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.


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