0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.