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If $W=\{x \in \mathbb R^4|x_3=x_1+x_2,x_4=x_1-x_2\}$ show that $W$ is or is not a subspace

I would imagine that vector $x = (a,b,c,d)$ and to show that something is a subspace it has to be closed under addition and scalar multiplication

Closed under multiplication

$(kx) \in W$, therefore, it is closed under scalar multiplication

Closed under addition

let $y \in W$ as well where $y=(a_1,b_1,c_1,d_1)$, then $x+y=(a+a_1,b+b_1,c+c_1,d+d_1)$ and $x-y=(a-a_1,b-b_1,c-c_1,d-d_1)$

if we let $\alpha=x+y$ and $\beta=x-y$, then $\alpha+\beta=2X$ which is true

Since it is closed both under addition and multiplication I conclude it is a subset.

But I feel that I am making an error in this verification and would like some help

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  • $\begingroup$ You might elaborate, along the lines of the answers below, on how $x \in W \Rightarrow kx \in W$. $\endgroup$ Commented Jul 1, 2020 at 3:17

2 Answers 2

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I think for addition you have to do this: given $x=(x_1, x_2, x_1+ x_2, x_1-x_2)$ and $y=(y_1, y_2, y_1+ y_2, y_1-y_2)$. Then $$x+y=(x_1+y_1, x_2+y_2, x_1+x_2+y_1+y_2, x_1-x_2+y_1-y_2)\\=(x_1+y_1, x_2+y_2, (x_1+y_1)+(x_2+y_2), (x_1+y_1)-(x_2+y_2))\in W$$

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  • $\begingroup$ ohhh i see i interpreted x1-4 as separate vectors $\endgroup$
    – John Rawls
    Commented Jul 1, 2020 at 3:17
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You haven't shown closure under addition. I'm going to change your terminology.

Assume $a=(a_1, a_2, a_3, a_4), b= (b_1, b_2, b_3, b_4) \in W$. Then by the definition of $W$, we have:

$$a_3=a_1+a_2, a_4=a_1-a_2, b_3=b_1+b_2, b_4=b_1-b_2.$$

Note that $a+b=(a_1+b_1, a_2+b_2, a_3+b_3, a_4+b_4)$. Then

$$(a_1+b_1)+(a_2+b_2)=(a_1+a_2)+(b_1+b_2)=a_3+b_3$$ $$(a_1+b_1)-(a_2+b_2)=(a_1-a_2)+(b_1-b_2)=a_4+b_4.$$

Therefore $a+b \in W$.

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