Different proof for $W \le V \to \dim(W) \le \dim(V)$

Question

Larson Edwards Falvo - Elementary Linear Algebra

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Different approach:

Let $\dim(V) = n$. If $\dim(W) > n$, then a basis for $W$ has more than $n$ vectors, contradicting either of the theorems below. QED

Is that wrong? Am I doing some kind of catch-22 here?

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Answer 1

Using Theorem 4.10 will work very well. It shows that a basis of $W$ can't be bigger than a basis of $V$.

Theorem 4.11 doesn't really put you in a much better position than you started in. You have to know that you can extend a basis of $W$ to a basis of $V$. If you know that then you don't need the theorem anyway.

Answer 2

Both proofs are correct. They are actually using the same principle. The proof above is using a direct method and the second one is using contradiction, the argument is exactly the same.

They are both implicitly using the fact that if $ w_1, w_2, \cdots w_n $ are elements of $W$ then necessarily $span(\{ w_1, w_2, \cdots, w_n \}) \subseteq V $