0
$\begingroup$

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
  2. Be closed under addition
  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 $\mathbb{R}^3$.

$\endgroup$
1
  • $\begingroup$ You can just prove its basis is linearly independent (is a basis) or not. $\endgroup$ Commented Nov 13, 2016 at 18:41

1 Answer 1

2
$\begingroup$

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.

Remark about your approach, to improve clarity, when we want to show counter example, sometimes specifying the values of the variables would make things clearer.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .