Inner product space functional analysis 2

(zero operator) let T:X----->X be a bounded linear operator on a complex inner product space X.If =0 for all x belonging to X show that T=0.
show that this does not hold in case of a real inner product space

Let x be any element of X. Then

0 =
= + + +
= +
= -i + i
= i( - );

0 =
= + + +
= + .

From the first equation, we have = . Combining this with the second,

= - = - .

We conclude that = 0, so Tx=0. Since x is arbitrary, this proves that T=0.


For the real case, let X = R^2 and let T be rotation by 90 degrees. Then clearly Tx is orthogonal to x for all x, but T is not zero.