Math Help - Zero Operator

1. Zero Operator

Dear Colleagues,

Could you please help me in solving the following problem:
Let $T:X\longrightarrow X$ be a bounded linear operator on a complex inner product space $X$. If $\langle Tx,x \rangle =0$ for all $x\in X$, show that $T=0$.
Moreover, this is does not hold in the case of real inner product space.

Regards,

Raed.

2. Originally Posted by raed
Dear Colleagues,

Could someone solve the problem.

Regards.
I'll help you with the problem. Note first that if $\langle u,x\rangle=\langle v,x\rangle$ for all $x\in V$ (where [tex]V{/math] is our inner product space) then $u=v$ (indeed, take $x=u-v$ and subtract). From this we get that if $\langle T(x),y\ranlge=0$ for all $x,y\in V$ then $T=0$ (indeed, note that $\langle T(x),y\rangle=0=\langle 0,y\rangle$...conclude from the first part). Now, to our actual problem let $x=zv+u\;\; z\in\mathbb{C}$ and compute that our assumption gives $0=z\langle T(u),v\rangle+\overline{z}\langle T(v),u\rangle$. Pick $z$ cleverly (don't think too hard) to create a system of equations that will reduce this problem to the last one (note that you'll have to pick one of the values of $z$ to be non-real otherwise you'll have disproved the second part). For the second part take your inner product space to be $\mathbb{R}^2$ with the usual inner product and recall that the inner product of orthogonal vectors is orthogonal...so what if a transformation sent a vector to its orthogonal vector.

3. Originally Posted by raed
Dear Colleagues,

Could you please help me in solving the following problem:
Let $T:X\longrightarrow X$ be a bounded linear operator on a complex inner product space $X$. If $\langle Tx,x \rangle =0$ for all $x\in X$, show that $T=0$.
Moreover, this is does not hold in the case of real inner product space.
Hint: Apply the polarisation identity $\displaystyle \langle Tx,y\rangle = \frac14\sum_{k=0}^3 i^k\langle T(x+i^ky),x+i^ky\rangle.$

Note that this is an intrinsically complex relation. In the real case the result fails. For example, in two-dimensional space a rotation through a right angle takes every vector to an orthogonal vector.

Edit. Sorry, didn't see drexel28's comment.