Question

Problem 2. Find linear endomorphism f:R→ Rsuch that f has no proper non-trivial invariant sub- spaces, but f = f of does have a proper non-trivial invariant subspace.

Answer 1

- ② Define + : 182 IR? by Hliny) -(-y,m). let & FIR, U, V FIR² = ua (wild valve, va) aute= (auity, duetve). i flautu) ztl (Quitur, auqta)) --- fragt vez), « u17V81) -= (-Qua-U2, au, tv) = taur ,aui) + (-42, 14) Ixt-Uq, "i) + (-02, 04) af (w) + f(u) i f is linear map from a from 1R² tolr? let USIR? be any proper invariant subspace of fa since dim 18 = 2 ) dim u al cuis propear] . a uzruy = {cula CERS where ire. & is generated by one element Now vis f innamont sukspace to) cu = f(u) a EU - » f(u) = xu [:U=< u> I - Let us urne) for no me & € IR > tuz, up) = (qui, auz)

=) ottle an, abu2 = dua-dur. dług = Luz - . and up= auzi ... 4.2.-ołuy Hi (146?) = 0 of u, zo 250 - U=0 Uzo because the dim 0.1 : tat-o. a) ... ? = -1 of LER which is not - possible =) f has no proper inuam aut risspace in ire. Nous B3 fef( (.)) = +1Hfm,y)) = f(ty, on}] = Eni-y) .Lt V=${(1; 0)) *(0,0)| RB food let VEV = va allo) for some a EIR fof (v). fof( (,0)) --- f ( FCCQ,0))) T iflorat) (4) a fof. (4) Ev -190)=-4. = fot (v) av: tis in variant under fof 4 dim = 1 2) v is pro der in lamant non-trinial Trebspace of tot in in G you can also take a loud en -