Orthogonal to a range

Question

This question is from a proof that if a linear operator T has an eigenvector, then so does T*(adjoint of T). The following part is what I don't get:

v(an eigenvector of T) is orthogonal to the range of T*-h'I. So T*-h'I is not onto and hence is not one-to-one.

I don't see how you can advance from finding a vector that is orthogonal to a range space to concluding that the transformation is not onto.

Answer 1

Do we have $Tv=hv$ ? If yes, then

$<T^{\star}x-\overline{h}x, v>=<T^{\star}x,v>-\overline{h}<x,v>=<x,Tv>-\overline{h}<x,v>=<x,hv>-\overline{h}<x,v>=\overline{h}<x,v>-\overline{h}<x,v>=0$,

hence $v$ is orthogonal to the range of $T^{\star}-\overline{h}I$.

Since $v \ne 0$, $v$ is not in the range of $T^{\star}-\overline{h}I$