Thread: vector spaces & subspaces

Determine whether W is a subspace of V




my solution:
First step: W is nonempty because it contains the zero polynomial (a=b=c=0)
Second step: Let

and

Then


So p(x) + q(x) is also in W (because it has the right form). Similarly, if k is a scalar, then



so kp(x) is in W.

Thus, W is a nonempty subset of that is closed under addition and scalar multiplication. therefore, W is a subspace of

Was just wondering if this was all good, becuase the initial condition of the set stated above (a*b*c=0) confused me a bit

p(x) + q(x) will be in W if and only if (a+d)(b + e)(c + f) = 0. W contains all polynomials of second degree such that the product of the coefficients is equal to zero. Similiary, for kp(x) to be in W, we must have (ka)(kb)(kc) = 0. You have to check this before you can go on and say that p(x) + q(x) and kp(x) are in W.

Hope this helps!

Originally Posted by james12

Determine whether W is a subspace of V




my solution:
First step: W is nonempty because it contains the zero polynomial (a=b=c=0)
Second step: Let

and

Then


So p(x) + q(x) is also in W (because it has the right form).

What do you mean "it has the right form"? What "form" are you talking about? Surely it is V because it is a quadratic but to be in W it must also have the product of its coefficients equal to 0 (which is the same as saying at least one of its coefficients is 0. Consider , . Are they both in W? What is their sum? Is it in W?

Similarly, if k is a scalar, then



so kp(x) is in W.

Thus, W is a nonempty subset of that is closed under addition and scalar multiplication. therefore, W is a subspace of

Was just wondering if this was all good, becuase the initial condition of the set stated above (a*b*c=0) confused me a bit

Ahh it's all clear now, thanks for the help guys, much appreciated!!