# Proof that a set of vectors is not in a subspace

I'm attempting to prove whether a set of vectors, $H = \left\{\begin{bmatrix}2x \\ 3y^2 \\ x^2-y\end{bmatrix} : x, y \in \mathbb{R}\right\}$ is or is not a subspace of $\mathbb{R}^3$. I think I have the basics correct, but I feel as if my proof leaves a lot to be desired.

My textbook tells be that for $H$ to me a subspace of $\mathbb{R}^3$, it must

1. Contain the zero vector
3. Be closed under scalar multiplication

Intuitively, I'm pretty sure that while it contains the zero vector it is not closed under addition. Here's my attempt at a proof, please let me know where I'm correct/mistaken/unnecessarily verbose/redundant etc.

let $\bf{A} = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} + \begin{bmatrix}y_1 \\ y_2 \\ y_3\end{bmatrix} = \begin{bmatrix}x_1 + y_2 \\ x_2 + y_2 \\ x_3 + y_3\end{bmatrix}$ where ${\bf{x},{y}} \in H$

Since the vectors $\bf{x, y}$ are in set $H$,

$x_3 = \frac{x_1}{2}^2 - \frac{x_2}{3}^\frac{1}{2}$

$y_3 = \frac{y_1}{2}^2 - \frac{y_2}{3}^\frac{1}{2}$

And if vector $\bf{A}$ is in set $H$, then

$x_3 + y_3 = (\frac{1}{2}(x_1 + y_1))^2 - (\frac{1}{3}(x_2 + y_2))^\frac{1}{2}$

Substituting $x_3$ and $y_3$,

$\frac{x_1}{2}^2 - \frac{x_2}{3}^\frac{1}{2} + \frac{y_1}{2}^2 - \frac{y_2}{3}^\frac{1}{2} \neq (\frac{1}{2}(x_1 + y_1))^2 - (\frac{1}{3}(x_2 + y_2))^\frac{1}{2}$

Therefore ${\bf{A}} \notin H$, and $H$ is not closed under addition. Therefore $H$ is not a subspace of $\mathbb{R}^3$.

• You can just prove its basis is linearly independent (is a basis) or not. – Jacob Wakem Nov 13 '16 at 18:41

A quicker alternative, $\begin{bmatrix} 0 \\ 3 \\ -1 \end{bmatrix} \in H$ but $\begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix} \notin H$ as the second coordinate has to be nonnegative. Hence it is not closed under scalar multiplication.