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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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You don't get to choose $T$. – Qiaochu Yuan Sep 14 '12 at 6:26
$T$ is given and is linear. It maps from $\mathbb{R}^3$ (which has a basis $(1,0,0), (0,1,0), (0,0,1)$) to $\mathbb{R}$. The question is to show how you would compute $a,b,c$ in terms of evaluating $T$ at specific points, and, of course, to ensure that $T$ actually equals the resulting form. – copper.hat Sep 14 '12 at 6:27
But if I had a choice, I would choose my $T$ to be Darjeeling, first cut. – copper.hat Sep 14 '12 at 6:30
up vote 3 down vote accepted

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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Thanks for the reply. I understand what you did, but I'm still confused about the question. I'm interpreting it as "show that there exist scalars a, b, and c such that [definition]." It doesn't seem like there is any constraint there. Maybe I'm just being stupid right now. – anon Sep 14 '12 at 6:55
The question is: I give you any linear transformation $T : \mathbb{R}^3 \rightarrow \mathbb{R}$. You have to find $a, b, c$ such that $T(x,y,z) = ax + by + cz$. In other words, you have to show that no matter which $T$ I give you, you can find $a,b,c$ (depending on that T) to make $T(x,y,z) = ax + by + cz$. – William Sep 14 '12 at 6:58
If I set $a$, $b$, and $c$ arbitrarily, then I modify the behavior of the $T$ you provide. It's still a little confusing because: how do I know what the behavior of $T$ is? Does that make sense? I'm going to reread this in the morning. – anon Sep 14 '12 at 7:16
@Anon No, I give you $T$, you need to find the $a$, $b$, and $c$ that works for my $T$. You don't get to change or pick the $T$. – William Sep 14 '12 at 7:31
Ok thank you so much. I think I understand it. – anon Sep 14 '12 at 7:34

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...

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