Question

Linear ALGEBRA!!

Answer 1

There's the problem.

Answer 2

First of all to show that a set of vectors is a basis for a vector space, you must show that
1) It spans the vector space
2) It is linearly independent

Answer 3

its just mindless row reductions

Answer 4

well I mean how can I show that it spans the vector space.. First of all doesn't B already span R^3?

Answer 5

Two show that a set of vectors spans the vector space, you need to show that any vector can be expressed as a linear combination of the set.

Answer 6

It's pretty easy to show that for B..

a(1) + b(x) + c(x^2) + d(x^3)

can describe any polynomial of 3 degrees or less with different values of a, b, c, and d.

Answer 7

B' just requires a bit of algebra to re-write it in this same form.

Answer 8

Okay.. So what would the notation be? This looks really easy but the notation is mad confusing!

Answer 9

"Two show that a set of vectors spans the vector space, you need to show that any vector can be expressed as a linear combination of the set."

^ and to do that means you have to set up a matrix, row reduce it and show that it has a solutions for all values of the parameters.

Answer 10

Its one of the things that involves so much typing, its very hard to help over the net with these sorts of questions

Answer 11

I dont see any matrices.. or do I use this one:
[ 1 0 0 0]
| 0 1 0 0]
0 0 1 0
0 0 0 1

Answer 12

elec, I'm pretty sure I showed it, barring some fancy set notation defining a, b, c, and d. No matrices required.

Answer 13

Just for part A right now. If you can demonstrate the dimension of the vector space, for a set of vectors to be a basis it must simply have the same cardinality as the dimension of the vector space and be linearly independent.

Answer 14

Row reduction will help you establish linear independence

Answer 15

If it reduces to the identity matrix its linearly independent

Answer 16

Okay hold on. Which matrix do I reduce?

Answer 17

Set up a matrix with each basis element as a column, then reduce.

Answer 18

I think its the language of linear algebra, that the asker is having trouble with.

Answer 19

Yes, I mean I can see that u can prove that:
ax^3+bx^2+cx+d is spanned or whatever they call it by {1,x,x^2,x^3} isn't that like the standard basis.. but how do I prove that?

Answer 20

\[ax^2 +bx^2 +cx +d = \lambda (1) + \theta (x) + \rho(x^2) + \beta(x^3) \]

Answer 21

compare coeffiecents

Answer 22

lyamda =a
theta = b
rho = c
beta = d

Answer 23

its kinda obvious.

Answer 24

Remember that vector spaces are closed under addition and scalar multiplication.

Answer 25

If x^2 is in the set, it can produce ax^2 for any a.

Answer 26

I just saw how my professor proved the first one, he basically said that for:
ax^3+bx^2+cx+d = 0 for all x and for all a,b,c,d that are real numbers, the only way that can be true is if:
a=b=c=d=0. Is there a similar proof for part 2?

Answer 27

Okay alchemista never mind that. How can I prove that:
a) B is linearly independent
b) span(B) = V

Answer 28

http://www.twiddla.com/572800
You can draw here..

Answer 29

The basic idea of linear independence is that you cannot express one vector as linear combination of the others.

Answer 30

That's why your prof talked about this:
av_1 + av_2 + ... +av_n = 0
Where the set is linearly independent if the only way to yield 0 is if a_1 to a_n is 0

Answer 31

Well what is that one vector? And what are the other vectors that's what I'm not getting. And also you guys kept saying that u can row reduce a matrix, but I don't see any matrices.

Answer 32

There is a bijection from any vector space to F^n. You can setup a matrix by going back and forth from the vector space to F^n.

Answer 33

I don't even know what a bijection is.

Answer 34

I always have trouble understanding abstract stuff. Can you give me one example?

Answer 35

ok one second

Answer 36

The first is the standard basis. So that would look like an identity matrix if placed in a matrix.

Answer 37

okay.

Answer 38

someone help me please .

Answer 39

I'm just going to write out the matrix for the second one.

Answer 40

Give me a second.

Answer 41

sorry tabbeyx we're busy try posting it as a new thread.

Answer 42

Okay alchemista, thank you so much for the help, i have a final in two days.

Answer 43

\[\left[ \begin {array}{cccc} 1&0&-1&0\\0&1&0&-3
\\ 0&0&3&0\\ 0&0&0&5\end {array}
\right] \]

Answer 44

Ohhh I see how you're doing this..

Answer 45

So I have to show that this matrix is linearly independent?

Answer 46

If it has rank 4 then its linearly independent. In short if it reduces to the identity.

Answer 47

its the vectors that are linearly independent not the matrix (just a clarification)

Answer 48

oh I see so every single column inside the matrix must be basic..

Answer 49

Last thing is how do I find change of basis matrix?

Answer 50

and also I remember there were two conditions:
First the matrix must be full rank
Second Span(B) = V how do I show that it spans V? Do i have to rewrite it as a combination of the standard Basis?

Answer 51

By the theory if a set of vectors is linearly independent and contains the same number of vectors as the dimension of the vector space it is a basis. A basis spans.

Answer 52

Proving the concept of dimension for bases is a little tricky and involves a small lemma.

Answer 53

Errr dimension for vector spaces. But I mean the proof involves showing that all bases must have the same cardinality.

Answer 54

Okay forget the proof.
For B' I showed that rank = 4 so it's linearly independent, but there are two conditions. B' must be lin indep and span(B') = V how do I prove the 2nd one?

Answer 55

That's what I just said. The vector space is dimension 4, the set contains 4 vectors, it spans.