# Thread: range of linear transformation

1. ## range of linear transformation

Let $\displaystyle L : R^4 \rightarrow R^3$ be defined by $\displaystyle L(x,y,z,w)=(x+y,z+w,x+z)$ Find a basis for range L.

What I did:
$\displaystyle { \left[\begin{array}{c}1\\0\\1\end{array}\right] \left[\begin{array}{c}1\\0\\0\end{array}\right] \left[\begin{array}{c}0\\1\\1\end{array}\right] \left[\begin{array}{c}0\\1\\0\end{array}\right] }$ spans range L

but rref of the coefficent matrix has a row of zeros so it's not a spanning set.

2. Originally Posted by Jskid
Let $\displaystyle L : R^4 \rightarrow R^3$ be defined by $\displaystyle L(x,y,z,w)=(x+y,z+w,x+z)$ Find a basis for range L.

What I did:
$\displaystyle { \left[\begin{array}{c}1\\0\\1\end{array}\right] \left[\begin{array}{c}1\\0\\0\end{array}\right] \left[\begin{array}{c}0\\1\\1\end{array}\right] \left[\begin{array}{c}0\\1\\0\end{array}\right] }$ spans range L

but rref of the coefficent matrix has a row of zeros so it's not a spanning set.
I may be crazy but isn't the image $\displaystyle \mathbb{R}^3$? Namely, isn't $\displaystyle \ker L=\left\{(x,-x,-x,x):x\in\mathbb{R}\right\}=\text{span}_\mathbb{R} \{(1,-1,-1,1)\}$ so that $\displaystyle \dim \text{im}(L)=4-1=3$ etc.