Zero Operator

**raed** · April 1st 2011, 11:55 AM

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.

**Drexel28** · April 3rd 2011, 11:13 PM

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.

**Opalg** · April 4th 2011, 12:23 AM

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.

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