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Define $T: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ by $T(x,y,z) = (2y + z, x-z)$. Find $kerT$ and $rangeT$

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.


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Oh whoops, $\mathbb{R}^2$ sorry! –  user1520427 Aug 23 '12 at 8:10
The range is just the subset of the codomain that has elements mapped to it. For example, the exponential function $\exp: \mathbb{R} \to \mathbb{R}$ has codomain $\mathbb{R}$ but its range is only the set of the positive real numbers. –  Ragib Zaman Aug 23 '12 at 8:12

1 Answer 1

up vote 4 down vote accepted

$(2y+z,x-z)=x(0,1)+y(2,0)+z(1,-1)$. 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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Cool thanks. There was a comment about my kernel having the wrong dimensions, so I have one more question. Is it correct that the dimension of the kernel should match the dimension of the domain's basis? –  user1520427 Aug 23 '12 at 8:28
I think I meant the dimension of an element in the kernel. So in this case $dim(\mathbb{R}^3) = 3$ so the elements in the kernel should have 3 variables ie $(x,y,z)$? –  user1520427 Aug 23 '12 at 8:34

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