Prove that $\mathbb{R^3}$ is not a vector space Consider $\mathbb{R^3}$ with the usual addition $+$ of vectors, but with scalar multiplication $\otimes$ defined by:
$k$ $\otimes$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ = $\begin{bmatrix} x \\ ky \\ 2kz \end{bmatrix}$, $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ $\in$ $\mathbb{R^3}.$
Show that $\mathbb{R^3}$ together with the operations + and $\otimes$ is not a vector space.
My attempt:
A property of a vector space is $(-1)$ $\otimes$ $\mathbf{u}$ = $\mathbf{-u}$
So, applying this would result to:
$(-1)$ $\otimes$ $(x, y, z)$ = $(x, -y, -2z)$.
To fulfill the property, this must be equal to $\mathbf{-u}$.
So, $(x, -y, -2z)$ = $-(x, y, z)$ = $(-x, -y, -z)$.
But, $(x, -y, -2z)$ $\ne$ $(-x, -y, -z)$ for all $x$.
Hence, it is not a vector space.
Is my attempt enough to prove it isn't a vector space? Or at least, correct? Should I show further proof?

An idea to simplify things: it's easy to show that in any vector space if must be that $\;1\cdot v= v\;$, for any (supposed) vector $\;v\;$ , but here
$$1\cdot\begin{pmatrix}x\\y\\z\end{pmatrix}:=\begin{pmatrix}x\\y\\2z\end{pmatrix}\neq\begin{pmatrix}x\\y\\z\end{pmatrix}$$
and thus we're done.
