If $V$ is a vector space and $v_1, v_2, . . . , v_n \in V$
span $V$, and $u_1, u_2, . . . , u_m ∈ V$ are linearly independent, then $m\le n$.
Use this to prove that if $V$ has dimension $n$ and $u_1, u_2, . . . , u_n \in V$ are linearly independent
then $u_1, u_2,\le, u_n$ form a basis of $V$.
Do I prove that $V$ has a basis with n elements? Not really sure how to approach this proof.
Best Answer
You need to show that $u_1,\dotsb,u_n$ span $V$. The easiest way to do this is to suppose that there is a vector $v\in V$ which is not in the span of $u_1,\dotsb,u_n$. Then $u_1,\dotsb,u_n,v$ is a linearly independent set. Then apply your result on the size of linearly independent sets. This will lead to a contradiction and therefore no such $v$ can exists