How to show a matrix can't be written as exponential?

Question

How can I show the matrix $$A = \left( \begin{array}{c c} -2 & 0 \\ 0 & -1 \\ \end{array} \right)$$

can't be written as $A = exp(a)$?

I've tried to write A like

$$A = \left( \begin{array}{c c} 2 & 0 \\ 0 & 1 \\ \end{array} \right) \left( \begin{array}{c c} -1 & 0 \\ 0 & -1 \\ \end{array} \right) = BC$$ B and C can be reached by the exponential map: $$ BC = exp(b)exp(c)$$ $$ b = \left( \begin{array}{c c} 0 & \pi \\ -\pi & 0 \\ \end{array} \right) $$ $$ c = \left( \begin{array}{c c} \log(2) & 0 \\ 0 & 0 \\ \end{array} \right) $$

Using the Backer-Campbell-Hausdorff formula the trace of $a'=b+c+\frac{1}{2}[b,c]+...$ seems to be different from log(2) (which must be the trace of the ipotetically matrix $a$) but I've computed only the first three orders using: $$ [b,c] = \log(2) b$$

Answer 1

Perhaps you would be interested in this link? By this method, we know that the matrix would have to be complex to cope with the logarithms of negative numbers.

Answer 2

Note that if $A = \exp(B),$ then $B$ must commute with $A$, hence must also be diagonal. Let $B = [b_{ij}]$. Then $\exp(b_{11}) =-2$ and $\exp(b_{22}) = -1.$ Therefore $b_{11}$ and $b_{22}$ are not real.

Answer 3

You should note that every invertible complex matrix is the $exp$ of some complex matrix. This should have told straight away that there exists an $a$ (complex matrix) with the property you seek.