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Need help understanding Linear algebra proofs (and linear algebra in general) |
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| Sep5-12, 01:40 AM | #1 |
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Need help understanding Linear algebra proofs (and linear algebra in general)
Hey all,
I am trying to get a head start on Linear Algebra before i start taking classes in a couple weeks. I am about to go into my second year undergraduate and all i have behind my belt is calculus (single varialbe, multivariable, and vector analysis (curl, divergence, etc)). I am having a ton of trouble wrapping my head around stuff like vector spaces. Spans are especially difficult for me to understand because i can't visualize them. I understand what is said when i read the book, but it doesn't help me with go further because of the way it's presented. I am using Axler's Linear Algebra Done Right to study from and here is an excerpt from chapter 2 about the lemme i can't understand. It is used for this proof How can I adjust my way of thinking as to not get bogged down by the the strange new world of linear algebra? It has new notation, new ideas, and new things that i simply can't understand!!! |
| Sep5-12, 03:50 AM | #2 |
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Hey Sysman and welcome to the forums.
The easiest way to think of it is as follows: a space will have some sort of basis. The basis will consist of so many linearly independent "vectors" and the right linear combination of said basis vectors can be used to describe any point in that space. Usually we look at bases like the standard ones (i,j,k in 3D) but these are not the only ways (they don't even have to be orthogonal but there are reasons why we use orthogonal bases). So the dimension of the space is equal to the minimum number of linearly independent vectors you need to describe the space. Now you can get a subset of all vectors in the vector space and create a "basis" for this set (i.e. reduce it to a set of vectors where all elements of our subset can be written as a linear combination of some of the vectors in the subset). You may end up finding that you have instances where you can construct a basis that has a less number of vectors than the basis for the whole space because a lot of the vectors are linearly dependent with the other ones. In the above case, this is why you get the inequality as opposed to the equality. To understand this as an example imagine you have (1,2,3,4) (2,4,6,8) (3,6,9,12): you know that all vectors in this space are linearly combinations of (1,2,3,4) so you only need one vector to describe the information in this set. But of course in the whole space of R^4 you need four linearly independent vectors, but you happen to have chosen a special subspace where they are all linearly dependent. |
| Sep5-12, 08:25 AM | #3 |
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If you have two vectors, then as long as those two vectors aren't collinear, the span is a plane through the origin. Every point on that plane is a linear combination of the two vectors. The span of a set of vectors defines a vector subspace--every vector that lies in the span is a linear combination of other vectors in the span. |
| Sep5-12, 08:27 AM | #4 |
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Need help understanding Linear algebra proofs (and linear algebra in general)
Is that the linear algebra done right that refuses to use determinants?
If so try swapping to Gilbert Strang's Introduction to Linear Algebra, it's a much nicer transition from no rigour to easy rigour. |
| Sep5-12, 03:20 PM | #5 |
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Recognitions:
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| Sep6-12, 12:47 AM | #6 |
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| Sep7-12, 10:51 AM | #7 |
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I have that (linear algebra done right) book too and I love it, but I have to admit that it is probably not suitable for some beginners in linear algebra.
On the other hand, the book by G. Strang is too wordy to my liking even though it covers many topics that one needs to learn on introductory level. |
| Sep7-12, 01:19 PM | #8 |
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one problem is your book, not written for a first course.
you might try: http://www.math.brown.edu/~treil/papers/LADW/book.pdf and maybe also: http://www.amazon.com/gp/offer-listi...condition=used |
| Sep7-12, 04:27 PM | #9 |
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| Sep14-12, 01:48 PM | #10 |
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Hey Sysman,
I took linear algebra a couple years ago but have a pretty good handle on it. Wanna add me on Skype and I'll try to answer any questions you have? My handle is Claymangs07. Best, Mariogs PS We used Knop's Linear Algebra: A First Course, and I thought it was great. |
| Sep16-12, 09:51 PM | #11 |
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