Is there a real matrix A such that (Exponential of matrices)

Question

Is there a real matrix A such that

$$\exp(A) = \begin{bmatrix} -\alpha & 0 \\ 0 & -\beta \end{bmatrix}, \text{ where }\alpha,\beta>0?$$

(Hint): In two dimensions the exponential matrix of

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

is given by:

$$\exp(A) = e^{\delta}\cos(\Delta)\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + e^{\delta}\frac{\sin(\Delta)}{\Delta}\begin{bmatrix} \phi & b \\ c & -\phi \end{bmatrix},$$

where $\delta=\dfrac{a+d}{2}$, $\phi=\dfrac{a-d}{2}$, $\Delta=\sqrt{\phi^2+bc}$.

Does this exercise only reduce the multiplication of the matrices given in the tip?

Answer 1

$$ A = \left( \begin{array}{cc} 0 & \pi \\ - \pi & 0 \end{array} \right) $$

The point is this: by straightforward summing, we can find, for real number $t,$ given

$$ B = \left( \begin{array}{cc} 0 & t \\ - t & 0 \end{array} \right) \; \; , $$

we get

$$ e^B = \left( \begin{array}{cc} \cos t & \sin t \\ - \sin t & \cos t \end{array} \right) \; \; . $$

That is, even powers of $B$ are diagonal, and odd powers of $B$ are off-diagonal, in fact skew symmetric. We are looking for $$ e^B = I + B + \frac{1}{2} B^2 + \frac{1}{6} B^3 + \frac{1}{24} B^4+ \frac{1}{120} B^5+ \frac{1}{720} B^6 + \cdots $$ We get the cosines on the diagonal from $$ I + \frac{1}{2} B^2 + \frac{1}{24} B^4+ \frac{1}{720} B^6 + \cdots $$ and the sines skew-symmetric from $$ B + \frac{1}{6} B^3 + \frac{1}{120} B^5 + \cdots $$

Answer 2

Let $A = PJP^{-1}$ be the complex Jordan normal form of $A$. Then, we have $\exp(A) = P\exp(J) P^{-1}$.

The Jordan block structure is (up to permutation) unique. Hence, $\exp(J)$ is also diagonal and (up to permutation) has the diagonal $-\alpha, -\beta$. In particular $J$ is also diagonal with non real valued diagonal component.

As $A$ is real, it follows both eigenvalues are not real, conjugated to each other, and distinct. Let $\lambda = x + yi$ denote an eigenvalue. Than, we have $$ \exp(\lambda) = \underbrace{\exp(x)}_{>0} \exp(yi), $$ which is real and negative if and only if $\exp(yi) = -1$, that is $y=\pi \bmod 2\pi$. Using the same logic on $\bar \lambda$, we obtain that $\alpha$ needs to be $\beta$.

So there is a solution $A$ if and only if $\alpha = \beta$. In that case $A$ is given by $$ A = \log(\alpha) I + \log(-I), $$ where $\log(-I)$ is multi-valued.

Aside:

• A canonic choice for $\log(-I)$ is given in @WillJagy's answer.

• Once you established $\alpha=\beta$, you can just interpret $A$ as a complex number $z = -\alpha + 0i$ and the question reduces to the complex logarithm.

Answer 3

The actual recipe given in Teschl is this, page 63. See that there are explicit notes about what to do when the square root will be of a negative quantity, $$ \cosh i \Delta = \cos \Delta \; , $$ $$ \frac{\sinh i \Delta}{i \Delta} = \frac{\sin \Delta}{\Delta} \; . $$

Using the symbols below, namely matrix $A,$ then $\delta = \frac{a+d}{2} \; , \; $ $\gamma = \frac{a-d}{2} \; . \; $ With real $t \neq 0,$ $$ \gamma^2 + bc < 0 \; , \; \; \mbox{LET} \; \; t = \sqrt { \;- \left( \gamma^2 + bc\right) \; \; } $$ THEN $$ e^A = e^\delta \cos t \; I \; + \; \frac{e^\delta \sin t}{t} \; \left( \begin{array}{cc} \gamma & b \\ c & -\gamma \end{array} \right) $$

Uses $$ \frac{1}{2ict} \left( \begin{array}{cc} \gamma - it & \gamma + it \\ c & c \end{array} \right) \left( \begin{array}{cc} \delta - it & 0 \\ 0 & \delta + it \end{array} \right) \left( \begin{array}{cc} -c & \gamma + it \\ c & - \gamma + it \end{array} \right) = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) $$