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I'm given the following question:

Consider the operation on a vector that rearranges the components in increasing order. For example: $T(3, 2, 6) = (2, 3, 6)$. Is this a linear mapping?

Now, can I assume this is NOT a linear mapping because of the following fact:

$T(3, 2, 6) = T(6, 2, 3) = (2, 3, 6)$?

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There is nothing that forces a linear mapping from being one-to-one; the fact that this mapping is not one-to-one does not, in and of itself, imply that it is not linear. In order to show it is not linear, you would have to produce two specific vectors $(a,b,c)$, $(x,y,z)$ such that $T(a+x,b+y,c+z)\neq T(a,b,c)+T(x,y,z)$, or a specific vector $(a,b,c)$ and a specific scalar $\alpha$ such that $T(\alpha a,\alpha b, \alpha c)\neq \alpha T(a,b,c)$. – Arturo Magidin Feb 13 '12 at 3:49
Hint: Now, what is $T(9,4,9)$? (Evaluate it in two ways.) – David Mitra Feb 13 '12 at 3:50
Given the data, David's Approach seems to be the answer! :) – user21436 Feb 13 '12 at 5:29
up vote 1 down vote accepted

What you've done so far does not suffice to show that $T$ is not linear.

To show that $T$ is not linear, you have to show that one of the following two properties does not hold:

$\ \ \ \ $1) $T(a{\bf x})= aT({\bf x})$ for all scalars $a$ and all vectors $\bf x$.

$\ \ \ \ $2) $T({\bf x}+ {\bf y})=T({\bf x})+T({\bf y})$ for all vectors $\bf x$ and $\bf y$.

You've noticed $$\tag{ 3 }T(3,2,6)=T(6,2,3)=(2,3,6).$$

Now notice, by the definition of $T$, that $$T\bigl( (3,2,6)+(6,2,3)\bigr)=T(9,4,9)=(4,9,9).$$ But from $(3)$ $$ T(3,2,6)+T(6,2,3)=(2,3,6)+(2,3,6)=(4,6,12)\ne(4,9,9). $$

So, 2) fails and, thus, $T$ is not linear.

Note that you need only verify that one of the two properties fails. Here, you could have shown that property 1) fails instead: $T(1,2,3) =(1,2,3)$ but $$T(-1,-2,-3)=(-3,-2,-1)\ne(-1)T(1,2,3).$$

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All were great answers above, I just was not sure how to apply that theorem T(x+y) = T(x) + T(y) in this specific case. Thank you. – intervade Feb 13 '12 at 14:18

No you cannot assume that.

A linear map can map different vectors to the same vector.

Since $(3,2,6)$ and $(6,2,3)$ are linearly independent, there is a linear map which maps them both to $(2,3,6)$

One approach hint: Try to disprove the property that $T(\alpha x) = \alpha T(x)$

Different approach hint: Consider what $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ get mapped to, and see what you can say about the whole map.

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Consider $T((1,0,0)+(0,1,0))$.

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