Is the kernel of the adjoint operator equal to the kernel of the operator ($\ker (A)=\ker (A')$)?

Question

I am in a middle of a proof where I asked myself about the following:

Is the kernel of the adjoint operator equal to the kernel of the operator ($\ker (A)=\ker (A')$)?

Theorem:Let $X,Y$ be Banach spaces and $A$ an linear bounded operator. The closure of the image is $\overline{Im\: }A=\{y\in Y:f(y)=0,\forall f\in Y'$ such that $A'f=0$}. $(A'f)(x)=f(A(x))$ is the adjoint operator.

$Im(A)=\ker(A')^\bot$

So I think that is straightforward the following identity:

$\ker (A)=Im(A)^\bot=\ker(A')$

So $\ker (A)=\ker (A')$

Question:

Are these moves valid? Is the kernel of the adjoint operator equal to the kernel of the operator ($\ker (A)=\ker (A')$)?

Answer 1

Even if $X=X'=Y=Y'$, it is not true that $\ker A'=\ker A$. For instance take $X=\mathbb C^2$, and $$ A=\begin{bmatrix} 0&1\\0&0\end{bmatrix}. $$ Then under the usual identifications one has $$ A'=\begin{bmatrix} 0&0\\0&1\end{bmatrix} $$ In this setting, the equality that holds is that $$ \ker A=\text{ran}\,(A')^\perp. $$ In the example above, one has $\text{ran}\,A\subset \ker A$.

Answer 2

This is not true. $\ker(A)$ is a subset of $X$, while $\ker(A')$ is a subset of $Y'$.

Where you initially seem to go off track is in the equality $\ker (A)=Im(A)^\bot$. Again, these are subsets of different spaces, hence cannot be equal.