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We have the transformation $f: \mathbb{R}^3 \to \mathbb{R}^2$ where $$f(v) = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 4 & 1 \\ \end{bmatrix} *V $$

What is the domain and codomain? Is this injective or surjective? Is this bijective?

I think the domain is the Matrix and the codomain is $V$. Is this thinking correct? Thanks

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up vote 1 down vote accepted

The domain is 3-dimensional and the codomain 2-dimensional. In this situation no linear transformation can ever be injective, but it could be surjective.

If the situation were reversed, where the codomain had higher dimension than the domain, then no linear transformation could be surjective. Bijective linear functions can only exist when domain and codomain have the same dimension (in the finite-dimensional case).

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Thank you that is very helpful. So to prove this is surjective I need to solve f(v)= w where f: V -> W. How can I do this? Also, if I say f: V-> W, what is f(v')? Thanks! – Allen Miller Apr 20 '13 at 5:43
To prove it is surjective you need to prove that for every $w\in W$, you can find a $v\in V$ such that $f(v)=w$. To do this, start with Brian's observation, note that $f(e_2), f(e_3)$ give his vectors. So, given $w\in W$, represent it using his basis for $\mathbb{R}^2$, then you can calculate which $v\in V$ will get there. There are details for you to work out. – vadim123 Apr 20 '13 at 13:40
I wrote out my equation and then after simplifying I got: a+ 2b =x & 2a+4b+ c=y. I'm not sure where to go from there. – Allen Miller Apr 20 '13 at 17:40

Domain: $R^3$ Codomain: $R^2$

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That was actually my first thought, but it seemed too easy. Can you walk me through showing this is injective. In my notes I need to show V is not equal to V'. What is V'? – Allen Miller Apr 20 '13 at 5:38

Observe that $\begin{bmatrix} 2 \\ 4 \end{bmatrix},\begin{bmatrix} 0\\1\end{bmatrix}$ span $\mathbb{R}^2$. Therefore, $f$ is surjective.

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