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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.

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  • $\begingroup$ Hello, there are some problems with your notation. I don't know why you wrote $\beta = a_{1}v_{1}, a_{2}v_{2}, \dots, a_{n}v_{n} = 0$. In math, when writing proofs, you should write them as though you are telling a story. Here is how I would have started your proof out: "We know since $\beta$ is a linearly independent set that if $a_{1}v_{1} + a_{2}v_{2} + \dots + a_{n}v_{n} = 0$, then it must be that $a_{1} = a_{2} = \dots = a_{n} = 0$." See how the way I started the proof allows the reader to understand where I am going, and what I understand linearly independent to mean? $\endgroup$
    – layman
    Apr 15 '15 at 21:58
  • $\begingroup$ In contrast, the first line of your proof is just math with no words, and the math seems incorrect to me, and it's hard to understand how you are starting because the way you wrote it isn't like a story. Sorry for rambling. $\endgroup$
    – layman
    Apr 15 '15 at 21:59
  • $\begingroup$ No, you are right. I wrote that wrong. $\endgroup$
    – EmaLee
    Apr 15 '15 at 22:00
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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.

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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 $0$.

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