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Be $F(x, y) = 3x + 4y$. I have to prove it's a linear appliction.

I am confused about the way to proceed. Is this the right way to proceed?

$$F(x_1 + x_2, y_1 + y_2) = 3(x_1 + x_2) + 4(y_1 + y_2) = 3x_1 + 4y_1 + 3x_2 + 4y_2 = F(x_1, y_1) + F(x_2, y_2)$$

Sorry for this silly question, I just have to acquire more confidence I guess...

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  • $\begingroup$ Your approach is correct, but you also need to check if $f(ax, ay) = a f(x,y)$. $\endgroup$
    – PierreCarre
    Commented May 4, 2023 at 20:47

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It is nearly correct. There's a step missing right at the start. It should begin with $$ F\bigl((x_1,y_1)+(x_2,y_2)\bigr)=F(x_1+x_2,y_1+y_2)=\cdots $$ And, of course, you should also prove that $$ F\bigl(\lambda(x,y)\bigr)=\lambda F(x,y). $$

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  • $\begingroup$ Ah yes, the scalar part I forgot! (But that is easier). Thank you! $\endgroup$
    – Heidegger
    Commented May 4, 2023 at 20:48
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Linearity means, fancily, additivity and homogeneity (of the order $1$ in the exponent). Supposing you have a function, this implies that $F(\vec{v}+\vec{w})=F(\vec{v})+F(\vec{w})$ and $F(a\vec{v})=aF(\vec{v})$. Essentially, just that whatever you've got has the properties that correspond to some linear space. For example, the derivative operator is linear since $D(f+g)=Df+Dg$ and $D(af)=aDf.$

Additivity: $$L(x+y)=L(x)+L(y)$$ Homogeneity: $$L(ax)=aL(x).$$

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