# Prove that if $L_1, L_2 \in L (R^2, R^2)$ then $(L_1 \oplus L_2) \in L (R^2, R^2)$ also.

We are given two linear functions $$L_1 L_2 \in L (\mathbb R^2, \mathbb R^2)$$ which we define addition as $$(L_1 \oplus L_2)$$ to be a map $$(L_1 \oplus L_2) : R^2 \to R^2$$ given by the formula, $$(L_1 ⊕ L_2)(x, y) = L_1(x, y) + L_2(x, y)$$

Example, suppose $$L_1$$ is the liner map $$L_1(x, y) = (2x + y, y)$$ and $$L_2$$ is the linear map $$L_2(x, y) = (x + y, 2x)$$. Then, $$(L_1 \oplus L_2) = (3x+2y, 2x+y)$$.

So now going back to the title, prove that if $$L_1, L_2 \in L (\mathbb R^2, \mathbb R^2)$$ then $$L_1 \oplus L_2) \in L (\mathbb R^2, \mathbb R^2)$$ also.

Please correct me if I'm wrong, or show me how this is done.

What I am assuming we do is, say that the $$\oplus$$ is ordinary component addition, using the example for the $$x$$ (from $$x$$,$$y$$) value of the linear map $$(L_1 \oplus L_2)$$ we will get $$3x+2y$$ which is just $$(2x+y + x+y)$$ which I restate is ordinary addition. Because we do this for both the $$x$$ and $$y$$ value of the result, our result is therefor still in $$\mathbb R^2$$. or $$(3x+2y, 2x+y)$$ is still in $$(x, y)$$ format.

Yes, your argument is a good start. The complete proof is actually not so difficult. You don't even need to argue about the individual components. Instead use the linearity of $$L_1,L_2$$ directly. Also, don't forget that linearity has to satisfy two conditions.

Def: A map $$L \colon \mathbb{R}^2 \to \mathbb{R}^2$$ is called a linear function, if

1) $$\forall v_1,v_2 \in \mathbb{R}^2 \colon L(v_1+v_2)=L(v_1)+L(v_2)$$

2) $$\forall v \in \mathbb{R}^2 , \lambda \in \mathbb{R} \colon L(\lambda \cdot v)= \lambda \cdot L(v)$$

Claim: If $$L_1, L_2$$ are linear functions, then $$L_1 \oplus L_2$$ is also a linear function.

Proof: We need to show the two properties of linear functions.

1) Let $$v_1, v_2 \in \mathbb{R}^2$$. Then \begin{align*} (L_1 \oplus L_2)(v_1+v_2) &=L_1(v_1+v_2)+L_2(v_1+v_2) \\ &= L_1(v_1)+L_1(v_2)+L_2(v_1)+L_2(v_2) \\ &= L_1(v_1)+L_2(v_1)+L_1(v_2)+L_2(v_2) \\ &= ( L_1 \oplus L_2) (v_1) +( L_1 \oplus L_2)(v_2), \end{align*} where we used that $$L_1,L_2$$ satisfy property 1) of being linear.

2) Let $$v \in \mathbb{R}^2, \lambda \in \mathbb{R}$$. Then \begin{align*} (L_1 \oplus L_2) (\lambda \cdot v) &= L_1(\lambda \cdot v)+L_2(\lambda \cdot v) \\ &= \lambda \cdot L_1(v)+\lambda \cdot L_2(v) \\ &= \lambda \cdot \left( L_1(v)+L_2(v) \right) \\ &= \lambda \cdot (L_1 \oplus L_2) (v), \end{align*} where we used that $$L_1, L_2$$ satisfy property 2) of being linear.

This completes the proof.