Some examples of linear maps are 2x2 matrices with real coefficients, that perform rotation, reflection, scaling, shear, squeeze, and projection

 

The drift I'm getting is that if the derivatives are constant the function is linear. This means that apparently affine functions are also linear since the derivative of a constant vanishes.

I'm pretty sure that is not the case.

f(x) = mx + b

f(x1+x2) = m·x1 + m·x2 + b ≠ f(x1) + f(x2) = m·x1 + m·x2 + 2b
f(α·x) = αmx + b ≠ α·f(x) = αmx + αb

From Wikipedia: http://en.wikipedia.org/wiki/Linear_equation#Connection_with_linear_functions

A linear equation, written in the form y = f(x) whose graph crosses the origin (x,y) = (0,0), that is, whose y-intercept is 0, has the following properties:

and

where a is any scalar. A function which satisfies these properties is called a linear function (or linear operator, or more generally a linear map). However, linear equations that have non-zero y-intercepts, when written in this manner, produce functions which will have neither property above and hence are not linear functions in this sense. They are known as affine functions.

 

No. affine is not linear in terms of vector spaces.

For this problem look at bilinear forms

http://en.wikipedia.org/wiki/Bilinear_form

 

I'm pretty sure that is not the case.

f(x) = mx + b

f(x1+x2) = m·x1 + m·x2 + b ≠ f(x1) + f(x2) = m·x1 + m·x2 + 2b
f(α·x) = αmx + b ≠ α·f(x) = αmx + αb

From Wikipedia: http://en.wikipedia.org/wiki/Linear_equation#Connection_with_linear_functions

A linear equation, written in the form y = f(x) whose graph crosses the origin (x,y) = (0,0), that is, whose y-intercept is 0, has the following properties:

and

where a is any scalar. A function which satisfies these properties is called a linear function (or linear operator, or more generally a linear map). However, linear equations that have non-zero y-intercepts, when written in this manner, produce functions which will have neither property above and hence are not linear functions in this sense. They are known as affine functions.

I'm Ok with that analysis, but with respect to the multivariable case would it be the case that planes going through the origin also represent linear functions? That is, Let D = 0 from my previous example and let:

Ax + By + Cz = 0, for A, B, C belonging to ℝ