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Let $W_1$ and $W_2$ be subspaces of vector space $V$. Prove that If $W_1 \subseteq W_2$ then $\dim(W_1) \le \dim(W_2)$.

My proof: Let $v_1, ...,v_n$ be base vectors of vector space $W_1$ and $W_1 \subseteq W_2$. Then $\dim(W_1) = n$. From the assumptions we have that $v_1, ..., v_n \in W_2$ and because they are base vectors in $W_1$ they are lineary independent. Therefore if $\operatorname{span}(v_1, ...,v_n) = W_2$ then $\dim(W_2) = n = \dim(W_1)$. Otherwise there exists vector $v_{n+1} \in W_2$ such that $v_{n+1}$ is not linear combination of $v_1, ...,v_2$ and therefore $\dim(W_2) \ge n+1$. Hence $\dim(W_1) \le \dim(W_2)$.

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    $\begingroup$ Looks ok, but you could've just left it at $v_1,\ldots,v_n$ being linearly independent in $W_2$, hence $\dim (W_2)$ is at least $n$. $\endgroup$
    – take008
    Aug 19, 2019 at 13:58

2 Answers 2

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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 $W_2$.

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you can just say: Let $\{w_1, \ldots, w_n\}$ be a base for $W_1$, and let $\{w_1, \ldots, w_m \}$ be a base for $W_2$.

Because $W_2 \subseteq W_1 \to Sp(W_2) \subseteq Sp(W_1) \to \{w_1, \ldots, w_m \} \subseteq \{w_1, \ldots, w_n \}$, hence $m \le n$

(if $W_1 \subseteq W_2 \to m=n$, and $W_1 = W_2$).

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