# Prove that $W = \{(x,y,z,r) : r=5x+2y+7z\}$ is a subspace of $\mathbb{R}^4$

Prove that $W = \{(x,y,z,r) : r=5x+2y+7z\}$ is a subspace of $\mathbb{R}^4$

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## 3 Answers

To show that a set $W$ is a vector subspace of $\mathbb{R}^4$, it suffices to show that:

1. $0\in W$
2. $W$ is closed under addition and
3. $W$ is closed under scalar multiplication.

1 is easy to show.

For 2,

If $a=\left(\begin{matrix}a_{1}\\a_{2}\\a_3\\ 5a_1+2a_2+7a_3\end{matrix}\right), \quad b=\left(\begin{matrix}b_{1}\\b_{2}\\b_3\\ 5b_1+2b_2+7b_3\end{matrix}\right)\in W$ then $a+b=\left(\begin{matrix}a_{1}+b_1\\a_{2}+b_2\\a_3+b_3\\ 5(a_1+b_1)+2(a_2+b_2)+7(a_3+b_3)\end{matrix}\right)\in W$

Something similar for 3.

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 Turns out that jim deleted his answer. – Cameron Buie Mar 6 at 14:52

Elements of $W$ are the solutions to the (homogeneous) equation $5x+2y+7z-r=0$. So, you can prove (or use) the general fact: the solutions of a homogeneous linear system form a subspace.

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Hint: Prove that $f\colon \mathbb{R}^4\to \mathbb{R},(x,y,z,r)\mapsto 5x+2y+7z-r$ is a linear application. What's its kernel?

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