- #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
I'm pretty sure you mean, S = {1 + x2, x + x3}Homework Statement
Given S = (1+x2, x +x3
It would be helpful for you to state the complete problem. My guess is that it is two parts: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
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>.I scanned it in. Its problem # 15
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.Linear independence mean the members of a set of vectors are independent of each 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.None is a multiple of the 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.
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.
Yes.
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.