Exercise 24

Here is an alternative proof of this section’s main result, Theorem 2.4. First is an example, then a lemma, then the theorem.

• Express this vector from $\mathbb{R}^{3}$ a as a linear combination of members of the basis.

  $$B=\langle\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}1\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 0\\ 2\end{pmatrix}\rangle\qquad\vec{v}=\begin{pmatrix}1\\ 2\\ 0\end{pmatrix}$$

• In that combination pick a basis vector with a non-zero coefficient. Alter $B$ by exchanging $\vec{v}$ for that basis vector, to get a new sequence $\hat{B}$. Check that $\hat{B}$ is also a basis for $\mathbb{R}^{3}$.

• (Exchange Lemma) Assume that $B=\langle\vec{\beta}_{1},\dots,\vec{\beta}_{n}\rangle$ is a basis for a vector space, and that for the vector $\vec{v}$ the relationship $\vec{v}=c_{1}\vec{\beta}_{1}+c_{2}\vec{\beta}_{2}+\cdots+c_{n}\vec{\beta}_{n}$ has $c_{i}\neq 0$. Prove that exchanging $\vec{v}$ for $\vec{\beta}_{i}$ yields another basis for the space.

• Use that, with induction, to prove Theorem 2.4.