A question about subspaces Okay, here's the problem:
Suppose that {$u_1,u_2,u_3$} forms a basis for a vector space V. Show that if 
$$v_1=u_1+2u_3$$$$v_2=u_1+2u_2+3u_3$$$$v_3=u_2-u_3$$ then {$v_1,v_2,v_3$} forms a basis for V.
So what I've decided so far is that if $M=\begin{bmatrix}
1&1&0\\ 
0&2&1\\ 
2&3&-1 
\end{bmatrix}$ and $U=\begin{bmatrix}
u_1&u_2&u_3
\end{bmatrix}$
then the vectors $v_1,v_2,v_3$ are the columns of $
UM$. So I just need to show, I guess, that the columns of $UM$ are linearly independent. $M$ is invertible; I'm pretty sure that's important. If I had a theorem that said that the product of two matrices with linearly independent columns has linearly independent columns then I'd be set, but I don't believe I do (I would if I knew that U was square). Am I going in a good direction here, or is there a smarter way to do this?
 A: Let $v \in V$ be arbitrary.
Because $u_1,u_2,u_3$ are a basis you know that $v=a_1 u_1+a_2 u_2+a_3 u_3$,
but by solving the linear system you have that
$$u_1=\frac{1}{3} (5 v_1 - 2 v_2 + 4 v_3)$$
$$u_2=\frac{1}{3} (-v_1 + v_2 + v_3)$$
$$u_3=\frac{1}{3} (-v_1 + v_2 - 2 v_3)$$
Therefore you know that $v \in \text{span}\{v_1,v_2,v_3\}$, because $v$ was arbitrary it follows that  $\{v_1,v_2,v_3\}$ spans your vector space, and because all bases of a vector space have the same size you found a basis of $V$.
A: Your reasoning is also correct. Here's the argument.
If $\{v_1,v_2,v_3\}$ is linearly independent then this forms a basis. 
So, let us suppose that $\alpha_1 v_1+\alpha_2 v_2+\alpha_3 v_3=0$. 
Then $(\alpha_1+\alpha_2)u_1+(2\alpha_2+\alpha_3)u_2+(2\alpha_1+3\alpha_2-\alpha_3)u_3=0$. 
Since $\{u_1, u_2, u_3\}$ forms a basis, we have 
$\begin{eqnarray*}\alpha_1+\alpha_2=0\\ 2 \alpha_2+\alpha_3=0\\ \text{ and }\\ 2\alpha_1+3\alpha_2-\alpha_3=0\end{eqnarray*}$
The above system has the only solution $\alpha_1=\alpha_2=\alpha_3=0$ iff the coefficient matrix $M$ is invertible.
