Linear Algebra and Its Applications, Exercise 2.6.19

Exercise 2.6.19. Let $V$ be the vector space consisting of all cubic polynomials of the form $a_0 + a_1x + a_2x^2 + a_3x^3$ and let $S$ be the subset of $V$ consisting of only those cubic polynomials for which $\int_0^1 p(x) \,dx = 0$ . Show that $S$ is a subspace of $V$ and find a set of basis vectors for $S$ .

Answer: For $S$ to be a subspace it must be closed under both vector addition and scalar multiplication. First, suppose $p$ is a member of $S$ and consider the vector $cp$ for any scalar $c$ . By the constant factor rule in integration we then have

$\int_0^1 cp \,dx = c \int_0^1 p \,dx = c \cdot 0 = 0$

So $cp$ is also a member of $S$ and $S$ is closed under scalar multiplication.

Second, suppose $q$ is also a member of $S$ and consider the vector $p+q$ . By the sum rule in integration we have

$\int_0^1 (p+q) \,dx = \int_0^1 p \,dx + \int_0^1 q \,dx = 0+0 = 0$

So $p+q$ is also a member of $S$ and $S$ is closed under vector addition.

Since $S$ is closed under both scalar multiplication and vector addition it is a subspace of $V$ .

We now attempt to find a set of basis vectors for $S$ . If $p$ is in $S$ then it is in the form $a_0 + a_1x + a_2x^2 + a_3x^3$ . Taking the indefinite integral of $p$ we have

$\int p \,dx = \int (a_0 + a_1x + a_2x^2 + a_3x^3) \,dx$

$= \int a_0 \,dx + \int a_1x \,dx + \int a_2x^2 \,dx + \int a_3x^3 \,dx$

$= a_0x + \frac{1}{2}a_1x^2 + \frac{1}{3}a_2x^3 + \frac{1}{4}a_3x^4 + C$

where $C$ is a constant.

The definite integral of $p$ over the interval from 0 to 1 is then

$(a_0 \cdot 1 + \frac{1}{2}a_1 \cdot 1^2 + \frac{1}{3}a_2 \cdot 1^3 + \frac{1}{4}a_3 \cdot 1^4 + C)$

$- (a_0 \cdot 0 + \frac{1}{2}a_1 \cdot 0^2 + \frac{1}{3}a_2 \cdot 0^3 + \frac{1}{4}a_3 \cdot 0^4 + C)$

$= a_0 + \frac{1}{2}a_1 + \frac{1}{3}a_2 + \frac{1}{4}a_3$

Since $p$ is a member of $S$ we then have

$\int_0^1 p \,dx = a_0 + \frac{1}{2}a_1 + \frac{1}{3}a_2 + \frac{1}{4}a_3 = 0$

We can create a set of basis vectors for $S$ by constructing vectors $\begin{bmatrix} a_0&a_1&a_2&a_3 \end{bmatrix}^T$ meeting this criterion. For the first basis vector we arbitrarily set $a_0 = 1$ and $a_2 = a_3 = 0$ . We then have

$a_0 + \frac{1}{2}a_1 + \frac{1}{3}a_2 + \frac{1}{4}a_3 = 1 + \frac{1}{2}a_1 = 0$

$\rightarrow \frac{1}{2}a_1 = -1 \rightarrow a_1 = -2$

So our first basis vector is $\begin{bmatrix} 1&-2&0&0 \end{bmatrix}^T$ corresponding to the polynomial $1 - 2x$ .

For the second basis vector we arbitrarily set $a_1 = 1$ and $a_0 = a_3 = 0$ . We then have

$a_0 + \frac{1}{2}a_1 + \frac{1}{3}a_2 + \frac{1}{4}a_3 = \frac{1}{2} + \frac{1}{3}a_2 = 0$

$\rightarrow \frac{1}{3}a_2 = -\frac{1}{2} \rightarrow a_2 = -\frac{3}{2}$

So our second basis vector is $\begin{bmatrix} 0&1&-\frac{3}{2}&0 \end{bmatrix}^T$ corresponding to the polynomial $x - \frac{3}{2}x^2$ . Note that this vector is linearly independent of the first basis vector since it includes a term in $x^2$ that the first vector lacks.

For the third basis vector we arbitrarily set $a_2 = 1$ and $a_0 = a_1 = 0$ . We then have

$a_0 + \frac{1}{2}a_1 + \frac{1}{3}a_2 + \frac{1}{4}a_3 = \frac{1}{3} + \frac{1}{4}a_3 = 0$

$\rightarrow \frac{1}{4}a_3 = -\frac{1}{3} \rightarrow a_3 = -\frac{4}{3}$

So our third basis vector is $\begin{bmatrix} 0&0&1&-\frac{4}{3} \end{bmatrix}^T$ corresponding to the polynomial $x^2 - \frac{4}{3}x^3$ . Note that this vector is linearly independent of the first and second basis vectors since it includes a term in $x^3$ that those vectors lack.

The subspace $S$ can have at most three basis vectors. (If $S$ had four basis vectors then since they would be linearly independent they would span $\mathbb{R}^4$ and we would have $S = \mathbb{R}^4$ . But this is not the case.) The following vectors can thus serve as a basis for $S$ :

$\begin{bmatrix} 1 \\ -2 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 1 \\ -\frac{3}{2} \\ 0 \end{bmatrix}\qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ -\frac{4}{3} \end{bmatrix}$

corresponding to the polynomials $x^2 - \frac{4}{3}x^2$ , $x - \frac{3}{2}x^2$ , and $x^2 - \frac{4}{3}x^3$ respectively.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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2 Responses to Linear Algebra and Its Applications, Exercise 2.6.19

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