## 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

1. Victor says:

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