Consider $$V = V_1 \times V_2 = \{({v_1,v_2})~|~ {v_1} \in V_1,{v_2} \in V_2\}$$ of two vector spaces $V_1,V_2$ over the same field $F$. (Note $V$ is a vector space as well). If we do the identification $$V_1 = V_1 \times \{{0_{V_2}}\} = \{({v_1,0_{V_2}})~|~{v_1} \in V_1\}$$

$$V_2 = \{{0_{V_1}}\} \times V_2 = \{({0_{V_1},v_2)}~|~{v_2} \in V_2\}$$

Show that $V= V_1 \oplus V_2$.

Please help me check if my logic for showing that the intersection is the $0$ vector is correct. And i am confused in this particular question, what is the zero vector anyway?? By how they define it, $0_{V_1}$ is the zero vector of $V_1$, $0_{V_2}$ is the zero vector of $V_2$, but in the end i showed that the intersection is $(0_{V_1},0_{V_2})$ which isnt exactly same to $0_{V_1}$ or $0_{V_2}$

Let $x \in V_1 \cap V_2$, then $x \in V_1$ and $x \in V_2$. Hence $x = (v_1,0_{V_2})$ for some $v_1 \in V_1$ and $x = (0_{V_1},v_2)$ for some $v_2 \in V_2$. Since both $x$ is in the intersection, the $x$ must be a vector in the form of both $V_1,V_2$. So $$x = (v_1,0_{V_2}) = (0_{V_1},v_2)$$ for some $v_1 \in V_1, v_2 \in V_1$.

However $v_1$ is forced to be equals to $0_{V_1}$ only and $v_2$ is forced to be equals to only $0_{V_2}$ as well because the simplest way is to say that the $0$ vector of any vector space is unique.

Or can i also say suppose not, the $v_1$ inside the intersection does not satisfy $v_1 \neq 0_{V_1}$, then the problem will be that $(v_1,0_{V_2})$ will definitely not be in $V_2$ since all elements in $V_2$ must exhibit the form of $(0_{V_1},v_2)$ where $0_{V_1}$ is fixed. And subsequently it contradicts the fact that $(v_1,0_{V_2})$ is in $V_2$ as well.

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    $\begingroup$ By the identification, $(0_{V_1},0_{V_2}) = 0_{V_1} = 0_{V_2}$. $\endgroup$ – Kenny Lau Aug 28 '17 at 12:14
  • $\begingroup$ The zero vector is indeed $(0_{V_1}, 0_{V_2})$. $\endgroup$ – Shaun Aug 28 '17 at 12:22

The zero of the vector space $V$ is the element $(a,b)\in V=V_1\times V_2$ that is the identity element of the group $(V,+)$, i.e such that

$$\forall (x,y)\in V,(a,b)+(x,y)=(x,y).$$

But $(a,b)+(x,y)=(a+x,b+y)$ so you see that $a=0_{V_1}$ and $b=0_{V_2}$. Hence the zero of $V$ is $(0_{V_1},0_{V_2})$.

Now, under your given identification, your argument is correct and proves that $V_1\cap V_2=\{0\}$, but you can simplify it: you said that, as $x\in V_1$ and $x\in V_2$, then $x=(v_1,0_{V_2})$ for some $v_1\in V_1$ and $x=(0_{V_1},v_2)$ for some $v_2\in V_2$. After that, you don't need to argue again that $x$ belongs to both of them. You have $x=(0_{V_1},v_2)$ and $x=(v_1,0_{V_2})$. Thus $(0_{V_1},v_2)=(v_1,0_{V_2})$. But you know that

$$\forall a,b,c,d,(a,b)=(c,d)\iff (a=c\quad\text{and}\quad b=d).$$

This means that $v_1=0_{V_1}$ in particular. Then $x=(v_1,0_{V_2})=(0_{V_1},0_{V_2})=0_V$.

Note that all you showed is that $V_1\cap V_2=\{0\}$. But $V=V_1\oplus V_2$ means two things: $V_1\cap V_2=\{0\}$ and $V=V_1+V_2$. It shouldn't be hard.


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