Let $\mathcal{F}_1$ denote the set of continuous real-valued functions $f(x)$ on the interval $0 \leq x \leq 1$ that meet the auxiliary constraints $f(0)=0$ and $f(1)=1$. Show that $\mathcal{F}_1$ is not a vector space over $\mathbb{R}$ with respect to the natural rules of vector addition and scalar multiplication that were introduced in Exercise 1.6.
- Span: If $\mathbf{v}_1, \ldots, \mathbf{v}_{\mathbf{k}}$ is a given set of vectors in a vector space $\mathcal{V}$ over $\mathbb{F}$, then
$$
\operatorname{span}\left\{\mathbf{v}_{\mathbf{1}}, \ldots, \mathbf{v}_{\mathbf{k}}\right\}=\left\{\sum_{j=1}^k \alpha_j \mathbf{v}_j: \alpha_1, \ldots, \alpha_k \in \mathbb{F}\right\} .
$$
In words, the span is the set of all linear combinations $\alpha_1 \mathbf{v}_1+\cdots+$ $\alpha_k \mathbf{v}_k$ of the indicated set of vectors, with coefficients $\alpha_1, \ldots, \alpha_k$ in $\mathbb{F}$. It is important to keep in mind that $\operatorname{span}\left\{\mathbf{v}_1, \ldots, \mathbf{v}_{\mathbf{k}}\right\}$ may be small in some sense. In fact, span $\left\{\mathbf{v}_{\mathbf{1}}, \ldots, \mathbf{v}_{\mathbf{k}}\right\}$ is the smallest vector space that contains the vectors $\mathbf{v}_{\mathbf{1}}, \ldots, \mathbf{v}_{\mathbf{k}}$. The number of vectors $k$ that were used to define the span is not a good indicator of the size of this space. Thus, for example, if
$$
\mathbf{v}_1=\left[\begin{array}{l}
1 \\
2 \\
1
\end{array}\right], \mathbf{v}_2=\left[\begin{array}{l}
2 \\
4 \\
2
\end{array}\right] \text { and } \mathbf{v}_3=\left[\begin{array}{l}
3 \\
6 \\
3
\end{array}\right] \text {, }
$$
then
$$
\operatorname{span}\left\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\right\}=\operatorname{span}\left\{\mathbf{v}_1\right\} .
$$
To clarify the notion of the size of the span we need the concept of linear dependence.
- Linear dependence: A set of vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ in a vector space $\mathcal{V}$ over $\mathbb{F}$ is said to be linearly dependent over $\mathbb{F}$ if there exists a set of scalars $\alpha_1, \ldots, \alpha_k \in \mathbb{F}$, not all of which are zero, such that
$$
\alpha_1 \mathbf{v}_1+\cdots+\alpha_k \mathbf{v}_k=\mathbf{0} .
$$
Notice that this permits you to express one or more of the given vectors in terms of the others. Thus, if $\alpha_1 \neq 0$, then
$$
\mathbf{v}_1=-\frac{\alpha_2}{\alpha_1} \mathbf{v}_2-\cdots-\frac{\alpha_k}{\alpha_1} \mathbf{v}_k
$$
and hence
$$
\operatorname{span}\left\{\mathbf{v}_1, \ldots, \mathbf{v}_k\right\}=\operatorname{span}\left\{\mathbf{v}_2, \ldots, \mathbf{v}_k\right\} .
$$
Further reductions are possible if the vectors $\mathbf{v}_2, \ldots, \mathbf{v}_k$ are still linearly dependent.
- Linear independence: A set of vectors $\mathbf{v}_1, \ldots, \mathbf{v}_{\mathrm{k}}$ in a vector space $\mathcal{V}$ over $\mathbb{F}$ is said to be linearly independent over $\mathbb{F}$ if the only scalars $\alpha_1, \ldots, \alpha_k \in \mathbb{F}$ for which
$$
\alpha_1 \mathbf{v}_1+\cdots+\alpha_k \mathbf{v}_k=\mathbf{0}
$$
are $\alpha_1=\ldots=\alpha_k=0$. This is just another way of saying that you cannot express one of these vectors in terms of the others. Moreover, if $\left\{\mathbf{v}_1, \ldots, \mathbf{v}_k\right\}$ is a set of linearly independent vectors in a vector space $\mathcal{V}$ over $\boldsymbol{F}$ and if
$$
\mathbf{v}=\alpha_1 \mathbf{v}_1+\cdots+\alpha_k \mathbf{v}_k \quad \text { and } \quad \mathbf{v}=\beta_1 \mathbf{v}_1+\cdots+\beta_k \mathbf{v}_k
$$
for some choice of constants $\alpha_1, \ldots, \alpha_k, \beta_1, \ldots, \beta_k \in \mathbb{F}$, then $\alpha_j=\beta_j$ for $j=1, \ldots, k$.