Let S = { (a+2b)x 2 + bx - (a+b) : a,b in R}. Show that S is a subspace of P 2 .
ID: 3143642 • Letter: L
Question
Let S = { (a+2b)x2 + bx - (a+b) : a,b in R}. Show that S is a subspace of P2. Find a basis for S.
NOTE: I've attempted the question myself (using my lecturer's method) and I got a basis of: B = {v1, v2} = { x2-1, 2x2+x-1 } Is this a correct basis? I'm not sure because I've read a definition that: "The number of elements in any basis is the dimension of the vector space". However, my basis only has two vectors and dim(S) = 3?
Furthermore, I know that A = {1, x, x2} is the standard basis of S. However the number of elements in A doesn't equal the number of elements in my basis B, even thought the theorem states that any two bases for a vector space must have the same number of elements :/
Explanation / Answer
(a+2b)x^2+bx-(a+b)=0
ax^2+2bx^2+bx-a-b=0
a(x^2-1)+b(2x^2+x-1)=0
which gives basis {(x^2-1),(2x^2+x-1)}
So your answer is correct.
Notice that S is subspace of P2 not equal to P2.
A subspace may not have all the elements of P2 and hence may not need all elements of standard basis of P2.
That's why you are getting basis of dimension 2 not 3.
For example:
set {p(t) :p(2) = 0} is subspace of P2.
Which is spanned by {(t-2),(t^2-4)} and has only two elements means dimension =2
while we know that P2 has standard basis {1,t,t^2} having dimension=3
They are not equal.
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