the set of all polynomials of degree less than or equal to 3 is a vector space.the dimension of subspace of all polynomials p3 such that p(x)=0 is
a) 0 b) 1 c) 2 d) 3
Fix a positive integer n.
Note that dim(Pn)= n + 1, since {1, x, x^2, ..., x^n} is a basis for Pn.
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Tweaking notation a bit... fix a scalar a (instead of x).
If we want the dimension of the subspace S = {p(x) in P3 : p(a) = 0},
note that p(x) = (x - a) * q(x) for some q(x) in P2.
This is due to counting the degrees of the polynomials.
Thus, S = {(x - a) * q(x): q(x) is in P2}.
Since dim(P2) = 3, we conclude that S also has dimension 3.
I hope this helps!
Comments
Fix a positive integer n.
Note that dim(Pn)= n + 1, since {1, x, x^2, ..., x^n} is a basis for Pn.
-----------------------------
Tweaking notation a bit... fix a scalar a (instead of x).
If we want the dimension of the subspace S = {p(x) in P3 : p(a) = 0},
note that p(x) = (x - a) * q(x) for some q(x) in P2.
This is due to counting the degrees of the polynomials.
Thus, S = {(x - a) * q(x): q(x) is in P2}.
Since dim(P2) = 3, we conclude that S also has dimension 3.
I hope this helps!