Question

What does it mean to check if two vectors, say Cos[Pix] and Sin[Pix], are linear independent in C[0,1]?

Answer 1

Deos that mean when x=0 and 1?

Answer 2

i believe C is the complex plane

Answer 3

So C[0,1] is simple saying no imaginary numbers?

Answer 4

if vectors are independant, then they are not scalar multiples of each other, nor can they be created from a combination of other vectors in the set

Answer 5

the C[0,1] is confusing without some context. What does this notation represent in your tesxt?

Answer 6

I know they are linear independent because the wronskain determinant shows them to be Pi but say you have a Wronskian determinant for two vectors with a non-zero solution and an X.

Answer 7

This is exactly written from text: For each of the following. show that the given vecotrs are linearly independent in C[0,1]: it then lists examples a, b, c, and d.

Answer 8

W[x^3/2, x^5/2]=5/2x^2-3/2x^3

Answer 9

That above answer shows it to be linearly independent but only for certain x values. I figuredd C[0,1] had something to do with checking certain x values

Answer 10

whatcha think? I think it has something to do with the vector space or the constants c1, c2

Answer 11

still trying to determine what "in C[0,1]" is refering to. Any ideas phi?

Answer 12

must have to do with x [0,1]

Answer 13

\[W\begin{vmatrix}
cos(\pi~x)&sin(\pi~x)\\
-\pi~sin(\pi~x)&\pi~cos(\pi~x)
\end{vmatrix}=\pi~(cos^2(\pi~x)+sin^2(\pi~x)=\pi\]

Answer 14

Okay I found an example showing C[-infinity,infinity] for W[1,x,x^2,x^3]=12 making it linearly independent for all x. So example above is linearly independent for C[0,1] and all x.

Answer 15

it's strange

Answer 16

i take it this is not a first year linear algebra question :)

I keep reading about the wronskian being applied of an interval [a,b] and it relating to a differential equation setup

Answer 17

Let \(f_1\) and \(f_2\) be solutions to the diffyQ: \(y''+p(t)y'+q(t)=0\). Then W(\(f_1,f_2\)) is either zero for all t, or never zero

Answer 18

http://ltcconline.net/greenl/courses/204/ConstantCoeff/linearIndependence.htm

Answer 19

No, it is actually. I did a google search before coming here and only found diff. eq stuff but we haven't gone into any of that.

Answer 20

Yeah i was on that page earlier

Answer 21

well, if its just first year stuff, then

Two vectors are linearly independent if \(c_1x+c_2y=0\) only if c1=c2=0

Answer 22

and i believe the C[0,1] is stating, within the interval from 0 to 1

Answer 23

\[c_1~cos(x~\pi)+c_2~sin(x~\pi)=0\]

http://www.wolframalpha.com/input/?i=cos%28pi+x%29%2Bsin%28pi+x%29%3D0

it appears that if we let c1=c2=1, that we get cos(pi x) + sin(pi x) = 0 between the interval 0 to 1

Answer 24

I see but if we are assuming C[0,1] has to do with c1, and c2 then the interval [0,1] for x no longer is what they are asking for?...correct my if im wrong.

Answer 25

C[0,1] im reading as an interval from x=0 to x=1

or simply the values of cos(t), and sin(t) from 0 to pi

along that interval we see that sin(t) = -cos(t) , when t = 3pi/4 . Since along the interval, we encounter an instance such that the one vector is a scalar multiple of the other ... id say that they cannot be classified as independant.

But thats real fuzzy since i took this stuff awhile ago

Answer 26

i dont have my textbook handy to reference tho

Answer 27

i see, for now I will stick with the wronskian determinant that says they are independent but I like the reasoning. Thanks for your time! If I get any answers I will post them soon. Have a good day! (:

Answer 28

good luck ;) and ill most likely have time to review this more over the weekend to see if my memory has failed me lol

Answer 29

http://books.google.com/books?id=jbJDUFZ27yMC&pg=PA102&lpg=PA102&dq=linearly+independent+in+an+interval&source=bl&ots=Ty-SE0TMt2&sig=gqDbq0JYFptFheyR7__zLzcp4jA&hl=en&sa=X&ei=nyZDUYL1HYiS9QTLxIHADA&ved=0CDIQ6AEwATgK#v=onepage&q&f=false

around page 102 i found this. Which i believe corroborates my memory :)

Answer 30

thanks amistre64