My Math Forum Problem involving adjoint and orthogonal complement

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 October 15th, 2013, 11:33 AM #1 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 Problem involving adjoint and orthogonal complement Let $A:E \rightarrow F$ be a linear mapping. E,F are finite-dimension inner product spaces. Show that: 1) $Ker (A^*)=Im(A)^\bot$ 2) $Im(A^*)=Ker(A)^\bot$ 3)$Ker(A)=Im(A^*)^\bot$ 4) $Im(A)=Ker(A^*)^\bot$ Where: $A^*$ is the adjoint of A. The simbol $\bot$ denotes the ortogonal complement. For exemple: $Im(A)^\bot$ is the set whose elements are vectors which are orthogonal to all vectors of $Im (A)$. I managed to prove (1) this way: 1) $v\in Ker(A^*)\Rightarrow A^*v=0\Rightarrow =0, \forall u\in E \Rightarrow =0, \forall u\in E \Rightarrow v\in Im (A)^\bot$. Similarly to the reciprocal implication. Please, help me to prove (2), (3) and (4).
 October 16th, 2013, 01:45 AM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Problem involving adjoint and orthogonal complement note1: Ortogonal complement is a symmetrical. So 4) is obvious. note2: A**=A. So 3) follows from 1) if you apply 1) to B = A*. By the way the direct proof is total similar to your clear proof, no tricks needed.
 October 16th, 2013, 04:26 AM #3 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 Re: Problem involving adjoint and orthogonal complement Good morning, Csak. After reading your post I've tried again: (3) follows from $(A^*)^*=A$, because $Ker(A)=Ker(A^*)^*=Im(A^*)^\bot$ (4) follows from $(A^\bot)^\bot$, because $Im (A)=Im(A^\bot)^\bot = Ker(A^*)^\bot$. And about (2)?
 October 16th, 2013, 07:23 AM #4 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Problem involving adjoint and orthogonal complement 2) follows from 3). How?
 October 16th, 2013, 09:09 AM #5 Member   Joined: Mar 2013 Posts: 71 Thanks: 4 Re: Problem involving adjoint and orthogonal complement I have no idea, but I propose the following direct proof Obs. $N(A)=Ker(A)$: $v \in Im(A^*) \Rightarrow \exists w \in F$ such that $A^*w=v$. But $x \in N(A) \Rightarrow Ax=0$. Then, $lt;v,x=>= ==0 \Rightarrow v \in N(A)^\bot$. On the other hand, $v \in N(A)^\bot \Rightarrow==0, \forall x \in N(A)$. But $x \in N(A) \Rightarrow Ax=0$. Then, $\forall w \in F,===0$. This implies $lt;x,A^*w=>= \Rightarrow A^*w=v \Rightarrow v \in Im(A^*)$. Is that correct?
 October 16th, 2013, 10:57 AM #6 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Problem involving adjoint and orthogonal complement Simply take the ortogonal complement of 3) and you get the 2).

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