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Problem pg.196 #8 and pg.207 #34
#8, pg.196: Determine if the set of all polynomials in $\mathbb{P}_n$ such that $\vec{p}(0)=0$ is a subspace of $\mathbb{P}_n$.
Solution: We will attempt to apply the subspace criterion. Let $S$ denote the set of all polynomials $\vec{p}$ in $\mathbb{P}_n$ such that $\vec{p}(0)=0$. First, it is obvious that $S$ is a subset of $\mathbb{P}_n$. The zero vector of $\mathbb{P}_n$ is the zero polynomial which is clearly in $S$. Let $\alpha,\beta$ be scalars and let $\vec{p}_1$ and $\vec{p}_2$ be in $S$ (hence $\vec{p}_1(0)=0$ and $\vec{p}_2(0)=0$). We see that $\vec{p}=\alpha \vec{p}_1 + \beta \vec{p}_2$ is in $S$ because we may compute
$$\begin{array}{ll}
\vec{p}(0) &= \alpha \vec{p}_1(0) + \beta \vec{p}_2(0) \\
&= 0 + 0 \\
&= 0.
\end{array}$$
Hence $S$ is a subspace of $\mathbb{P}_n$.
#34, pg.207: Define $T \colon C[0,1] \rightarrow C[0,1]$ as follows: for $\vec{f} \in C[0,1]$ let $T(\vec{f})$ be the antiderivative of $\vec{F}$ of $\vec{f}$ such that $\vec{F}(0)=0$. Show that $T$ is a linear transformation and describe the kernel of $T$.
Solution: We must show that given $\vec{f}_1$ and $\vec{f}_2$ in $C[0,1]$ and scalars $\alpha,\beta$ the following formula holds:
$$T(\alpha \vec{f}_1 + \beta \vec{f}_2)=\alpha T(\vec{f}_1) + \beta T(\vec{f}_2).$$
Recall that the following formula for any integral holds:
$$\int \alpha f_1(x) + \beta f_2(x) dx = \alpha \int f_1(x) dx + \beta \int f_2(x)dx.$$
Notice that writing
$$\vec{F}(x)=\left( T(\vec{f}) \right)(x)=\int_0^x f(t) dt$$
is the antiderivative with the property that
$$\vec{F}(0) = \int_0^0 f(t)dt \stackrel{\mathrm{calculus \hspace{2pt} fact}}{=} 0.$$
We are confident in saying the antiderivative with this property because all antiderivatives are vertical shifts of each other -- think of the "$+C$" from calculus! Now compute
$$\begin{array}{ll}
T(\alpha \vec{f}_1 + \beta\vec{f}_2) &= \displaystyle\int_0^x \alpha \vec{f}_1(t) + \beta \vec{f}_2(t) dt \\
&\stackrel{\mathrm{calculus \hspace{2pt} facts}}{=} \alpha \int_0^x \vec{f}_1(t)dt + \beta \int_0^x \vec{f}_2(t) dt \\
&= \alpha T(\vec{f}_1) + \beta T(\vec{f}_2),
\end{array}$$
proving that $T$ is a linear transformation.
The kernel of $T$ is the trivial vector space $\{\vec{0}\}$. This is because the only function whose integral is constant is the zero function (other constant functions don't work! for example $\int 5 dx = 5x+C$) -- this can be seen by noticing that the derivative is the inverse of integration and the kernel of the derivative operator is the set of constant functions (we did this in class).
