Find basis of an image

Question

Let $ \psi : \mathbb{R}^4 \rightarrow \mathbb{R}^3$ be a linear transformation described by a formula $$\psi ([x_1,x_2,x_3,x_4])=[x_1+x_3+x_4, -x_2-x_4,x_1+x_2+x_3+2x_4].$$ Find basis of image $\psi(U)$ subspace described by an equation $U\subset\mathbb{R}^4: \ x_1-x_2+2x_3-x_4=0$. What to do with this?

Answer 1

Find a basis of $\;U\;$ and check what $\;\psi\;$ does to it, say:

$$U=\left\{\;(x_2-2x_3+x_4\,,\,\,x_2\,,\,\,x_3\,,\,\,x_4)\;:\;x_2,x_3,x_4\in\Bbb R\;\right\}=$$

$$=\text{Span}\,\{\;(1,1,0,0)\;,\;\;(-2,0,1,0)\;,\;\;(1,0,0,1)\;\}$$

So, for example:

$$\psi(1,1,0,0)=(2\,,\,\,-1\,,\,\,2)$$

and etc.