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

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.

-

## 2 Answers

I'm afraid your task is impossible as written. The following pair of matrices have matching trace and determinant, but different exponentials.

$$X=\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix},$$

$$Y=\begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}.$$

The difference is not enormous, because

$$X=\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$$ and $$Z=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ commute, $XZ = ZX = a Z,$ so we do get $$e^Y = e^{X+Z} = e^X e^Z = e^Z e^X,$$ which you do not get when commutativity of a pair fails.

-
Thanks, that's what I suspected. –  Jimmy R Feb 9 '13 at 20:20
@JimmyR, for 2 by 2 the characteristic polynomial is quadratic, the roots are explicit using the quadratic formula, so finding the Jordan Normal Form can be done without decimal approximation, either by hand or machine. See math.stackexchange.com/questions/80324/sina-where-a-is-a-matrix/… about the gerbils. –  Will Jagy Feb 9 '13 at 20:26

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.

-