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

  • 5
    $\begingroup$ 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


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 $W_2$.


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 $W_1 = W_2$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.