# Counter example for infinite dimensional vector space

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?

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

• 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 :/ – Supantho Raxit Sep 12 at 8:51
• @SupanthoRaxit You can verify this from the definition of $T^{*}$: $\langle T^{*}x, y \rangle = \langle x, Ty \rangle$. – Kavi Rama Murthy Sep 12 at 8:55
• 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? – Supantho Raxit Sep 12 at 9:20
• @SupanthoRaxit I have added some details. – Kavi Rama Murthy Sep 12 at 9:27
• Thanks so much! It's all clear now!! – Supantho Raxit Sep 12 at 9:30