Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 3 (Spans, bases and finite support)

Put $U=⟨u_1,\dots ,u_k⟩$. Taking all coefficients zero puts the zero vector in $U$. If $s={\sum }_i{\alpha }_i{u}_i$, $t={\sum }_i{\beta }_i{u}_i$ lie in $U$ and $a,b\in \mathbb{R}$, thenThus $U$ is closed under vector addition and scalar multiplication, contains the zero vector, and is therefore a linear subspace of $V$. This is precisely why the linear span is a linear subspace.

Suppose ${\sum }_{i=1}^{k+1}{\alpha }_i{u}_i=0$. If ${\alpha }_{k+1}\ne 0$, thencontrary to the hypothesis. Hence ${\alpha }_{k+1}=0$. The assumed linear independence of $u_1,\dots ,u_k$ then gives ${\alpha }_1=\cdots ={\alpha }_k=0$. Therefore $u_1,\dots ,u_{k+1}$ are linearly independent. Adding a vector outside the previous linear span creates no nontrivial relation.

Write $v={\sum }_{i=1}^{k+1}{\alpha }_i{u}_i$. Since $v\notin ⟨u_1,\dots ,u_k⟩$, necessarily ${\alpha }_{k+1}\ne 0$. Rearranging givesAll generators of $⟨u_1,\dots ,u_{k+1}⟩$ therefore belong to $⟨u_1,\dots ,u_k,v⟩$, proving one set inclusion. Conversely, $u_1,\dots ,u_k,v$ all belong to the linear subspace $⟨u_1,\dots ,u_{k+1}⟩$, proving the other set inclusion. ThusThis linear span exchange does not require linear independence of the original vectors.

For $\varphi (\alpha )={\sum }_{i=1}^k{\alpha }_i{u}_i$, its image of a linear map is exactly $⟨u_1,\dots ,u_k⟩$. Consequently $\varphi$ is surjective if and only if $u_1,\dots ,u_k$ are a spanning set for $V$.

If the vectors are linearly independent and $\varphi (\alpha )=\varphi (\beta )$, then ${\sum }_i({\alpha }_i-{\beta }_i)u_i=0$, so ${\alpha }_i={\beta }_i$ for all $i$. Hence $\varphi$ is injective. Conversely, if $\varphi$ is injective, any relation ${\sum }_i{\alpha }_i{u}_i=0$ says $\varphi (\alpha )=\varphi (0)$ and thus $\alpha =0$, proving linear independence.

Combining both equivalences givesA basis is therefore exactly a choice of vectors giving existence and uniqueness of coordinates, expressed by a bijection.

Take any finite selection of distinct indices $i_1,\dots ,i_r$ and a relationThe coordinate at $i_j$ on the left is exactly ${\alpha }_j$, while the zero vector has coordinate zero. Hence ${\alpha }_j=0$ for each $j$. Every finite subset of $E$ is therefore linearly independent, which is the definition of linear independence for an infinite set. Thus $E$ is linearly independent in the vector space of finitely supported sequences.

Every $v\in V$ is a finitely supported sequence, so for a finite set of indices $S$,Thus $E$ is a spanning set. If $F$ contained $v\notin E$, the displayed expression would give a nontrivial relationamong the distinct vectors of the finite subset $\{v\}\cup \{e^{(i)}:i\in S\}$ of $F$: the coefficient of $v$ is $1$. This contradicts the linear independence of $F$. It also covers $v=0$, whose singleton is not linearly independent. Hence $F=E$. In particular, $E$ is both a basis and a maximal linearly independent subset of $V$.