Question

Prove that the nullspace and range of a normal operator are unchanged by taking any positive power of the operator

Answer 1

sont Let A be a normal operator de Case 1: If A in non singular e » i ken (A) = likes (AK): Zero spaces - thus. we are clone in this case d i "Here. Ker (A) = Nullspace (A) at Kernel of A Case 2: If it is singular i Theorem D . If A is a normal operator then * Kes (A) - Key (a*) + ing (A) is or thogonal to Ken (4) - ing (A*) is one thogonal to Ker(A*) A is normal is normal @

- dimo I since ing (A*) I kelam) 8 dim (ing a") + dim (ker A) so we have kes(A* )=(ing A*) ken (A) == ing (A*) - Also ker LA) = (img alt ker (A) te ing (A)

Cloin! Kes A" L BOLT - Let OfXe Ken (A2) and y be arbitrary ☆ 0 : < Avely> : <xuls*yy ! ied An E (ing A*) ! Also for any metaix B we have ima (B) = (ker B*) to . thes #x € (ing.tu) n (Ker 4“). - ing at senden An 20 3 x. € Ker(A) » Ker A2 C Ker (A) . It is obvious that Ker (A) ç Ker(A2) this In the _ now | ken (A) same spirit we we will show that = Ker (42) let that ker (AK) = Kerlt) ker CARH) = ker(A)

- let XE Ken Cart) & y be arbituary о оо са с ескеба е 0.- Takt ly> = KAKI Ag> 5 6x е ( img x )* n(ing n) = karte) перед Di Aku n t : 0 Ken LAK) - so Ker(A) { induction) & №b а | кел ден = be

we know that dim ling (A))+dim ker(A)) = dim (V) this from here we can directly say that Ting (AR) = ing (A) |