Proof verification: if $W_1 \subseteq W_2$ then $\dim(W_1) \le \dim(W_2)$

Well, this is a consequence of the famous Steinitz exchange lemma:

If $\{v_{1},\dots ,v_{m}\}$ is a set of $m$ linearly independent vectors in a vector space V, and $\{w_{1},\dots ,w_{n}\}$ spans $V$, then $m\leq n$ and after reordering $\{v_{1},\dots ,v_{m},w_{m+1},\dots ,w_{n}\}$ spans $V$.

Here you take $\{v_{1},\dots ,v_{m}\}$ as a basis of $W_1$ and $\{w_{1},\dots ,w_{n}\}$ as a basis of $W_2$, where $W_1$ is a subspace of $W_2$.

Tags:

Linear Algebra

Vector Spaces

Proof Verification

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