Linear algebra-Basis of a linear map

[manuel325]

Homework Statement

Let $L: R^{2} → R^{2}$ be a linear map such that $L ≠ O$ but$L^{2} = L \circ L = O.$
Show that there exists a basis {$A$, $B$} of $R^{2}$ such that:

$L(A) = B$ and $L(B) = O.$​

The Attempt at a Solution

Here's the solution my book provides :

Well I have two questions:
1.Why do they say that $aA+bB=O$?. I mean I don't understand the solution from that point until the end (Why the solutions $a=0$ and $b=0$ are enough to prove the existence of that basis??May someone please explain??
Thanks in advance . Any help would be appreciated

 

[krome]

The solution says IF $aA+bB=0$, THEN $a=b=0$. That is what it means for the vectors $A$ and $B$ to be linearly independent. Vectors in a basis must be linearly independent.

 

[manuel325]

The solution says IF $aA+bB=0$, THEN $a=b=0$. That is what it means for the vectors $A$ and $B$ to be linearly independent. Vectors in a basis must be linearly independent.

Thanks but why $O=L(aA+bB)=aL(A)$?? could you please explain what they do there, please??

 

[krome]

Thanks but why $O=L(aA+bB)=aL(A)$?? could you please explain what they do there, please??

$L$ is a linear map, which means $L(aA+bB) = aL(A) + bL(B)$.

 

[manuel325]

$L$ is a linear map, which means $L(aA+bB) = aL(A) + bL(B)$.

Thanks