Finding range of a linear transformation
Define $T: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ by $T(x,y,z) = (2y + z, x-z)$. Find $\mbox{ker}(T)$ and $\mbox{range}(T)$
I could find the kernel easy enough, and ended up getting $\{(-2x, x, -2x) : x \in \mathbb{R}\}$ but I don't really know how the get the range. In this case isn't the range effectively just the set with elements satisfying the equation $T$? I'm not really sure what the question is wanting to be honest. Some help would be great.
Thanks
$\endgroup$ 21 Answer
$\begingroup$$(2y+z,x-z)=x(0,1)+y(2,0)+z(1,-1)$. Since $(0,1)$ and $(2,0)$ span $\mathbb R^2$, the range is $\mathbb R^2$. To find the kernel, set $(2y+z,x-z)=(0,0)$ so that we have $z=x=-2y$. This gives the kernel to be $\{(-2y,y,-2y):y\in\mathbb R\}$ which is what you have obtained correctly. Note the kernel is simply the line passing through the origin with direction $(-2,1,-2)$.
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