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Let $V$ be a vector space in $F$, and suppose that $V$ has a finite spanning set $S=\{v_1,\ldots,v_n\}$. Show that if $T=\{u_1,\ldots,u_m\}$ is a linearly independent subset of V, then $m\leq n$. (We are not assuming $T$ is a subset of $S$)

I have tried going about this using contradiction, but I'm unclear as to whether that is enough or how to start from there. Thanks in advance

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    I think the title of this question can be improved...2012-09-13

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Hint: Since $S$ is a spanning set, you can write $u_i=\sum\limits_{j=1}^n\lambda_{ij}v_j$ for some $\lambda_{ij}\in F$. If $m>n$, can you find a linear dependence among these elements?

Hint 2: We have $u_1=\sum\limits_{j=1}^n\lambda_{1j}v_j$ and $u_2=\sum\limits_{j=1}^n\lambda_{2j}v_j$. If $\lambda_{21}\neq 0$, then $u_1-\frac{\lambda_{11}}{\lambda_{21}}u_2=0v_1+x_2v2+\cdots + x_nv_n$ while otherwise $u_2=0v_1+x_2v_2+\cdots + x_nv_n$ for some $x_2,\ldots,x_n\in F$. Either way, we've found a linear combination of the first $2$ elements of $T$ which eliminates $v_1$. Proceeding in this manner, what do we get using the first $n+1$ elements of $T$?

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    @tkrm I edited to include another hint.2012-09-13
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You can select a basis from $S$ and you can extend $T$ a basis. These basises have the same length.