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

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 $$\dim(W_1) \le \dim(W_2)$$.

• 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 $n$. Aug 19, 2019 at 13:58

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 $$W_2$$.
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 $$W_1 = W_2$$).