# Find a basis of U

Let $$U = \{p∈P_4(\mathbb R):p′′(4)=0\}$$. I showed it was a subspace of $$P_4(\mathbb R)$$ by showing its closed under scalar multiplication and addition. I have to find a basis. So I found

$$B=\{1,x,-12x^2+x^3,-96x^2+x^4\}$$

My question is how to prove it's a basis. I know I have to prove it's linearly independent and spans $$U$$. I proved it's linearly independent. I know to prove it spans $$U$$ I have to prove

span$$(B)⊂U$$ and $$U⊂$$span$$(B)$$

The first one I understand since span$$(B)$$ is a linear combination and U is a subspace, therefore closed. The second one I'm stuck on how to do. Any help would be appreciated.

• You can show that dim U<=4. Then the linear independence of B implies it is a basis of U. – GaryChanCCO Feb 12 '19 at 6:31
• @DragunityMAX I understand that argument because dim of subspace is <= dim of vector space and an independent list with correct length is a basis. I kind of want to practice the span part though since i get lost sometimes with it – sweets Feb 12 '19 at 6:48

Let $$u=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0 \in U$$, then $$2a_2+24a_3+192a_4=0$$

Now we want to find $$c_1,c_2,c_3,c_4$$ such that $$c_1(1)+c_2(x)+c_3(-12x^2+x^3)+c_4(-96x^2+x^4)=u$$

Obviously, the only possible choice is $$c_1=a_0,c_2=a_1,c_3=a_3,c_4=a_4$$

Now $$c_1(1)+c_2(x)+c_3(-12x^2+x^3)+c_4(-96x^2+x^4)-u=-12a_3x^2-96a_4x^2-a_2x^2$$ $$=-\frac{1}{2}(2a_2+24a_3+192a_4)x^2=0$$

So $$U\subset span B$$

• So we're trying to find coefficients such that our basis would look like an arbitrary element in U. I understand that. Can I ask how you came up with c1=a0,c2=a1,c3=a3,c4=a4? Also why you subtracted u in the fourth line? Sorry, this is my first linear alg class...and also my first proof based class – sweets Feb 12 '19 at 7:33
• oh wait is it c1=a0,c2=a1,c3=a3,c4=a4 because those are the degrees of each part of basis B? and in basis B, x^2 doesn't appear as the deg and hence thats why c3=a3? @DragunityMAX – sweets Feb 12 '19 at 7:38
• The choice of c's is because the degrees of each part of B are different. Only term -12x^2+x^3 contains x^3 so you must choose c3=a3. The reason for c4=a4 is similar. I subtracted u in the fourth line to show the combination is equal to u ( z=u equivalent to z-u=0 ). @sweets – GaryChanCCO Feb 12 '19 at 8:53
• I get it now! Thank you very much – sweets Feb 12 '19 at 17:41