Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 1

The vectors are linearly independent ifThey span $V$ if every $v\in V$ is a linear combination $v={\sum }_{j=1}^n{a}_j{v}_j$ with real coefficients. They form a basis of $V$ if they are linearly independent and span $V$. Equivalently, every $v\in V$ has exactly one such representation: existence is the spanning set condition, and subtracting two representations shows that uniqueness is precisely linear independence.

The Steinitz exchange lemma states that if $u_1,\dots ,u_r$ are linearly independent and $w_1,\dots ,w_s$ span a vector space $V$, then $r\le s$ and some $r$ of the $w_j$ can be replaced by $u_1,\dots ,u_r$ while preserving the spanning set property.

Here is the exchange argument. Suppose that after $k$ replacements, a spanning set iswhere the remaining $w_j$ have been relabelled if necessary. Express $u_{k+1}$ as a linear combination of this family. At least one remaining $w_j$ has a nonzero coefficient; otherwise $u_{k+1}$ belongs to the span of $u_1,\dots ,u_k$, contradicting linear independence. Relabel this entry $w_{k+1}$. Solving for $w_{k+1}$ shows that it lies in the span ofThe new family therefore still spans $V$. Starting at $k=0$ repeats this step until all $r$ independent vectors have been inserted. If $r>s$, after $s$ steps the next $u_{s+1}$ would lie in the span of $u_1,\dots ,u_s$, a contradiction. Thus $r\le s$, and the construction proves the replacement assertion as well.

Put $w_1=v_1$ and $w_j=v_j-v_1$ for $j\ge 2$. Since $v_1=w_1$ and $v_j=w_j+w_1$, the two families have the same linear span. Moreover,If the $v_j$ are linearly independent, a vanishing expression forces $a_j=0$ for $j\ge 2$ and then $a_1=0$. Conversely,so linear independence of the $w_j$ forces all $b_j=0$. Thus both the spanning set condition and linear independence are equivalent for the two families, proving the desired equivalence of bases.

Write $u=(a,b{)}^T$, $v=(c,d{)}^T$ and $\Delta =ad-bc$. If $xu+yv=0$, thenEliminating $y$ gives $\Delta x=0$, and eliminating $x$ gives $\Delta y=0$. Hence $\Delta \ne 0$ implies linear independence.

For the converse, suppose $\Delta =0$. If $u=0$, the family already contains the zero vector and is linearly dependent. If $a\ne 0$, the equality $ad=bc$ gives $v=(c/a)u$. If $a=0$ but $u\ne 0$, then $b\ne 0$ and $bc=0$ forces $c=0$, so $v=(d/b)u$. Both cases give a nontrivial linear relation. ThereforeThis is the two-dimensional determinant criterion, proved directly without assuming that criterion.

Subtracting the first vector from each of the other two preserves the basis property by iii. The resulting family isIts last two vectors belong to the plane with first coordinate zero. By b, they are linearly independent there precisely whenIn that case their two-dimensional span is the whole plane, and adjoining $e_1$ gives a basis of ${\mathbb{R}}^3$. If the displayed expression vanishes, a nontrivial linear relation among those two vectors prevents a basis. Thus