[Math] Computing exponential of a $2\times 2$ matrix using only its trace and determinant.

Question

I want to compute the exponential of an arbitrary $2\times 2$ matrix over $\mathbb{R}$ only using its trace and determinant. I've shown that for a traceless matrix $A$ there is the following formula:
$$
\exp(A)=(\cos\sqrt{\det A})\mathbb{I}+\frac{\sin\sqrt{\det A}}{\sqrt{\det A}}A
$$
Suppose $X$ is an arbitrary matrix, say $X=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then $X$ can be written as a sum:
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}=\underbrace{\begin{pmatrix} \frac{a-d}{2} & b \\ c & -\frac{a-d}{2}\end{pmatrix}}_{T}+\underbrace{\frac{a+d}{2}\mathbb{I}}_{S}
$$
Notice that $TS=ST$, so $\exp(X)=\exp(T)\exp(S)$. Since $T$ is traceless I can use the formula and write $\exp(T)$ in terms of its determinant, but my final result needs to depend on determinant and trace of $X$ only, so
$$
\det T=\frac{a-d}{2}\cdot\frac{d-a}{2}-bc=-\frac{a^2+d^2}{2}+ad-bc=-\frac{a^2+d^2}{2}+\det X
$$
This is where I'm stuck, as I want to write $a^2+d^2$ in terms of $\det X=ad-bc$ and $\mathrm{tr} X=a+d$, but cannot do that.

Answer 1

This is in fact possible for a 2x2 matrix and the factorization you are after is:

$\det T=-1/4\, \left( a-d \right) ^{2}-bc=ad-bc-1/4\, \left( a+d \right) ^{2}=\det X-1/4\,(\mathrm{tr} X)^{2}$.

Alternatively, note that for a 2x2 matrix the characteristic equation can be written explicitly in terms of the determinant and the trace (http://en.wikipedia.org/wiki/Characteristic_polynomial):

$\det (X-\lambda\mathbb{I})=\det X-(\mathrm{tr} X) \lambda+\lambda^{2},$

and in general:

$\det T=\det (X-1/2(\mathrm{tr} X)\mathbb{I}),$

so letting $\lambda=1/2(\mathrm{tr} X)$ gives the same result. In the nxn case you would need more than $\det X$ and $\mathrm{tr} X$ to fix such relationships; you would need n independent invariants so you could consider using the eigenvalues of $X$.

If two matrices share the same determinant and trace, they may not necessarily share the same exponential, as your formula also features an explicit dependence on the matrix itself.