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- Mar 31st 2010, 09:02 AMbenjamin872easy? proving a linear function
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- Mar 31st 2010, 09:11 AMDefunkt
It should be clear that this function is not linear; take, say, $\displaystyle x=1, y=2$ then $\displaystyle f(x+y) = x+y+3 = 1+2+3 = 6 \neq f(x) + f(y) = x+3+y+3 = 1+3+2+3 = 9$

and also, if we take, say, $\displaystyle \alpha = 2$ then $\displaystyle f(\alpha x) = \alpha x + 3 = 2 * 1 + 3 = 5 \neq \alpha f(x) = \alpha(x+3) = 2(1+3) = 8$ - Mar 31st 2010, 12:55 PMHallsofIvy
- Mar 31st 2010, 01:04 PMbenjamin872
ok but then il try $\displaystyle f(x)=2x+4 $ say for x=1 and y=2

i find f(x+y) not equal to f(x)+f(y) which makes 10 not equal 14. so this is not linear also? can someone give me an example of a linear function that works.

......... wait so the these properties cannot be satisfied if there is a constant, but i thought a linear function could be plotted y=mx+b im confused - Apr 1st 2010, 03:17 AMHallsofIvy
Those are two different meanings of the word "linear". In algebra, a function of the form f(x)= mx+ b is called "linear" because its graph is a straight line.

But Linear Algebra uses a more stringent definition for "linear transformation"- we must have f(x+ y)= f(x)+ f(y) and f(ax)= af(x). As I said above, the only "linear" functions in that sense, from R to R are of the form f(x)= ax for some number a.