Real logarithm of a real matrix?

Question

What is the real logarithm of \begin{equation} \begin{pmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{pmatrix}? \end{equation}

I know it exists via this theorem (found here):

A real matrix has a real logarithm if and only if it is invertible and each Jordan block belonging to a negative eigenvalue occurs an even number of times.

(If you know of a general algorithm for this then that would be very helpful).

Answer 1

Let $$ A = \pmatrix{ -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 } $$ There exists a matrix $P$ such that $P^{-1}AP = B$, where $B$ is given by $$ B = \pmatrix{ -I & I\\ 0 & -I } $$ Where $I$ here is the $2 \times 2$ identity matrix.

Now, starting from the other end, we define $$ J = \pmatrix{0&-1\\1&0}, \quad X = \pmatrix{\pi J & I\\0 & \pi J} $$ we then have $$ e^X = \pmatrix{e^{\pi J} & e^{\pi J}\\0 & e^{\pi J}} = \pmatrix{-I & -I\\0 & -I} $$ Now, there exists a matrix $Q$ such that $Qe^{X}Q^{-1} = B$.

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All together, we have $$ A = PBP^{-1} = PQe^{X}Q^{-1}P^{-1} \implies\\ A = \exp[(PQ)X(PQ)^{-1}] $$ Thus, $(PQ)X(PQ)^{-1}$ will be a matrix logarithm of $A$.