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I think I need to be using induction in this case. i prove the subspace with dimension 0, then assume there is a subspace of k dimensions and then prove that k+1 holds under addition/scalar multiplication aswell.

any hints/advice?

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    $\begingroup$ $V$ has a basis $(e_1,\ldots,e_n)$. Define $U_i=span\{e_1,\ldots,e_i\}$. Then $\dim U_i=i$. $\endgroup$ – Dietrich Burde Oct 19 '16 at 21:48
  • $\begingroup$ @DietrichBurde: Please don't answer in the comment section. $\endgroup$ – Martin Argerami Oct 19 '16 at 22:50
  • $\begingroup$ I didn't know it was a duplicate. And if you knew it was, why give a hint? Regarding answers in the comment section, you are definitely not the only one, if it is any consolation to you; it is a huge problem with this site, that leaves thousands of questions unanswered. $\endgroup$ – Martin Argerami Oct 20 '16 at 13:39
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You don't need anything that complicated. You will obtain a subspace of dimension $k$ if you take the span of $k$ linearly independent vectors.

So, if you fix a basis $\{v_1,\ldots,v_n\}$, you can form subspaces $$ V_k=\text{span}\,\{v_1,\ldots,v_k\} $$ and you will have $\dim V_k=k$.

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