Consider a linear operator O acting on a Hilbert space H. If the dimension of H is finite, I have shown that: dim(ker O) = dim(ker O†). Where O† is the Hermitian conjugate of O.
Does this hold when H is infinite-dimensional? If not, is there a counterexample?
Best Answer
Let $H=\ell^{2}$ and $T((a_n))=(0,a_1,a_2,...)$. Then $ker (T)=\{0\}$ so $dim (ker(T))=0$. You can verify that $T^{*}((a_n))=(a_2,a_3,...)$ so $ker (T^{*})$ is one dimensional.
Let $(a_n),(b_n) \in \ell^{2}$. Then $ \langle (a_n), T(b_n) \rangle =\langle (a_n),(0,b_1,b_2,...) \rangle=(a_1)(0)+a_2b_1+a_3b_2+... =\langle (a_2,a_3,...), (b_n) \rangle$. Hence $T^{*}(a_n)=(a_2,a_3,...)$.