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I'm having a really hard time understanding how to figure out if a mapping is linear or not. Here is my homework question:

Determine which of the following mappings F are linear.

(a) $F: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $F(x,y,z) = (x, z)$

(b) $F: \mathbb{R}^4 \to \mathbb{R}^4$ defined by $F(X) = -X$

(c) $F: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $F(X) = X + (0, -1, 0)$

Sorry about my formatting. I'm not sure how to write exponents and the arrow showing that the mapping is from R^n to R^m. Any help is greatly appreciated!!

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Have you tried plugging in the definition of "linear"? –  Jonas Teuwen Feb 11 '11 at 1:15
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1 Answer 1

To check if a mapping is linear in general, all you need is verify the two properties.

  1. $f(x+y) = f(x) + f(y)$
  2. $f(\alpha x) = \alpha f(x)$

The above two can be combined into one property: $f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$

Edit

For instance, if we want to show say $F(x) = f(x_1,x_2,x_3) = x_1 - 4x_2 + x_3$ is linear, where $x = (x_1,x_2,x_3)$, then all you need to do is as follows.

\begin{align*} F(\alpha x + \beta y) & = f(\alpha x_1 + \beta y_1,\alpha x_2 + \beta y_2,\alpha x_3 + \beta y_3) \\ & = (\alpha x_1 + \beta y_1) - 4(\alpha x_2 + \beta y_2) + (\alpha x_3 + \beta y_3)\\ & = (\alpha x_1 - 4 \alpha x_2 + \alpha x_3) + (\beta y_1 - 4 \beta y_2 + \beta y_3)\\ & = \alpha (x_1 - 4 x_2 + x_3) + \beta (y_1 - 4 y_2 + y_3)\\ & = \alpha f(x_1,x_2,x_3) + \beta f(y_1,y_2,y_3)\\ & = \alpha F(x) + \beta F(y)\\ \end{align*}

Hence the above function is linear.

EDIT

As Arturo points out problem $c$ is not a linear map because of the constant hanging around. Such maps are called affine maps. Affine maps are those for which $f(x) - f(0)$ is a linear map.

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I know that this is the definition, but I'm having a hard time trying to figure out how to apply it. Can you show me an example with one or two of the functions from the problem I posted? –  user6886 Feb 11 '11 at 13:02
    
@Arturo: Ok. I have seen this argument before as well. Though I don't quite understand the rationale behind mapping $0$ to $0$. I guess it is more of a choice (?). Of course we could consider $g(x) = f(x) - f(0)$ and argue for linearity of $g$. –  user17762 Feb 11 '11 at 22:19
    
@Sivaram: It's not that you require the map to send $0$ to $0$; $f(a+b)=f(a)+f(b)$ requires the map to send $0$ to $0$ (just like a homomorphism of abelian groups). I know there are plenty of situations (e.g., diff. equations) where you consider such maps and call them "linear", but in the context of linear algebra you don't want to do that. Otherwise, all sorts of theorems need to include the condition "... and maps $\mathbf{0}$ to $\mathbf{0}$" for them to hold (e.g., representation of linear transformation as multiplication by coordinate matrices) –  Arturo Magidin Feb 11 '11 at 22:28
    
@Sivaram: Also, consider how someone having some confusion applying the condition $F(\alpha x+\beta y) = \alpha F(x)+\beta F(y)$ will react to a function which is being called linear and does not seem to satisfy that when you just plug vectors in... –  Arturo Magidin Feb 11 '11 at 22:30
    
@Arturo: I don't completely agree with the second comment. The condition for linearity could be reformulated as $F(\alpha x + \beta y) - F(0) = \alpha (F(x) -F(0)) + \beta (F(y) -F(0))$ to include such maps. But I agree this makes them messy. –  user17762 Feb 11 '11 at 22:38
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