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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.

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  • $\begingroup$ You can show that dim U<=4. Then the linear independence of B implies it is a basis of U. $\endgroup$ – GaryChanCCO Feb 12 '19 at 6:31
  • $\begingroup$ @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 $\endgroup$ – sweets Feb 12 '19 at 6:48
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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 $U\subset span B$

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  • $\begingroup$ 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 $\endgroup$ – sweets Feb 12 '19 at 7:33
  • $\begingroup$ 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 $\endgroup$ – sweets Feb 12 '19 at 7:38
  • $\begingroup$ 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 $\endgroup$ – GaryChanCCO Feb 12 '19 at 8:53
  • $\begingroup$ I get it now! Thank you very much $\endgroup$ – sweets Feb 12 '19 at 17:41

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