Arturo O. answered • 12/26/16
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You can solve this by inspection. Note that w = v, hence w = (0)u + (1)v, and therefore is spanned by u and v since we are able to express w as a linear combination of u and v.
Another way to look at this:
If w is in the span of u and v, then w = au + bv for a pair of scalars a and b. Let us test this by trying to find a and b.
Assume w = au + bv, and see if we get real solutions.
w = au + bv
2x2 + 4x + 5 = a(x2 + 2x + 3) + b(2x2 + 4x + 5)
Grouping like terms on the right hand side, we get
2x2 + 4x + 5 = (a + 2b)x2 + (2a + 4b)x + (3a + 5b)
The coefficients of like powers of x must be identical on both sides, so we get a system of equations:
a + 2b = 2
2a + 4b = 4
3a + 5b = 5
The second equation is just twice the first, so it is redundant and we are left with
a + 2b = 2
3a + 5b = 5
Solve for a and b.
a = 2 - 2b
3(2 - 2b) + 5b = 5
6 - 6b + 5b = 5
-b = -1
b = 1
a = 2 - 2b = 2 - 2(1) = 0
a = 0
This is what we got by inspection; w is spanned by u and v.
Claire H.
Thank you, this was great help. Seems rather simple now that you have shown me. Thanks again!
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12/27/16
Arturo O.
You are welcome, Claire.
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12/27/16
Claire H.
Sorry to bother you again, but I was just wondering for w to span u and v, must you obtain only one value for a and b (w = au+bv), or can there be infinitely many solutions? I guess in other words, can it be a linearly dependent system?
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12/31/16
Arturo O.
You said "for w to span u and v," but I am sure you meant "for w to be spanned by u and v." The vector w is linearly dependent on u and v, and the vectors u and v are linearly independent in this problem. If a linearly dependent vector w is expressed as
w = c1u + c2v,
the coefficients c1 and c2 are unique. Otherwise, u and v could not be linearly independent.
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12/31/16
Mark M.
12/26/16