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For a complete normed vector space $X$, we define the mapping
$$exp:mathcal{L}(X;X)rightarrow mathcal{L}(X;X)$$
$$exp A:=E+frac{1}{1!}A+frac{1}{2!}A^2+cdots + frac{1}{n!}A^n+cdots.$$
Here $E$ is identity.
Show that begin{pmatrix}
-1 & 0\
1&-1
end{pmatrix}
is not of the form $exp A$ for any $2times 2$ matrix $A$.

I appreciate any help!

Edit: I think in this problem $A$ is assumed to be complex.

Edit 2: Now I think that the matrix is of the form $exp A$ for some complex matrix $A$. For example $A=begin{pmatrix} pi i&0\ -1& pi iend{pmatrix}$.

Assume the given matrix is of the form $exp(A)$.
Put $A$ in its Jordan normal form: $A = P^{-1}TP$, with $T$ lower triangular.
Then $exp(A) = P^{-1}exp(T)P$. But since $exp(A)$ is obviously already presented into its Jordan normal form, and since the Jordan form is unique (up to the permutation of the Jordan blocks, that reduce to a trivial block here), we must have $P = E$, that is, $exp(A) = exp(T)$. Then observe that $exp(T)_{1,1} = exp(T_{1,1})$ to obtain a contradiction.

EDIT: there is a problem with this argument: it is not possible to assert that $P = E$. To fix this bug, you have to show first that $P$ must be lower triangular (seek $P$ lower triangular explicitly with unknowns), hence (!)
$$(P^{-1}exp(T)P)_{1,1} = exp(T)_{1,1}= exp(T_{1,1})$$.

Here is the beginning of an answer in the case where $A$ is not supposed to be real, that can probably work with some more work.
Assume that the given matrix is of the form $exp(A)$. It is well known that $A$ can be triangulated, that is $A = P^{-1}TP$ with $T$ lower triangular. Hence $exp(A) = P^{-1}exp(T) P$. Since the determinant and the trace of a matrix is invariant under matrix conjugation, you have

$det(exp(T)) = det(exp(A)) = 1$ and $Tr(exp(T)) = Tr(exp(A)) = -2$.

Solving, it follows that the two diagonal elements of $exp(T)$ are equal to $-1$. Now, these elements are the $exp$ of the diagonal elements of $T$ (hence $T$ cannot be real otherwise $Tr(T) > 0$ in contradiction with $Tr(T) = -2$). Thus, the diagonal elements of $T$ are both of the form $ipi + 2kpi$. Working out the powers of $T$, it should be possible to show that $exp(T)$ cannot have a 1 below the diagonal (not checked).

Let $U=begin{pmatrix}-1&0\1&-1end{pmatrix}=-I+N$, where $N^2=0$.

Proposition 1. $U$ cannot be written in the form $e^A$ where $A$ is real.

Proof. If $e^A=U$, then $-I+N=(e^{A/2})^2=B^2$ is a square of a real matrix $B$ that is not diagonalizable over $mathbb{C}$ (because $N$ is not diagonalizable). Then $B$ has a double eigenvalue $pm i$; thus $tr(B)=pm 2i$ is not real, a contradiction.

Proposition 2. $U$ is the exponential of a complex matrix.

Proof. That is true iff $det(U)not= 0$ and, here, $det(U)=1$.

A particular solution is given by the OP; $A=ipi I_2-N$.

Indeed $e^A=e^{ipi} I_2e^{-N}=-(I-N)=U$.