Is $\{(x,y,z) \in \mathbb{R}^3: 2x+3y+yz=0\}$ a linear subspace of $\mathbb R^3$

Question

I dont know what to do

Check if $$\big\{(x,y,z) \in \mathbb{R}^3: 2x+3y+yz=0 \big\}$$ is a linear subspace of $\mathbb{R}^3$.

so: $$z=\frac{-2x-3y}{y} = \frac{-2x}{y} -3$$

$$(x,y,z) = (x,y,\frac{-2x}{y} -3)$$

is that right? if yes what then?

Answer 1

It is not a linear subspace of $\mathbb{R}^3$! For instance, $(-1,2,1)$ belongs to it. But $2(-1,2,1)=(-2,4,2)$, which does not belong.

Answer 2

The linear subspaces of $\mathbb R^3$ are $\{0\}$ and lines and planes passing through origin.

$\begin{cases}f(x,y,z)=2x+3y+yz \\\\S=\{(x,y,z)\in\mathbb R^3\mid f(x,y,z)=0\}\end{cases}$

$f(0,0,0)=0\iff \{0\}$ belongs to the set $S$ so this condition is fulfilled.

But because of the term in $\mathbf{yz}$ which is not linear, this is neither a point nor a line nor a plane, so it cannot be a linear subspace.

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Let show in details why $f$ is not a linear function (in case you are not convinced by the above argument about $yz$).

• let $v=(x,y,z)\in S$ and let check if $\alpha v$ belongs to $S$.

$f(\alpha v)=2\alpha x+3\alpha y+\alpha^2 yz=\alpha\underbrace{(2x+3y+yz)}_{f(v)=0}-\alpha yz+\alpha^2 yz=\alpha(\alpha-1)yz$

This is not always $0$.

For instance $v=(0,1,-3)\in S$ and $\alpha=2$ then $f(v)=0$ but $f(\alpha v)=-6$

so $(0,2,-6)\notin S$

• another possibility is to check $f(u+v)\neq f(u)+f(v)$

$f(u)+f(v)-f(u+v)=2x+3y+yz+2a+3b+bc-2(a+x)-3(b+y)-(b+y)(c+z)=-bz-yc$

This is not always $0$.

For instance $u=(0,0,1)\in S$ and $v=(3,-2,0)\in S$ but $u+v=(3,-2,1)\notin S$

because $f(u+v)=-2$.