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I need help making progress on a linear algebra question.

Consider the subset $W = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}$ such that $x_1 + x_2 + x_3 = 2$

Does $x = \begin{bmatrix} 1 \\ 0\\ 1\\ \end{bmatrix} \in W?$

Does $u = \begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix} \in W?$

Does $v = \begin{bmatrix} 2\\ 1\\ -1\\ \end{bmatrix} \in W?$

My thoughts:

I believe that $x$ and $v$ are in in $W$ because they satisfy the condition. I believe $u$ is not in $W$ because it does not satisfy the condition.

I am not sure whether or not this is correct.

I also need to determine whether or not $W$ is a subspace of $\mathbb{R}^3$ which I do not know how to begin

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    You are right that $x,v \in W$ but $u \notin W$. Your thinking is correct for why as well. One way to see why $W$ is not a subspace is by taking your $x,v \in W$ and adding them to form $x+v$. Is it true that $x+v \in W$? – 2017-02-20
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    no it is not true – 2017-02-20
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    Then $W$ cannot be a subspace, for it is not closed under addition. It isn't closed under scalar multiplication as well (this is easy to check by showing $2x \notin W$). But the answer below, that $0 \notin W$ is a very strong reason why $W$ is not a subspace. – 2017-02-20

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A vector $(x_1, x_2, x_3)$ would be in $W$ if and only if $x_1+x_2+x_3=2$. So, as you said, $x\in W$, $u\notin W$ and $v\in W$.

Now, check your definition of subspace. Every subspace must contain the zero vector, which is $(0,0,0)$ in this case. But it doesn't satisfy the equation $x_1+x_2+x_3=2$, so $W$ isn't a subspace.

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    You're welcome~ – 2017-02-20