Is the field of invertible complex linear operators algebraically closed? I am reading Axler's Linear Algebra Done Right and he proved every invertible linear complex operator has a square root. Following the prove one can show that they have any n-th root, hence I wonder if the stronger algebraically closed condition fulfills too. I have absolutely no idea how to tackle such problem.
In other words, let $T$ be an invertible operator in $\mathcal L(V)$, $P$ be a complex polynomial, is there an invertible operator $S$ such that $P(S)=T$.
 A: The answer is no.
Take $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Then $A$ isn't a square : there's no matrix $B$ such that $B^2=A$.
Indeed, such a matrix would commute with $A$. But
$$\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\0&0\end{pmatrix}$$ while
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&a\\0&c\end{pmatrix}$$
Therefore if such a matrix existed, it would be upper triangular (i.e. $c$ would be $0$).
But if its square is $A$ then it's diagonal must be $0$ as well (i.e. $a=d=0$).
But then, its square would be the null matrix, not $A$.
Edit : asking for invertible matrix doesn't change the answer.
Just take $A'=I_2+A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $P=X^2+1$. Then $$P(B)=A'\Longleftrightarrow B^2=A$$ which is impossible for the reasons explained above.
A: For invertible operators, this isn't even true for linear polynomials. For a simple counterexample, take $p(x) = x+1$ and let
$$T = \begin{pmatrix}1&1\\0&1\end{pmatrix}.$$
A: I feel obligated to complete Ayoub’s answer. The answer is indeed no, but it’s almost true (discounting the invertibility requirements added later, which are addressed in a comment) in the following sense: if the eigenvalues of $A$ each have one pre-image by $P$ which isn’t a root of $P’$, then it’s true.
Why? We can assume that $A$ is triangular by blocks, and all blocks have the diagonal of a scalar matrix. Clearly, we only need to solve for each block, and thus assume that $A=\alpha I+N$, $N$ nilpotent upper triangular and $\alpha \in P((P’)^{-1}(\mathbb{C}^{\times}))$. Write $\alpha=P(\beta)$ with $P’(\beta)\neq 0$, we search for a solution of the form $B=\beta I+N’$ with $N’$ nilpotent and upper triangular.
Then the equation can be re-written $P_1(N’)=N$, where $P_1$ is a complex polynomial such that $P_1(0)=0 \neq P_1’(0)$.
It can be shown by induction (ie find an explicit recursive formula on the coefficients) that there is a formal power series $f$ such that $P_1(f(T))=T$ and $f(0)=0$. Thus, if $f(T)=Q(T)+T^{n+1}g(T)$, where $g$ is a formal power series and $Q$ is a polynomial, $T^{n+1}|P_1(Q(T))-T$.
Thus by Cayley-Hamilton $N’=Q(N)$ works.
