Prove Independence

  • #1
judahs_lion
56
0

Homework Statement


Given S = (1+x2, x +x3

And augment S to form a Basis S' of P3


The Attempt at a Solution



0 + 0x + 0x2 + 0x3 = a(1+x2)+b(x +x3)

= a + ax2 + bx + bx3
 

Answers and Replies

  • #2
judahs_lion
56
0
Isn't S dependent? X + X3 = (1 + X2)X
 
  • #3
36,213
8,194

Homework Statement


Given S = (1+x2, x +x3
I'm pretty sure you mean, S = {1 + x2, x + x3}
And augment S to form a Basis S' of P3


The Attempt at a Solution



0 + 0x + 0x2 + 0x3 = a(1+x2)+b(x +x3)

= a + ax2 + bx + bx3
It would be helpful for you to state the complete problem. My guess is that it is two parts:
a) Prove that the functions in S = {1 + x2, x + x3} are linearly independent.
b) Augment S to a set S' that is a basis for P3.

For a, how is linear independence defined? From your work above, I'm not sure that you know. The definitions for linear independence and linear independence are similar, and there is a subtlety that students often don't grasp.
For b, have you learned about the Gram-Schmidt process?
 
  • #4
judahs_lion
56
0
I scanned it in. Its problem # 15
 

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  • #5
judahs_lion
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Linear independence mean the members of a set of vectors are independent of each other. None is a multiple of the other.
Haven't gotten to Gram-Schmidt process
 
  • #6
36,213
8,194
I scanned it in. Its problem # 15
Prob. 15 is almost identical to prob. 13. The polynomials in P3 are essentially the same as vectors in R4. For example, 1 + 2x2 <---> <1, 0, 2, 0>.
 
  • #7
36,213
8,194
Linear independence mean the members of a set of vectors are independent of each other.
This isn't the definition, and besides, a definition of a term ought not use the same term in the definition. Look in your book and see how it defines linear independence.
None is a multiple of the other.
This is a necessary condition for linear independence, but it is not sufficient. For example, consider the set {<1, 0, 0>, <0, 1, 0>, <1, 1, 0>}. None of these vectors is a multiple of any other vector in the set, yet these vectors are not linearly independent.
 
  • #8
judahs_lion
56
0
This isn't the definition, and besides, a definition of a term ought not use the same term in the definition. Look in your book and see how it defines linear independence.
This is a necessary condition for linear independence, but it is not sufficient. For example, consider the set {<1, 0, 0>, <0, 1, 0>, <1, 1, 0>}. None of these vectors is a multiple of any other vector in the set, yet these vectors are not linearly independent.

Thanks for pointing out that so , what i need to do is a transformtion as i did in the attached. Then the rest is just as problem 13?
 

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  • #10
judahs_lion
56
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Yes.

But the second part. Augment S to form a basis S' for P3 , that would still be in the form of a polynomial?
 
  • #11
36,213
8,194
You can use the augmented basis you found in #13, and "untransform" the vectors to get the other two polynomials you need for a basis for P3.
 
  • #12
judahs_lion
56
0
You can use the augmented basis you found in #13, and "untransform" the vectors to get the other two polynomials you need for a basis for P3.

Ok, to verify 13 is done properly?
 

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  • #13
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Sure, those vectors are linearly independent, one of many possible sets of four vectors that span R^4.
 
  • #14
judahs_lion
56
0
S' = {1+x2, x+x3, 1, x }
 
  • #15
judahs_lion
56
0
Would Reducing the Matrix (1, 0; 0,1; 1, 0; 0,1) to (1,0; 0,1; 0,0; 0,0) have been another way to prove independence?
 
  • #16
36,213
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Yes, but with just two vectors, that's overkill. Two vectors are linearly independent as long as neither one is a multiple of the other. If you have three vectors, though, it's not as obvious. I gave you an example of this in another thread.
 

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