Let $T$ be bounded linear functional on a Banach Space $X$. I wonder if ker$(T)$ is isomorphic is ker$(T^*)$, where $T^*$ is the Banach space dual of $T$. Note that $X$ need not be finite dimensional.
Let me discuss the source how did I come to this question. Murphy's book "$C^*$-algebras and Operator Theory", he says that:
Theorem: Let $T$ be compact operator on a Banach Space $X$ and $\lambda \in \mathbb{C}\setminus\{0\}$. Then $T-\lambda I$ is Fredholm and of index zero.
The next theorem easily concludes that it is Fredholm:
Theorem: Let $T$ be compact operator on a Banach Space $X$ and $\lambda \in \mathbb{C}\setminus\{0\}$. Then
Ker($T-\lambda I$) is finite dimensional.
Im$(T-\lambda)$ is closed and finite codimensional. In fact, $\frac{X}{T-\lambda I}$ and ker$(T^*-\lambda I)$ have same dimension.
Now since $T$ is compact, then $T^*$ is compact. So Ker$(T^*-\lambda I)$ is again finite dimensional.
So, $T-\lambda I$ is Fredholm. But to show that we need show that index is zero. for that we need to show that nul($T)$=def$(T)=$ dim $\frac{X}{T-\lambda I}=$ Ker$(T^*-\lambda I)$.
This is already false for Hilbert spaces. A typical example would be given by the unilateral shift. That is, let $X=\ell^2(\mathbb N)$, and $$\tag1 T(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots). $$ As it is well-known, $X^*=\ell^2(\mathbb N)$ and $$\tag2 T^*(x_1,x_2,\ldots)=(x_2,x_3,\ldots). $$ Then $\ker T=\{0\}$ while $\ker T^*=\mathbb C\,(1,0,0,\dots)$. That is, $$ \dim\ker T=0,\qquad\qquad\dim\ker T^*=1. $$ The same example works for any $\ell^p(\mathbb N)$, $1\leq p<\infty$. The dual is now $\ell^q(\mathbb N)$ with $q=p/(p-1)$, and the adjoint of $T$ is still as in $(2)$.
By considering $T^n$ we get examples where $\dim\ker T^n=0$ and $\dim\ker (T^*)^n=n$. And using direct sums with zero of the appropriate dimension plus switching roles if needed, one can see that for any $m,n\in\{0\}\cup\mathbb N$ there exists $T\in B(\ell^2(\mathbb N))$ with $$ \dim\ker T=m,\qquad\qquad\dim\ker T^*=n. $$