# vector spaces & subspaces

• Jul 22nd 2010, 04:47 PM
james12
vector spaces & subspaces
Determine whether W is a subspace of V

$V=P_2$
$W = \left\{ a + bx + cx^2 : abc=0 \right\}$

my solution:
First step: W is nonempty because it contains the zero polynomial (a=b=c=0)
Second step: Let
$p(x) = a + bx + cx^2$
and
$q(x) = d + ex + fx^2$
Then
$p(x) + q(x) = \left( a+d \right) + \left( b+e \right)x + \left( c + f \right)x^2$

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

$kp(x) = ka + kbx + kcx^2$

so kp(x) is in W.

Thus, W is a nonempty subset of $p_2$ that is closed under addition and scalar multiplication. therefore, W is a subspace of $p_2$

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
• Jul 22nd 2010, 05:59 PM
buri
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!
• Jul 23rd 2010, 03:42 AM
HallsofIvy
Quote:

Originally Posted by james12
Determine whether W is a subspace of V

$V=P_2$
$W = \left\{ a + bx + cx^2 : abc=0 \right\}$

my solution:
First step: W is nonempty because it contains the zero polynomial (a=b=c=0)
Second step: Let
$p(x) = a + bx + cx^2$
and
$q(x) = d + ex + fx^2$
Then
$p(x) + q(x) = \left( a+d \right) + \left( b+e \right)x + \left( c + f \right)x^2$

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 $u= x^2+ 2$, $v= x^2+ 3x$. Are they both in W? What is their sum? Is it in W?

Quote:

Similarly, if k is a scalar, then

$kp(x) = ka + kbx + kcx^2$

so kp(x) is in W.

Thus, W is a nonempty subset of $p_2$ that is closed under addition and scalar multiplication. therefore, W is a subspace of $p_2$

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
• Jul 23rd 2010, 04:11 AM
james12
Ahh it's all clear now, thanks for the help guys, much appreciated!!