# V is a linear space of n dimensions, show that V has subspaces of every dimension between 0 and n. [duplicate]

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.

• $V$ has a basis $(e_1,\ldots,e_n)$. Define $U_i=span\{e_1,\ldots,e_i\}$. Then $\dim U_i=i$. – Dietrich Burde Oct 19 '16 at 21:48
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 $\dim V_k=k$.