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Offline PhyssionTopic starter

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Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231314#msg1231314
« on: April 21, 2016, 06:25:29 am »
Hello,

Code: [Select]
Check whether the set {x^2 − 5, 3x^2 + 1, x + 1} is a basis of the space of polynomials over Q of degree at most 2. Justify your answer (show all work).(Q being the rationals)

I'm really not sure how to approach this question - would really appreciate any assistance working through this. How do I check whether or not this is a basis?

Offline iancudorinmarian

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Re: Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231316#msg1231316
« Reply #1 on: April 21, 2016, 07:02:25 am »
Letís say those polynomiăls are a, b, c. You have to prove that there are three numbers u1, u2, u3 differemt from 0 that u1*a+u2*b+u3*c=0

On phone, really hard to type :/

iirc, there is a theorem that says "any n liniarily  independent vectors form a base"
« Last Edit: April 21, 2016, 07:05:39 am by iancudorinmarian »

Offline Jenkar

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Re: Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231318#msg1231318
« Reply #2 on: April 21, 2016, 07:54:23 am »
Since i'm bored in class :
We know that the dimension of the linear vector space of polynomials on R of degree <= 2 (named after R[X]) is 3.
Let us prove that the linear vector space of polynomials on Q of degree <= 2 (named after Q[X]) is a subspace of R[X].
- The "0" of R[X] is the null polynomial, 0. 0 is in Q[X].
- If a is in Q[X] and b is in Q[X], then a+b is in Q[X].
- Let k be in R, and a be in Q[X]. k * a is in Q[X] (we don't add any roots, and all the roots stay the same)
Therefore, Q[X] is a subspace of R[X].
Hence, the dim(Q[X]) <= dim(R[X]) = 3

Now, we have the vectors x^2-5 , 3x^2 + 1, x+1. Let us prove that they are linearly independant.
Let a,b,c be so that for all x, a(x^2-5) + b(3x^2 + 1) + c(x+1) = 0.
This gives us :
a+3b = 0
c = 0
-5a+b+c = 0
Hence :
a+3b = 0
-15a +3b = 0
c = 0
And so :
c=0
a+3b = 0
16a = 0
Which gives us a = b = c = 0, hence the vectors are linearly independent.
Since we have 3 independent vectors and we know that the dimension of the space is at most 3, we have that the dimension of the space is 3 and hence that the 3 vectors form a basis (see http://ltcconline.net/greenl/courses/203/Vectors/basisDimension.htm ).
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Offline iancudorinmarian

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Re: Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231321#msg1231321
« Reply #3 on: April 21, 2016, 08:08:39 am »
I messed up, Jenkar is right, they have to be 0 (a, b, c)

Offline PhyssionTopic starter

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Re: Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231323#msg1231323
« Reply #4 on: April 21, 2016, 09:00:49 am »
Thank you, Jenkar - I've just seen your solution after finding another method from a classmate. Really appreciate the time you put into this, I'm able to follow your explanation. The other method is below:



For the set {x^2 - 5, 3x^2 + 1, x + 1}, let:
a1 = x^2 - 5
a2 = 3x^2 + 1
a3 = x + 1;

And let:
b1 = x^2
b2 = x
b3 = 1

Thus:
b1 = 1(a1) + 3(a2)
b2 = 1(a3)
b3 = -5(a1) + 1(a2) + 1(a3)

This can then be put into a matrix with respect to a and b.

1 3 0
0 0 1
-5 1 1

Following the elementary row operations:
R2 <--> R3
R2 --> R2 + 5R1
R2 --> R2 - R3
R2 --> R2 / 16
R1 --> R1  - 3R2

We end up with the matrix in row-reduced echelon form, which, in this case, forms the identity matrix.

100
010
001

Because this ends up as the identity matrix, we know there must exist a single unique solution for any arbitrary polynomial in Q<=2, and all three terms are linearly independent, thus this set must form a basis.



I understand everything up until the final step - I'm not exactly sure why the identity matrix tells us this. Can anyone else confirm it for me?

Offline Jenkar

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Re: Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231327#msg1231327
« Reply #5 on: April 21, 2016, 09:37:14 am »
Short version : since you obtain that matrix by elementary row operations, that means that there is a combination of b1, b2 and b3 that is equal to a1, a combination of b1, b2, and b3 that is equal to a2, and a combination of b1, b2 and b3 that is equal to a3.

And (very important) vice versa.
Since you can build vectors that form a basis from your starting vectors, that means that your starting vectors form a generating set. This (without any assumptions about independance & dimensionality) does NOT show that the solution they generate is unique, just that they can generate one.
« Last Edit: April 21, 2016, 09:39:00 am by Jenkar »
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Re: Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231340#msg1231340
« Reply #6 on: April 21, 2016, 12:48:12 pm »
In a more general sense you can show a set of vectors is a basis if you can generate a known basis (1, x, x^2 for example) from them and the number of elements in the set matches the dimension (as long as the dimension is finite ;))
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Re: Linear algebra help needed - checking basis http://elementscommunity.org/forum/index.php?topic=61821.msg1231389#msg1231389
« Reply #7 on: April 21, 2016, 11:14:00 pm »
I can't remember the exact details (~15 yrs since I did this) but the fact that you get the identity matrix at the end means that those three polynomials (in one quantity or another) can describe any rational 2nd order function. (in this case, since x^2 is the highest power)  I don't remember the reason this is true; I think it's something to do with the definition of a basis.

 

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