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In the vector space $P_3$ of all $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, let $S$ be the subset of polynomials with $\int_0^1p(x)dx=0$. Verify that $S$ is a subspace and find a basis. I'm not sure where to start on this question. Can somebody help please? Thanks! |
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Hints: As usual, to show that $S$ is a subspace of $P_3$, prove that the following hold $\ \ $1) $S$ is non-empty. $\ \ $2) $p\in S$ implies $\alpha p\in S$ for all $\alpha\in\Bbb R$. $\ \ $3) $p_1,p_2\in S$ implies $p_1+p_2 \in S$. For 1), you just need to find any polynomial $p\in P_3$ with $\int_0^1 p(x)\, dx=0$. For 2) and 3), use the linearity of the definite integral. For a basis of $S$, note that the dimension of $S$ is at most 4. Since there are polynomials in $P_3$ that are not in $S$, the dimension of $S$ is in fact at most 3. Now try to find three independent vectors in $S$. A hint here: how can you modify $p(x)=x^n$ so that its definite integral over $[0,1]$ is 0? For example, take the polynomial $p(x)=x$. We have $\int_0^1 x\,dx={1\over2}$. So, $p$ is not in $S$... But, $x-{1\over2}$ is in $S$.
An alternate approach towards finding a basis of $S$, if you can't appeal to dimensionality arguments, is the following: Assume $p(x)=a+bx+cx^2+dx^3$ is in $S$. Then $$ 0=\int_0^1 p(x)\,dx = a+{b\over 2}+{c\over 3}+ {d\over4}. $$ It follows that $a=-{b\over 2}-{c\over 3}- {d\over4}$. Thus, we may write: $$\eqalign{ p(x)&=\textstyle \bigl(bx-{b\over 2}\bigr)+\bigl(cx^2-{c\over3}\bigr)+\bigl(dx^3-{d\over 4}\bigr)\cr &=\textstyle b\bigl(x-{1\over 2}\bigr)+c\bigl( x^2-{1\over3}\bigr)+d\bigl( x^3-{1\over 4}\bigr).\cr } $$ Using the above, you should be able to find identify three independent vectors in $P_3$ and argue that they span $S$. |
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Hint $\ $ The map $\rm\: L(f) = \int_0^1 f\:$ is an $\mathbb R$-linear map between the $\mathbb R$-vector spaces $\rm\:P_3$ and $\mathbb R$, hence its kernel $\rm\: K = ker\ L = \{f : Lf = 0\}$ forms a subspace of $\rm\:P_3$. If you haven't yet learned this general fact then you can prove it now with no greater effort than that required for your special case, but with the benefit that you now have a more general result which yields greater conceptual insight. To find a basis of $\rm\:K,\:$ note that for $\rm\:n\in \{1,2,3\},$ $\rm\: 0 \ne x^n - L(x^n)\in K,\:$ by $\rm\:L(r) = r\:$ for $\rm\:r\in \mathbb R$. But $\rm\:dim\ K \le 3,\:$ (why?) and these $3\:$ elements of $\rm\:K\:$ are independent, hence $\ldots$ |
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Take a general polynomila of P3, calculate its integral between 0 to 1, and the result equal to 0 gives you an homogeneous linear equation in the coefficients. Take for example a0=-a1/2-a2/3-a3/4, substitute in the initial polynomial, and you have a general polynomial of S. Descompose it as a linear combination of some polynomilas, where the coeficinets are the scalars in the combination. By this way you have : S=Span(...) and you know that span is a subspace of the space. and the polynomilas of the linear combination are a generator. |
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To verify it is a subspace, you must check the vector space axioms. You need to show that the $0$-vector is contained in it. You need to show if $a\in\mathbb{R}$, then $\int_0^1ap(x)\,dx=0$ for any polynomial $p(x)$ such that $\int_0^1p(x)\,dx=0$. And you must show for any two polynomials which integrate to zero, that their sum also integrates to zero. You should also check that the sum of two polynomials with degree at most $3$ also has degree at most $3$. |
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An engineer's non-rigorous approach: You need to check 3 things to confirm that a space is indeed a subspace:
(All these three can be condensed into one condition that $k_1A+k_2B$ lie in space if $A \& B$ do. But I like to think of it in the above 3 rules.) Now, select a polynomial $p_1$ as per the requirement such that it lies in the space. Such a polynomial exists because the $0$ vector itself suffices. Condition number one is ticked off. If $\int_0^1p_1(x)=0$, then for any value of $k$, $\int_0^1kp_1(x)=0$. Condition 2 is ticked off. Now, if $p_1$ and $p_2$ are in the space, then $\int_0^1p_1(x)+p_2(x)= \int_0^1p_1(x)+\int_0^1p_2(x) =0 $. Condition 3 is ticked off. |
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