# Thread: range of linear transformation

1. ## range of linear transformation

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

What I did:
${
\left[\begin{array}{c}1\\0\\1\end{array}\right]
\left[\begin{array}{c}1\\0\\0\end{array}\right]
\left[\begin{array}{c}0\\1\\1\end{array}\right]
\left[\begin{array}{c}0\\1\\0\end{array}\right]
}$
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 $L : R^4 \rightarrow R^3$ be defined by $L(x,y,z,w)=(x+y,z+w,x+z)$ Find a basis for range L.

What I did:
${
\left[\begin{array}{c}1\\0\\1\end{array}\right]
\left[\begin{array}{c}1\\0\\0\end{array}\right]
\left[\begin{array}{c}0\\1\\1\end{array}\right]
\left[\begin{array}{c}0\\1\\0\end{array}\right]
}$
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 $\mathbb{R}^3$? Namely, isn't $\ker L=\left\{(x,-x,-x,x):x\in\mathbb{R}\right\}=\text{span}_\mathbb{R} \{(1,-1,-1,1)\}$ so that $\dim \text{im}(L)=4-1=3$ etc.