Mistake in proof about vector spaces

The error is the following. Theorem 6.10(f) says that if you have a set that spans $V$, you can remove elements ("reduce") until you get a basis. In the proof of 6.11, the elements in the basis of $V$ may not belong to $W$, so the argument does not apply.

For example, maybe $V=\mathbb R^3$, and you take the canonical basis $\{(1,0,0),(0,1,0),(0,0,1)\}$. Then if $W=\{(x,y,z): x=y=z\}$, none of the three elements in the basis belongs to $W$.

The usual argument is that a basis of $W$ will be linearly independent, and so it will also be linearly independent in $V$, and it follows that $\dim W\leq \dim V$.

Your professor is correct.

The problem is with the word " reduced"

Let us look at an example.

Let W be the subspace of $\mathbb {R}^2$ spanned by $\{(1,1)\}.$

Let $ B=\{(1,0),(0,1)\}$ be a basis for $\mathbb {R}^2$

Then $B$ spans $W$, but we can not reduce $B$ to a basis for $W$

Maybe there is a difference in use of terminology here but I would say that the problem is with the statement "$B$ spans $W$".

The way I have always seen the term used is as follows.

• We define ${\rm span}(B)=\{\hbox{linear combinations of vectors in $B$}\}$.
• Then "$B$ spans $W$" means that ${\rm span}(B)$ equals $W$.

In the proof you have quoted, they appear to be saying that "$B$ spans $W$" means that ${\rm span}(B)$ contains $W$.

So it is a question of using the precise meaning of terminology. BTW, I would not agree with your professor that it is a "very subtle" error, IMHO it is a blatant error :)