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Let $L:Z->W$ be defined by $L(a,b)=b$ for all $a$ in $V$ and $b$ in $W$. I previously proved $Z$ is a vector space defined by $Z=V$x$W$. Now I want to prove $L(a,b)=b$ is a linear transformation

  1. So $L(a)$ is going to end up being the zero vector maybe? I'm not sure how else we'd end up with just $b$. But would $L(b)$ be the same as just $b$?

Such as $0+L(b)$? And then $L(b)$ just equals $b$?

Sorry I have a tendency to way, way overthink these proofs.

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    There's no such thing as $L(a)$ or $L(b)$. The "input" for this function is an ordered pair of vectors. You need to show that for any $(a,b), (c,d) \in Z$ and any $\lambda \in \Bbb F$, $L((a,b)+\lambda(c,d)) = L(a,b) + \lambda L(c,d)$. – 2017-02-27
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    @BobbieD I've fixed it now, I apologize for that. Can I ask what $F$ is? I haven't used this notation for anything before? Is it similar to $ \alpha \in \mathbb R$ – 2017-02-27
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    Vector spaces can defined to use any "field" $\Bbb F$ as their scalars. Since you didn't mention in the question that the field you're using is specifically the real numbers $\Bbb R$, I went with the more general $\Bbb F$. – 2017-02-27
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    I suspect you have confused the exercise. – 2017-02-27
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    So am I using $ L (a,b) + \lambda L(c,d)$ so I can prove that it equals $b+\lambda d$? @BobbieD – 2017-02-27
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    No. Your solution will look like this $$\begin{align}``L((a,b)+\lambda (c,d)) &= \cdots \\ &= \cdots [math] \\ &= \cdots \\ &= L(a,b) + \lambda L(c,d)\end{align}$$ Thus $L$ is a linear transformation." – 2017-02-27
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    Why aren't we using what was defined to us, i.e. $L(a,b)=b$? – 2017-02-27
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    You do use the definition (in two out of the only three steps in this extremely short proof). Alright, though. It looks like you're not getting what you're supposed to do so I'll just fill in the steps: $$\begin{align}L((a,b)+\lambda (c,d)) &= L(a+\lambda c,b+\lambda d) \\ &= b+\lambda d \\ &= L(a,b) + \lambda L(c,d)\end{align}$$ – 2017-02-27
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    I do have $L (a + \lambda c, b + \lambda d)$. So since $L$ was defined as $L(a,b)=b$ we can assume $L(a+\lambda c)$ is nothing? I apologizing that I'm not getting this and I feel like I'm asking dumb questions but like I said I tend to really overthink transformations. EDIT: I am looking over this again and it makes sense. I apologize I was really really overthinking and now it is so obvious. Wow I am so sorry. – 2017-02-27
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    No worries. I do want to reiterate, though, $L(a+\lambda c)$ is undefined. $L$ is a **two-argument** function. But if what you're asking about is "does the first argument really not matter?", then the answer is yes. The "output" of $L$ really does not depend on the first argument at all. – 2017-02-27

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