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Which of the following mappings are linear?

(a)$\varphi : \mathbb{R} \rightarrow \mathbb{R}^{2}, \varphi(x) = (1 - x, (1 - x)^{2}),$

(b)$\varphi : \mathcal{F}(\mathbb{R},\mathbb{R}) \rightarrow \mathbb{R}, \varphi(f) = f(1),$

(c)$\varphi : \mathbb{R}[t] \rightarrow \mathbb{R}[t], \varphi(f) = t \cdot f', $

(d)Let $V$ be the $\mathbb{R}$-Vectorspace of converging sequences $(a_{n})$ of real numbers. Let $\varphi \rightarrow \mathbb{R}, \varphi((a_{n})) = a^{3}_{1} +$ lim $a_{n}.$

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@user3711: what are you having trouble with? –  Arturo Magidin Nov 19 '10 at 21:52

2 Answers 2

up vote 10 down vote accepted

HINT: A mapping is said to be linear if $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$.

Verify whether the maps which you have given satisfy the above 2 conditions. If they satisfy then conclude that the map is linear. A simple example would be to consider $\varphi : \mathbb{R} \to \mathbb{R}$ defined by $\varphi(x)=cx$, note that $\varphi(x+y)=c(x+y)=cx+cy=\varphi(x)+\varphi(y)$ and $\varphi(ax)= a(cx)=a \varphi(x)$. Hence $\varphi$ is a linear map. With this example in mind, proceed for your question.

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Hmm so if I understand that correctly (a) is clearly not a linear mapping since: $\varphi(x+y)=(1-x-y,(1-x-y)^{2})\neq(2-x-y,(1-x)^{2}+(1-y)^{2})=\varphi(x)+\var‌​phi(y)$ is that correct? –  ghshtalt Nov 20 '10 at 0:22
with (b) i'm not sure if i totally understand the concept, is it that the function of a function is equalto the function's value at one? in this case i would like to simply write $\varphi(f+g)=(f+g)(1)=f(1)+g(1)=\varphi(f)+\varphi(g) and \lambda\varphi(f) = \lambda f(1) = \varphi(\lambda f) \Rightarrow$ (b) is a linear mapping?? –  ghshtalt Nov 20 '10 at 0:30
p.s i really appreciate that you responded so thoughfully and quickly :) –  ghshtalt Nov 20 '10 at 0:32
Yes, that is the correct interpretation of (b). –  JSchlather Nov 20 '10 at 2:20
ok cool so (b) is taken care of.. does that mean i got (a) wrong? any other clues for (c) and (d)? i am still having difficulty understanding them –  ghshtalt Nov 20 '10 at 8:21

To amplify Chandru1's response, it is important to note the difference in how we check that a map is linear and how we check that a map is not linear.

In order to show that a map $\varphi$ is linear, we need to show that for every $\mathbf{x}$ and every $\mathbf{y}$ $\varphi(\mathbf{x}+\mathbf{y})=\varphi(\mathbf{x})+\varphi(\mathbf{y})$; and that for every scalar $\alpha$ and every $\mathbf{x}$, we have $\varphi(\alpha\mathbf{x})=\alpha\varphi(\mathbf{x})$. This is often done at the level of "general formulas", checking that both expressions in each case lead to identical formulas. Thus, for example, to check that you function (b) is a linear transformation, we verify that both $\varphi(f+g)$ and $\varphi(f)+\varphi(g)$ lead to the same expression: $$\varphi(f+g) = (f+g)(1) = f(1)+g(1);\qquad \varphi(f)+\varphi(g) = f(1)+g(1);$$ and likewise with $\varphi(af)$ and $a\varphi(f)$: $$\varphi(af) = (af)(1) = a\cdot f(1);\qquad a\varphi(f) = a\cdot f(1).$$

However, what you write in your comment for (a) is not a good proof that the map $\varphi(x) = \bigl(1-x, (1-x)^2\bigr)$ is not linear. When you write something like: $$ \varphi(x+y) = \bigl( 1-x-y, (1-x-y)^2\bigr) \neq (2 - x -y, (1-x)^2+(1-y)^2\bigr) = \varphi(x)+\varphi(y)$$ you are actually asserting that the two expressions as never equal. But in many instances, students will write this even when the expressions may sometimes be equal, yet think this constitutes a proof that the map is not linear.

This is not the case. You must either explain why the two expressions actually are never equal (this happens in this case, because $1-x-y$ is never equal to $2-x-y$), or you must exhibit specific values of $x$ and $y$ for which the two expressions evaluate to different values. It is not enough for the formulas to "look different".

In order to show that the function is not linear, you must show that there exist specific $x$ and $y$ such that $\varphi(x+y)$ is not equal to $\varphi(x)+\varphi(y)$, or that there exist a specific $\alpha$ and a specific $x$ such that $\varphi(\alpha x)\neq \alpha\varphi(x)$. That is, the verification that the equality does not always hold is not done at the level of "the formulas don't look the same", but rather by saying: "here are specific values, and, look, they don't match". So to show that (a) is not linear, rather than working out the expressions and seeing that they don't seem to be identical (just because two things don't look identical doesn't mean they don't always take the same value; "$\sin^2(x)+\cos^2(x)$" looks very different from "$1$", but they always take the same value), you should find specific values of $x$ and $y$ for which the answers don't come out the same. For instance, you could show that if $x=y=1$, then $$\varphi(1+1) = \varphi(2) = \bigl( 1-2, (1-2)^2\bigr) = (-1,1),$$ but $$\varphi(1)+\varphi(1) = \bigl( 1-1, (1-1)^2\bigr) + \bigl(1-1, (1-1)^2\bigr) = (0,0)+(0,0)=(0,0),$$ and since $(-1,1)\neq (0,0)$, then $\varphi(1+1)\neq\varphi(1)+\varphi(1)$. This shows the map is not linear, not the fact that the formulas for $\varphi(x+y)$ and for $\varphi(x)+\varphi(y)$ don't seem to be identical.

Think about it this way: to show that not every student in your class has red hair, it is not enough to argue that it would be highly unlikely that everyone in a big class would have such an uncommon color of hair, that it never happens, that it doesn't look that way when one takes a quick look, etc. The sure way of showing that not every student in your class has red hair is to stand someone up and show that he/she does not have red hair, and then you're absolutely done.

The same is true for (d).

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right, that absolutely makes sense to me. i think the only thing i'm still not understanding is how to interpret (c) and (d) (without even trying to test if they are linear). i wasn't sure if the brackets around $t$ meant something in particular and i need to do some reviewing/learning of converging sequences, limits, etc... Thanks a lot for the response! –  ghshtalt Nov 24 '10 at 18:36
@user3711: The brackets in (c) are part of the notation. It means your vector space is the space of all polynomials with coefficients in $\mathbb{R}$ with indeterminate t. In (d), each vector is really a sequence of real numbers, which you can think of as an "infinite tuple". –  Arturo Magidin Nov 24 '10 at 18:38

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