a matrix, and determinants. It is important to see the bigger picture, to understand what a proof is doing and why it's doing it. But it is also important to be able to work through each step of a proof with mathematical rigor. Having one without the other is pretty much useless.

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#1 2007-08-18 13:15:07

help: linear algebra

#2 2007-08-19 00:44:23

Re: help: linear algebra

#3 2007-08-19 06:24:05

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