Let $T:V\to V$ be a diagonalizable linear map on a vector space $V$. Does it hold that $T^{**}:V^{**}\to V^{**}$ is diagonalizable? Edit:  I found my own answer.  But I will be happy to see other answers as well.

Question: Let $T:V\to V$ be a linear operator on a vector space $V$.  (a) Suppose that $T$ is diagonalizable. Does it hold that $T^{**}:V^{**}\to V^{**}$ is diagonalizable?  (b) If $V$ has a generalized eigenspace decomposition with respect to $T$, then does it also hold that $V^{**}$ has a generalized eigenspace decomposition with respect to $T^{**}$?

Here, $V$ is defined over a field $F$ and $V^*$ denotes the algebraic dual $\operatorname{Hom}_F(V,F)$ of $V$.  So, the double dual of $V$ is $V^{**}=(V^*)^*$.  The dual map $T^*:V^*\to V^*$ is defined by $T^*(f)=f\circ T$ for each $f:V\to F$.  Similarly, the double dual $T^{**}=(T^*)^*$ satisfies $$T^{**}\varphi(f)=(\varphi\circ T^*)(f)=\varphi(f\circ T)$$ for all $\varphi:V^*\to F$ and $f\in V^*$.
We say that $T$ is diagonalizable if $V$ has a direct sum decomposition into eigenspaces $V_\lambda$ of $T$
$$V=\bigoplus_{\lambda\in F} V_\lambda,$$
where $V_\lambda=\big\{v\in V:Tv=\lambda v\big\}$.  If $V_\lambda\neq \{0\}$, $\lambda$ is an eigenvalue of $T$.  
We can also say that $T$ is Jordanizable if $V$ has a direct sum decomposition into generalized eigenspaces $V^\lambda$ of $T$
$$V=\bigoplus_{\lambda\in F} V^\lambda,$$
where $V^\lambda=\big\{v\in V:(T-\lambda)^dv=0\text{ for some integer }d>0\big\}$.  If $V^\lambda\neq \{0\}$, $\lambda$ is a generalized eigenvalue of $T$.  A Jordan block of $T$ is a subspace $J$ of some $V^\lambda$ that is an indecomposable $F[X]$-submodule with the action $X\cdot v=Tv$ for all $v\in V$ that is not a subspace of a larger indecomposable $F[X]$-submodule of $V$.

Here is what is known.
If $V$ is finite dimensional, then we have a natural identification $V=V^{**}$, so $T$ and $T^{**}$ are basically the same operator under this identification.   Thus, the answers to both questions are positive.
However, if $V$ is infinite dimensional, things are a little cloudy.  I know that if $T$ is diagonalizable and has finitely many eigenvalues, then so does $T^{**}$.  Hence, the answer to (a) is yes in this case.  But what if $T$ has infinitely many eigenvalues?
We can say something similar for (b).  Suppose  $V$ can be decomposed into a direct sum of finitely many generalized eigenspaces of $T$.  If for each generalized eigenvalue $\lambda$, there exists a positive integer $d_\lambda$ such that any $(T-\lambda)^{d_\lambda}$ vanishes on the generalized eigenspace $V^\lambda$, then $V^{**}$ also can be decomposed into a direct sum of finitely many generalized eigenspaces of $T^{**}$.  But I don't have the answer if $T$ has infinitely many generalized eigenvalues or if $d_\lambda$ does not exist for some $\lambda$.
The idea of the proof of my partial results for (a) and (b) is that if $p(T)=0$ for some polynomial $p$, then $p(T^{**})=0$ as well.  But this proof does not work if $T$ has infinitely many (generalized) eigenvalue or if there exist arbitrarily large Jordan blocks (in terms of dimensions).  I do not expect that the answer to (a) is yes (and so this will answer (b) as well), but I cannot find a way to construct a counterexample.

In fact, let $V^{*m}$ denote the $m$th dual of $V$ and $T^{*m}$ the $m$th dual of $T$ (i.e., $V^{*m}=\big(V^{*(m-1)}\big)^*$ and $T^{*m}=\big(T^{*(m-1)}\big)^*$, with $V^{*0}=V$ and $T^{*0}=T$).  If $T$ satisfies the polynomial equation $p(T)=0$, then every $T^{*m}$ satisfies $p\big(T^{*m}\big)=0$.  Hence, if $T$ is diagonalizable with finitely many eigenvalues, then so are all $T^{*m}$.  If $V$ has a generalized eigenspace decomposition wrt $T$ with finitely many generalized eigenvalues and there is a universal upper bound on the dimension of every Jordan block, then $V^{*m}$ has a generalized eigenspace decomposition wrt $T^{*m}$.
 A: Answer for (b).
Let $V$ be the space of real sequences $(x_1,x_2,x_3,\ldots)$ with finitely many non-zero members.  Then, $V=\bigoplus_{k=1}^\infty\Bbb{C}e_k$, where $e_i$ is the sequence $(0,0,\ldots,0,1,0,0,\ldots)$ with $1$ at the $i$th entry and $0$ at all other places.  Hence, we can identify $V^*$ with $\prod_{k=1}^\infty \Bbb{C}e_k$.  In other words, for $f\in V^*$, we identify $f$ with $$(f_1,f_2,f_3,\ldots)=\big(f(e_1),f(e_2),f(e_3),\ldots\big).$$
In this way, $V\subseteq V^*$ and $e_1,e_2,e_3,\ldots$ are linearly independent elements of $V^*$.
Consider $T:V\to V$ to be the left shift operator
$$(x_1,x_2,x_3,\ldots)\mapsto (x_2,x_3,x_4,\ldots).$$
Then, $T$ has one generalized eigenvalue $0$, and so $V=V^0$.  We claim that $T^{**}:V^{**}\to V^{**}$ is not Jordanizable.  First, we find all eigenvalues of $T^{**}$.  
Suppose that $\lambda$ is an eigenvalue of $T^{**}$.  Then, there exists a non-trivial $\varphi\in V^{**}$ such that $T^{**}(\varphi)=\lambda\varphi$.  Hence,
$$\lambda\varphi(f)=T^{**}\varphi(f)=\varphi(f\circ T)$$
for every $f\in V^*$.  As before, write $f=(f_1,f_2,f_3,\ldots)$.  Then, $$f\circ T=(0,f_1,f_2,f_3,\ldots).$$
Hence,
$$0=\varphi(f\circ T)-\lambda \varphi(f)=\varphi(0,f_1,f_2,\ldots)-\lambda \varphi(f_1,f_2,\ldots)=\varphi(-\lambda f_1,f_1-\lambda f_2,f_2-\lambda f_3,\ldots).$$
If $\lambda \neq 0$, then for any $g=(g_1,g_2,g_3,\ldots)\in V^*$, there exist $f_1,f_2,f_3,\ldots\in\Bbb{C}$ such that $$g_1=-\lambda f_1,\ g_2=f_1-\lambda f_2,\ g_3=f_2-\lambda f_3,\ \ldots,$$
i.e.,
$$f_1=-\frac{g_1}{\lambda},\ f_2=-\frac{g_2}{\lambda}+\frac{g_1}{\lambda^2},\ f_3=-\frac{g_3}{\lambda}+\frac{g_2}{\lambda^2}-\frac{g_1}{\lambda^3},\ \ldots.$$ 
Thus, $\varphi(g)=0$ for all $g\in V^*$, which gives $\varphi=0$, contradicting the assumption that $\varphi$ is non-trivial.  Therefore, $\lambda=0$.
We have shown that $T^{**}$ has only one eigenvalue $0$.  Thus, we need to show that $V^{**}$ is not a generalized eigenspace corresponding to the generalized eigenvalue $0$.  We do this by picking a certain $\varphi$.  Let $B$ be a basis of $V^*$ extending $\{e_1,e_2,e_3,\ldots\}\cup\{E\}$, where
$$E=(1,1,1,\ldots).$$
Define $\varphi:V^{*}\to\Bbb{C}$ by linearly extending the relations
$$\varphi(e_k)=0,\ \varphi(E)=1,\ \text{and}\ \varphi(b)=0$$
for all $k=1,2,3,\ldots$ and $b\in B\setminus\{E,e_1,e_2,e_3,\ldots\}$.  If $\varphi$ is in the generalized eigenspace $\big(V^{**}\big)^0$, then $\big(T^{**}\big)^n\varphi=0$ for some positive integer $n$.  However, for any $n=1,2,3,\ldots$, we get
$$\big(T^{**}\big)^n\varphi(E)=\varphi(E-e_1-e_2-\ldots-e_n)=\varphi(E)-\varphi(e_1)-\varphi(e_2)-\ldots-\varphi(e_n)=1,$$
which is nonzero.  So, $\varphi$ is not in $\big(V^{**}\big)^0$, so $T^{**}$ is not Jordanizable.  Indeed, one can show that $\big(V^{**}\big)^0=V^0=V$, where we identify $V$ as a subspace of $V^{**}$ via the canonical injection $V\hookrightarrow V^{**}$.

Answer for (a).
Let $V$ be the same as before.  Now, $T:V\to V$ is defined to be
$$T(x_1,x_2,x_3,\ldots)=(x_1,2x_2,3x_3,\ldots).$$
Clearly, $T$ is diagonalizable with eigenvalues $1$, $2$, $3$, $\ldots$.  That is, $V=\bigoplus_{k=1}^\infty V_k$ with $V_k=\Bbb{C}e_k$ is the eigenspace corrsponding to the eigenvalue $k$ of $T$.  First, I claim that the only eigenvalues of $T^{**}$ are $1$, $2$, $3$, $\ldots$.
To show this, let $\lambda\in\Bbb{C}$ and $\varphi\in V^{**}$ such that $T^{**}\varphi=\lambda\varphi$.  For $f=(f_1,f_2,f_3,\ldots)\in V^*$, we have
$$f\circ T=(f_1,2f_2,3f_3,\ldots).$$
That is,
$$0=T^{**}\varphi(f)-\lambda\varphi(f)=\varphi\big((1-\lambda)f_1,(2-\lambda)f_2,(3-\lambda)f_3,\ldots\big).$$
If $\lambda\notin\Bbb{N}_1$, then for any $g=(g_1,g_2,g_3,\ldots)\in V^*$, there exist $f_1,f_2,f_3,\ldots\in\Bbb{C}$ such that
$$f_k=\frac{g_k}{\lambda-k}$$
for all $k$.  Thus, $\varphi(g)=0$ for all $g\in V^*$, making $\varphi=0$.  Thus, the only possible eigenvalues of $T^{**}$ are the positive integers.
We now identify the eigenspace of $T^{**}$ with eigenvalue $k\in\Bbb{N}_1$.  Observe that for $\varphi$ in the eigenspace $\big(V^{**}\big)_k$, we have
$$\varphi\big((1-k)f_1,(2-k)f_2,\ldots,-f_{k-1},0,f_{k+1},2f_{k+2},\ldots\big)=0.$$
For any $g=(g_1,g_2,g_3,\ldots)\in V^*$, we have
$$g=g_ke_k+\big((1-k)f_1,(2-k)f_2,\ldots,-f_{k-1},0,f_{k+1},2f_{k+2},\ldots\big),$$
where
$$f_j=\frac{g_j}{j-k}$$
for all $j\neq k$.  Thus,
$$\varphi(g)=\varphi(g_ke_k)=g_k\varphi(e_k).$$
Since $\varphi\neq 0$, we must have $\varphi(e_k)\neq 0$.  WLOG, $\varphi(e_k)=1$.  This shows that $\varphi$ is precisely the image of $e_k$ under the canonical embedding $V\hookrightarrow V^{**}$.  Thus, $\big(V^{**}\big)_k=\Bbb{C}e_k$ (if we identify $V$ as a subspace of $V^{**}$).
Now, $V^{**}$ is not equal to the direct sum $\bigoplus_{k=1}^\infty \big(V^{**}\big)_k=\bigoplus_{k=1}^\infty \Bbb{C}e_k=V$ simply because $V^{**}$ has an uncountable dimension, and $V$ is a countable dimensional vector space.  This proves that $T^{**}$ is not diagonalizable.  Indeed, we can also prove that $\big(V^{**}\big)^k=\big(V^{**}\big)_k=\Bbb{C}e_k$.  Thus, $T^{**}$ is not Jordanizable either.

Interestingly, it is possible that $T^*$ has no generalized eigenspaces at all in $V^*$, even if $V$ decomposes into the direct sum of generalized eigenspaces of $T$. 
The same $V$ as in the answers above works just fine, with $T$ being the left shift operator
$$(x_1,x_2,x_3,\ldots)\mapsto(x_2,x_3,x_4,\ldots).$$
It turns out that $T^*$ has no eigenvalues, so it also has no generalized eigenvalues.
However, if $V$ has a generalized eigenspace decomposition wrt $T$ with finitely many generalized eigenvalues and with a universal upper bound for the dimensions of all Jordan blocks, then the generalized eigenvalues $T^{*m}$  are precisely the generalized eigenvalues of $T$.  In such case, for a generalized eigenvalue $\lambda$ of $T$ with maximum Jordan block dimension $d_\lambda$, the maximum Jordan block dimension of $T^{*m}$ with the generalized eigenvalue $\lambda$ is also $d_\lambda$.  Therefore, the minimal polynomial of $T^{*m}$ is the same as the minimal polynomial of $T$.
