# Prove that V contains a subspace of each dimension

Let $V$ be an $n$-dimensional vector space over $\mathbb R$. Prove that $V$ has a subspace of dimension $r$ for each $0 \le r \le n$.

Is this as simple as saying that $V$ has a basis of $n$ elements, and then take $r$ elements from this basis and this will generate an $r$-dimensional subspace? Or am I missing something? It seems too simple.

Thanks

• you are correct. There is nothing difficult about this. Just make sure to give a good argument why taking $r$ elements from a basis will give you a subspace of dimension $r$. Also, notice my editing to your question to correctly type-set the maths. – Ittay Weiss Jun 2 '13 at 10:21
• @Wooster Your thoughts are correct. Could it be that that problem was set before you learned basis and all that? – Git Gud Jun 2 '13 at 10:21
• Okay, @GitGud, this is quite possible, I think it probably was! Thanks! – Wooster Jun 2 '13 at 10:22
• The question is interesting, if you're not allowed to use any basis. What you need to do then is somehow show that every vector space has a codimension 1 linear subspace and use this inductively. – HSN Jun 2 '13 at 13:35

For sake of having an answer, yes, if $\{v_1,v_2,\ldots,v_n\}$ is a basis of $V$, then $\operatorname{span}\{v_1,v_2,\ldots,v_r\}$ is an $r$-dimensional subspace of $V$ for every $0\le r\le n$ (with $\operatorname{span}\emptyset=\{0\}$ by convention).