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Let T: $\mathbb{R}^3\rightarrow \mathbb{R}$ be linear. Show that there exist scalars a, b, and c such that $T(x,y,z)= ax + by + cz$ for all $(x,y,z) \in \mathbb{R}^3$

Can I just say "you can pick $a=b=c=0$" or do I have to actually expand out $T(kx+ x', ky + y', kz + z')$ and verify that T is linear where $k\in \mathbb{R}$?

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    But if I had a choice, I would choose my $T$ to be Darjeeling, first cut.2012-09-14

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No. If $a = b = c = 0$, then $T(x,y,z) = 0$. But, it could happen that $T$ is a nonzero linear transformation.

To prove this, note that $(x,y,z)$ means $xe_1 + ye_2 + z e_3$ where $e_1$, $e_2$, and $e_3$ are basis for $\mathbb{R}_3$. You have by linearity

$T(x,y,z) = T(xe_1) + T(ye_2) + T(z e_3) = x T(e_1) + yT(e_2) + z T(e_3)$.

Letting $a = T(e_1)$, $b = T(e_2)$ and $c = T(e_3)$, you have proven the desired result.

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    Ok thank you so much. I think I understand it.2012-09-14
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You are supposed to show that there is a single triple $a,b,c$ such that $T(x,y,z)=ax+by+cz$ for all $x,y,z\in\mathbb R$.

Hint: Let $a=T(1,0,0)$. Similarly...