# Thread: In need of someone to check my work.

1. ## In need of someone to check my work.

So I just wanted to make sure I'm correct in my thinking for this problem.

I'm trying to find a basis for the subspace W = { x in $R^{3} | x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, 2x + y = 0$ }

With this, we know x = -y /2 so therefore, a basis to the subspace =

$\begin{bmatrix} -1/2 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

is my work correct?

2. Looks good.

This subspace consists of all vectors of the form $\left( \begin{array}{c} \frac{-1}{2}y\\y\\z\end{array}\right)=y\left( \begin{array}{c} \frac{-1}{2}\\1\\0\end{array}\right) +z\left( \begin{array}{c} 0\\0\\1\end{array}\right)$,

so the vectors $\left( \begin{array}{c} \frac{-1}{2}\\1\\0\end{array}\right)$ and $\left( \begin{array}{c} 0\\0\\1\end{array}\right)$ span the subspace. Since they are obviously linearly independent, they form a basis for the subspace.

3. Originally Posted by topspin1617
Looks good.

This subspace consists of all vectors of the form $\left( \begin{array}{c} \frac{-1}{2}y\\y\\z\end{array}\right)=y\left( \begin{array}{c} \frac{-1}{2}\\1\\0\end{array}\right) +z\left( \begin{array}{c} 0\\0\\1\end{array}\right)$,

so the vectors $\left( \begin{array}{c} \frac{-1}{2}\\1\\0\end{array}\right)$ and $\left( \begin{array}{c} 0\\0\\1\end{array}\right)$ span the subspace. Since they are obviously linearly independent, they form a basis for the subspace.
My thoughts exactly. Thanks a lot

4. By the way, if you don't like fractions you could also say that y= -2x so that
$\begin{pmatrix}1 \\ -2 \\ 0\end{pmatrix}$ and $\begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}$ are basis vectors.