Linear

Assume the existence of something ( $\diamondsuit$ ) that takes arguments such as ‘u’ and ‘v’, with scalars such as ‘a’ existing, and we have two binary operations, addition, +, and multiplication, *.

That something is linear if the following two statements are true:

$\diamondsuit(u+v)=\diamondsuit(u)+\diamondsuit(v)$

$\diamondsuit(au) = a\diamondsuit(u)$

Linear Operator

Let $\hat O$ be a linear operator on the functions f(x) and g(x).

$\hat O[f(x) +g(x)] = \hat Of(x) + \hat Og(x)$

$\hat O[af(x)] = a \hat Of(x)$

We proved Linear character in two steps, both times. Occasionally someone writes an equality that is only true if both of our statements are true, thus proving “linear” with a single equality–albeit, something that is harder to visually follow.

$\hat O[af(x) +bg(x)] = a \hat Of(x) + b \hat Og(x)$

Try to see the first test and then add in ‘a’ and ‘b’ so that they move out for the right side of the equality–as they do in the second test.

And we’ll show the idea one more time, using the “something”:

$\diamondsuit(au+bv)=a\diamondsuit(u)+b\diamondsuit(v)$

There is a saying, “constants move with impunity”. We already know that a constant moves outside of Integration and Differentiation from Calculus. Both Integration and Differentiation are Linear Operators.

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