Let V be a vector space and let Z be a subspace of V.

Prove that every basis for Z is a subset of a basis for V.

I was attempting to prove that every element of a basis for Z is also in a basis for V. But I'm stuck in my thoughts. Here is what I have,

Let B={v$_1$, v$_2$, ..., v$_n$} be a basis for Z and let A={u$_1$, u$_2$, ..., u$_n$}. Then a vector z $\in$ Z can be represented as a linear combination of basis B and some scalars a $_i$$\in$ F (i=( 1, 2, ... , n).

z = a$_1$v$_1$ + a$_2$v$_2$ + ... + a$_n$v$_n$

I know that since B and A are a basis for Z and V respectively then the linear combination of the basis is linearly independent and span(B)=Z and span(A)=V, but I'm not really sure use that. Thank you in advance for your help.

  • 2
    $\begingroup$ Start with a basis for $Z$, and show that you can "extend" it to a basis for $V$ by adding new elements in a certain way. $\endgroup$ – angryavian May 4 '17 at 6:38
  • $\begingroup$ Is $V$ finite-dimensional? $\endgroup$ – 5xum May 4 '17 at 6:39
  • $\begingroup$ Yes, V is finite dimensional. Although I think I got it after @angryavian comment. I completely forgot that I am able to do that. Thank you both for the quick replies $\endgroup$ – Niko L May 4 '17 at 6:43

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