0
$\begingroup$

I'm working on linear transformation and trying to answer :

Let E and F be two vector-spaces on $\mathbb{C}$ and $L:E \rightarrow F$ an application such as :

$\forall u,v \in \ E, L(u+v)=L(u)+L(v) $

$\forall u \ \in \ E, \forall \alpha \ \in \ \mathbb{R}, L(\alpha u)=\alpha L(u)$

Prove that $L$ is a linear transformation if and only if $L(iu)=iL(u)$

I have tried taking $u=(x1,y1)$ and $v=(x2,y2)$ but did not manage to prove it.

$\endgroup$
0

2 Answers 2

0
$\begingroup$

Note that $(a+ib)u=au+b(iu)$ (since $E$ is a vector space over $\mathbb{C}$) so we have ( form the first statement): $$ L((a+ib)u)=L(au)+L(b(iu)) $$ and, since $a,b \in \mathbb{R}$, from the second statement: $$ L((a+ib)u)=L(au)+L(b(iu))=aL(u)+bL(iu) $$ So $L$ is linear iff $L(iu)=iL(u)$

$\endgroup$
0
$\begingroup$

Hint: Take $\lambda \in \mathbb{C}$. Write $\lambda = a + bi$ (for $a$ and $b \in \mathbb{R}$), and use that to show $L(\lambda u) = \lambda L(u)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .