Adjoint Operator and Zero Transformation in Complex Inner Product Spaces

[holezch]

Homework Statement

V: complex inner product space with adjoint T*

Suppose that < T( x ) , x > = 0 for all x in V, then T is the zero transformation.

The Attempt at a Solution

< T( x ) , x > = < x, T*(x ) > = 0
0 = < x, 0 > = < 0, x >
< x, T*(x ) > = 0 = < x, 0 >

if < x , y > = < x, z> , then y = z
so T*(x ) = 0 for any x, which means T* is the zero transformation, which implies that T is the zero transformation..

is this okay? thanks

 

[hunt_mat]

Doesn't look quite right. Look at applying T to the basis elements $$<Te_{i},e_{j}>=0$$ and you should get that T is the zero operator. Another hint is write
$$Te_{i}=\sum_{k=1}^{N}\alpha_{ik}e_{k}$$

 

[holezch]

thanks, I'm less interested in the right answer, but I'd like to know why the steps I posted above were wrong? thanks

 

[Raskolnikov]

< T( x ) , x > = < x, T*(x ) > = 0
0 = < x, 0 > = < 0, x >
< x, T*(x ) > = 0 = < x, 0 >

if < x , y > = < x, z> , then y = z
so T*(x ) = 0 for any x, which means T* is the zero transformation, which implies that T is the zero transformation..

is this okay? thanks

<x , y> = < x , z> does not by itself imply y=z. Consider, for example, the Euclidean inner product. The vector x can be normal to two different vectors y and z, and so <x,y>=<x,z>=0, but y $$\neq$$ z. I hope this helps.

 

[holezch]

you're right, the implication in full is : if < x , y > = < x, z > for all x in V, then y = z
in this case, x may be arbitrarily chosen, but 'y' ( a.k.a T*( x ) ) is fixed for each x , so we cannot consider < x , y > = < x, z > for any x, since y is never the same vector
thanks for reading