Step $2.$ Now we show that $T$ is closed under vector addition. Suppose that $v_1$ and $v_2$ belong to

We want to show that $v_1+v_2$ also belongs to $T.$ To show this, firstly note that we know these vectors must be of the form

$v_1=s([3],[-1],[2])$ and $v_2=t([3],[-1],[2])$

for some $s,$tinR. So we have

$v_1+v_2=s([3],[-1],[2])+t([3],[-1],[2])=([3],[-1],[2])$

where $($in terms of $s$ and $t)$

$=$

Thus $v_1+v_2$ is of the form required, and so we have shown $T$ is closed under vector addition.