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 October 13th, 2010, 09:27 AM #1 Newbie   Joined: Dec 2008 From: Copenhagen, Denmark Posts: 29 Thanks: 0 A problem from functional analysis Hi I'm currently following a first course in functional analysis, and a problem from my weekly assignment is bothering me quite a bit. It would be greatly appreciated if anyone could give me a hint or maybe even go through the steps in the solution. The problem is given below. For a normed vector space $V$, we denote by $B(V)$ the space of bounded linear operators $T \, : \, V \rightarrow V$. (The problem) Let $V$ be a complex vector space with an inner product $(\cdot \, , \, \cdot)$ and assume that $T \in B(V)$. 1) Show that $(Tx,y)=0$ for all $x,y \in V$ if and only if $T$ is the zero operator. 2) Show next that $(Tx,x)=0$ for all $x \in V$ if and only if $T$ is the zero operator. 3) Are these results true if the vector space is a real vector space?
 October 13th, 2010, 04:00 PM #2 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 Re: A problem from functional analysis 1. and 2. The if parts are trivially obvious. For 1. If Tx = y ? 0 for some x, the (Tx,y) ? 0. Therefore only if. For 2. Simple example: 2 dimensional space with base vectors a and b, with (a,b)=0. Let Ta=b and Tb=-a, then (Tx,x)=0 for all x, therefore only if is false. I don't see any distinction between real and complex spaces. However you may have to use complex conjugates for complex spaces in the above analysis.
 October 15th, 2010, 01:42 PM #3 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 Re: A problem from functional analysis Correction: The analysis for 1. holds for both both real and complex vector spaces. The analysis for 2. is only for real vector spaces. I believe that 2. is true for complex vector spaces, but I haven't been able to work it out completely.
 October 16th, 2010, 06:00 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: A problem from functional analysis Suppose V is complex, and $\langle Tx,x\rangle=0$ for all $x\in V.$ Then for all $x,y\in V:$ $\langle T(x+ y),x+ y\rangle=0,$ and $\langle T(x+ iy),x+ iy\rangle=0.$ Expand these out, and you should be able to deduce that $\langle Tx,y\rangle=0$ for all $x,y\in V,$ and therefore (2) reduces to (1).
 October 20th, 2010, 05:53 AM #5 Newbie   Joined: Dec 2008 From: Copenhagen, Denmark Posts: 29 Thanks: 0 Re: A problem from functional analysis Thanks for responding both of you. I solved the problem in the following way (hopefully no typo's) 1) The if part is trivial. The only if part I show by contraposition. Let $T$ be any operater in $B(V)$ but not the zero operator. Then there exists $x \in V$ such that $Tx \neq 0$ and so we have $(Tx,Tx)= llTxll^2 \neq 0$. Choose $y=Tx \in V$ and then we have found $x,y \in V$ such that $(Tx,y) \neq 0$. Thus $T$ must be the zero operator. 2) Again the if part is trivial. The only if part I show by choosing again $T$ as any operator in $B(V)$ but not the zero operator. Then we know there exists $x,y \in V$ such that $(Tx,y) \neq 0$. Assume now that $(Tx,x)= 0 \, \forall x \in V$. Then we have  $0 = (T(x+iy),x+iy) = (Tx+iTy,x+iy) = (Tx,x) + (iTy,iy) + (Tx,iy) + (iTy,x) = \bar{i}(Tx,y) + i(Ty,x) = -i(Tx,y) + i(Ty,x) = i((Ty,x)-(Tx,y)) = \{from (*)\} = -2i(Tx,y) \neq 0 \qquad \qquad \text{contradiction}$ Thus $T$ must be the zero operator. 3) These results are not true if $V$ is a real vector space. This is showed by an example. Choose $V=\mathbb{R}^2$ and let $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be given by $T((a,b))= (-b,a)$ for $a,b \in \mathbb{R}$. Then $T$ is linear, bounded and not the zero operator. Equip $V$ with the standard inner product and then we have $(Tx,x)= 0$ for all $x \in \mathbb{R}^2$.

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