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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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    Given the data, David's Approach seems to be the answer! :)2012-02-13

3 Answers 3

1

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.2012-02-13
2

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.

2

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