Linear Dependence Lemma

theoremvector-spaces

Intuition

Formal Statement

Suppose $v_1,\dots ,v_m$ is a linearly dependent list in $V$. Then there exists $k\in \{1,\dots ,m\}$ such that $v_k\in \mathrm{span}(v_1,\dots ,v_{k-1})$. Moreover, $\mathrm{span}(v_1,\dots ,v_{k-1},v_{k+1},\dots ,v_m)=\mathrm{span}(v_1,\dots ,v_m)$.

Proof

Since the list is linearly dependent, there exist $a_i\in \mathbb{F}$ (not all $0$) such that, $0=a_1{v}_1+\cdots +a_m{v}_m$. Let $k$ be the largest element of $\{1,\dots ,m\}$ such that $a_k\ne 0$.

Then, $v_k=-\frac{a_1}{a_k}{v}_1-\cdots -\frac{a_{k-1}}{a_k}{v}_{k-1}$, so $v_k\in \mathrm{span}(v_1,\dots ,v_{k-1})$.

Now suppose $k$ is as above. Let $b_1,\dots ,b_{k-1}\in \mathbb{F}$ be such that $v_k=b_1{v}_1+\cdots +b_{k-1}{v}_{k-1}$. Suppose $u\in \mathrm{span}(v_1,\dots ,v_m)$, then $u=c_1{v}_1+\cdots c_m{v}_m=(c_1+c_k{b}_1)v_1+\cdots +(c_{k-1}+c_k{b}_{k-1})v_{k-1}+c_{k+1}{v}_{k+1}+\cdots +c_m{v}_m$, so $u\in \mathrm{span}(v_1,\dots ,v_{k-1},v_{k+1},\dots ,v_m)$.

Corollaries