1
$\begingroup$

I'm working on this exercize about linear transformation:

Let $E=\mathbb{R^N}$, $T:E \rightarrow E: \ (u_n)_{n\geq 0} \rightarrow (u_{n+1})_{n\geq 0}$

$S:E \rightarrow E: \ (u_n)_{n\geq 0} \rightarrow (v_{n})_{n\geq 0}$ with $v_0=0$ and for all $i \geq 1, v_i=u_{i-1}$

1)Prove T and S are linear transformation and tell if they are injective or surjective.

I prove T is a linear transformation using $T(\lambda (u_n)+(v_n))=\lambda T((u_n))+T(v_n))$ but how could we do for S according the the conditions ? How could we determine if they are injective or surjective ?

Thank you

$\endgroup$
1
$\begingroup$

$S((u_n)+(u'_n)) =S((u_n+u'_n)) = (w_n)_{n\geq0}$ where $w_n=0$ for and for $i>0$ $w_i=u_i+u'_i$. But then $(w_n)_{n\geq0}=(v_n)_{n\geq0}+(v'_n)_{n\geq0}$ where $v_0=0$ and $v_i=u_i$ for $i>0$ and $v'_0=0$ and $v'_i=u'_i$ for $i>0$. Thus $(w_n)_{n\geq0}=S(u_n)+S(u'_n)$. Thus $S((u_n)+(u'_n))=S(u_n)+S(u'_n)$. Show $S(\lambda(u_n)_{n\geq0})=\lambda S((u_n)_{n\geq0})$ the same way.

$\endgroup$

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.