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$f\colon (x,y,z) \mapsto (x,x,x,x)$

I know you use $f(u + Av) = f(u) + Af(v)$ but I'm not sure when you're going from $R^3$ to $R^4$ and you get all of one element in the transformation.

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up vote 2 down vote accepted

So let's just check what you told us to check ;-). For arbitrary $u = (x,y,z) \in R^3$ and $v = (x',y',z')\in R^3$ and $A \in R$ (not sure whether you mean the real numbers $\mathbb R$ or a ring $R$, but it makes no difference here) we will show $$ f(u + Av) = f(u) + Af(v) $$ by using just $f$'s definition. It holds \begin{align*} u + Av &= (x,y,z) + A(x',y',z')\\ &= (x+Ax', y+Ay', z+Az')\\ f(u + Av) &= f(x+Ax', y+Ay', z+Az')\\ &= (x+Ax', x+Ax', x+Ax', x+Ax')\\ &= (x,x,x,x) + A(x',x',x',x')\\ &= f(x,y,z) + Af(x',y',z')\\ &= f(u) + Af(v) \end{align*} So $f$ is $R$-linear.

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We have $a f(x,y,z) = a (x,x,x,x) = (ax, ax, ax, ax) = f(ax,ay,az)$ and

$f(x + x', y + y', z + z') = (x + x', x + x',x + x', x + x') = (x,x,x,x) + (x', x', x', x') = f(x,y,z) + f(x', y',z')$.

Hence $f$ is linear. Hope this helps.

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$f((x_1,y_1,z_1)+(x_2,y_2,z_2))=f(x_1+x_2,y_1+y_2,z_1+z_2)=(x_1+x_2,x_1+x_2,x_1+x_2,x_1+x_2)=(x_1,x_1,x_1,x_1) + (x_2,x_2,x_2,x_2)=f(x_1,y_1,z_1)+f(x_2,y_2,z_2)$

So can you prove f(av)=af(v), where a is a constant?

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So I'd have to prove both? f(Ax',Ay',Az') = Af(x',y',z') so (Ax', Ax', Ax', Ax') = A(x',x',x',x') = (Ax', Ax', Ax', Ax') Sufficient? – Becky Nov 28 '12 at 15:31
@Becky yes, you have to prove both, or you can prove $f(u + Av) = f(u) + Af(v)$ it's the same. However, I prefer to divide this problem into these two steps. – user42912 Nov 28 '12 at 15:36

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