Let $S,L\subset V$, both finite. If $\textrm{span}(S)=V$ and $L$ is LI, $|S| \geq |L|$. (Artin's proof and alternative proof.)

96 Views Asked by At

Proposition (3.16) of Artin's Algebra book (1st edition, page 92) claims that the order of a subset $L$ of a vector space $V$, which contains only linearly independent (LI) vectors can't be greater than the order of another subset $S$ that spans $V$.

The first part of my question is about the proof himself writes, and the second part is about a proof that I attempted to get.

Part 1: Artin's proof

Artin reduces the task of finding a dependence relation of the vectors in $L$ by finding out when the system of equations

$\sum_{i=1}^{m}\sum_{j=1}^{n}c_ja_{ij}v_i =0$, where $\sum_{j=1}^{n}c_ja_{ij}$ are the coefficients. In matrix form:

$$\begin{bmatrix} \vert & \vert& & \vert\\ v_1 & v_2 &...& v_m\\ \vert & \vert& &\vert\end{bmatrix} \begin{bmatrix} \sum_{j=1}^{n}c_ja_{1j} \\ \sum_{j=1}^{n}c_ja_{2j}\\ \vdots \\\sum_{j=1}^{n}c_ja_{mj}\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\\vdots\\0\end{bmatrix}$$

He states that this is a system of $m$ unknowns and $n$ equations, where $n$ is the order of $L$.

What I cannot understand is why this system has $n$ equations. What I can see, in not very abstract terms, is that the number of equations is the length of the vectors $v_i$ (i.e. the dimension of $V$).

Part 2: Alternative proof

Let $m$ be the order of $S$, so it contains $m$ vectors. Among these $m$ vectors, only a number $k\leq m$ will be LI.

By virtue of proposition (3.10) in the same book, the span generated by that subset of $k$ LI vectors (call it $\tilde{S}$) is equal to the span generated by the whole set $S$, namely the whole vector space $V$.

This means that any vector in $L\subset V$, is also in $\textrm{span}(\tilde{S})$,so it can be expressed as a linear combination of these $k$ independent vectors. No more than $k$ LI vectors can be obtained by combining $k$ LI vectors: can get the same set of $k$ vectors and then linear combinations of them. Then the order of $L$ can be at most the number of LI vectors in $S$, and thus no greater than the order of $S$ itself.

If possible I'd like to ask for a clarification of the last statement in the proof of part 1, and for a check of part 2 being correct.