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Let $T: \mathbb R^3 \to \mathbb R^2$ be given by $T(x,y,z) = (y,z)$.
Show that $T$ is linear.
Prove that the $\ker(T) = \mathrm{Span}(e1)$.

Could you help me?

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  • $\begingroup$ What are your thoughts on the problem so far? What have you tried? $\endgroup$ – Cameron Buie Nov 5 '13 at 15:17
  • $\begingroup$ I am not sure how can I find the ker(T). I am guessing that it should be sth like this : ker(T) = { (x,y,z) : y=z= 0}. About the linearity: i should prove that T((x,y,z) + (x1,y1,z2)) = (y+y1, z +z1) or something else? $\endgroup$ – user2013804 Nov 5 '13 at 15:20
  • $\begingroup$ You're absolutely right! $$\operatorname{ker}(T)=\{(x,y,z)\in\Bbb R^3:y=z=0\}.$$ Now you must show that this is precisely the subspace of $\Bbb R^3$ spanned by $e_1$. As for linearity, you must show that for any $\vec u,\vec v\in\Bbb R^3$ and any real scalar $\lambda,$ you have $$T(\vec u+\vec v)=T(\vec u)+T(\vec v)$$ (which is basically what you said) and that $$T(\lambda\vec u)=\lambda T(\vec u).$$ $\endgroup$ – Cameron Buie Nov 5 '13 at 15:23
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Do prove linearity, and as for the kernel observe that

$$(x,y,z)\in\ker T\iff T(x,y,z):=(y,z)=(0,0)\iff y=z=0\implies$$

$$\implies \ker T=\text{Span}\,\{(1,0,0)\}$$

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