Prove that the set of bases is linearly independent Suppose that $W$ and $W'$ are subspaces of the vector space $V$ with the property that $W\cap W'=\{0\}$, 
and suppose that $\beta$ is a basis for $W$ and $\beta'$ is a basis for $W'$.  Prove that the set $\beta\cup\beta'$ is linearly independent.
What I have/know so far:
Suppose $\beta$=($v_{1}$, $v_{2}$,..., $v_{n}$) and $\beta'$= ($v_{1}$, $v_{2}$,..., $v_{n}$). 
So $\beta\cup\beta'$= ($a_{1}v_{1}$, $a_{2}v_{2}$,..., $a_{n}v_{n}$)+ ($a'_{1}v_{1}$, $a'_{2}v_{2}$,..., $a'_{n}v_{n}$)=$0$. We know that $W\cap W'=\{0\}$ (I am told that this is important)... We must show that $a_{1}=a_{2}=....a_{n}= a'_{1}=a'_{12}=...a'_{n}=0.$
Not sure what else to do here.
 A: Suppose $\beta=\{v_1,\ldots, v_n\}$ and $\beta'=\{v_1',\ldots, v_m'\}$.  Note that these are sets of vectors, and they need not have the same size.  Suppose now we have a linear combination of $\beta\cup\beta'=\{v_1,\ldots,v_n,v_1',\ldots, v_m'\}$ that equals zero, i.e. some numbers $a_1, \ldots, a_n, a_1',\ldots, a_m'$ such that 
$$0=a_1v_1+\cdots + a_nv_n+a_1'v_1'+\cdots +a_m'v_m'$$
The trick here is to move all the terms from $W'$ to the other side, getting $$a_1v_1+\cdots +a_nv_n=-a_1v_1'-\cdots -a_m'v_m'$$
Now, the left hand side is a linear combination of $\beta$, hence is in $W$.  The right hand side is a linear combination of $\beta'$, hence is in $W'$.  But their intersection is only the zero vector, hence both LHS and RHS equals zero.  But now we have $a_1v_1+\cdots +a_nv_n=0$; since $\beta$ is independent in fact $a_1=a_2=\cdots=a_n=0$.  Similarly, $a_1'v_1'+\cdots+a_m'v_m'=0$;  since $\beta'$ is independent in fact $a_1'=\cdots=a_m'=0$.
Hence we have proved that all of the coefficients in our linear combination were in fact 0; hence $\beta\cup\beta'$ must be independent.
A: Let $\beta=\{v_1,\dots,v_s\},\beta'=\{v_1',\dots,v_t'\}$. Suppose $c_1v_1+\dots+c_sv_s+c_1'v_1'+\dots+c_t'v_t'=0$, then $c_1v_1+\dots+c_sv_s=-(c_1'v_1'+\dots+c_t'v_t')=0$ as the intersection of two spaces is $\{0\}$. Hence, all $c_i,c_i'$ are $0$.
