3
$\begingroup$

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?

$\endgroup$

1 Answer 1

4
$\begingroup$

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 $T^{*}(a_n)=(a_2,a_3,...)$.

$\endgroup$
5
  • $\begingroup$ Thank you so much! I'd really appreciate it if you could please explain why T∗((an))=(a2,a3,...). I am really new to this :/ $\endgroup$ Sep 12, 2019 at 8:51
  • 1
    $\begingroup$ @SupanthoRaxit You can verify this from the definition of $T^{*}$: $ \langle T^{*}x, y \rangle = \langle x, Ty \rangle$. $\endgroup$ Sep 12, 2019 at 8:55
  • $\begingroup$ thanks again! But I'm still at a loss as to how T* acts this way. Would you please be as kind to explain this in a bit more detail? $\endgroup$ Sep 12, 2019 at 9:20
  • 1
    $\begingroup$ @SupanthoRaxit I have added some details. $\endgroup$ Sep 12, 2019 at 9:27
  • $\begingroup$ Thanks so much! It's all clear now!! $\endgroup$ Sep 12, 2019 at 9:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .