Functions acting on matrix

Functions based on additions and multiplications can readily be defined for matrices. But many more functions are used. In particular, the function $e^x$ is of importance.

Exponential Function

There are two possible definitions of exponential function. Let a matrix $A$ be given. The matrix $e^A$ is then defined by $$ \begin{equation} e^A = \sum_{k=0}^{\infty} \frac{1}{k!} A^k. \end{equation} $$ where $A^0 = 1$. A second definition is $$ \begin{equation} e^A = \lim_{m \to \infty} \left(1 + \frac{1}{m} A \right)^m. \end{equation} $$

The proof that the two definitions (1) and (2) coincide is just like it goes for ordinary exponentials, by applying the binomial expansion on eq. (2).

Eq (2) allows us to derive an important relation= $$ \begin{equation} \text{Det}(e^A) = \lim_{m \to \infty} \left( \text{Det}\left( 1 + \frac{1}{m} A \right) \right)^m \end{equation} $$

We compute the determinant betweeen brackets by ignoring contributions of order $1/m$ or smaller. For instance, first consider a $2\times 2$ matrix. i.e.

$$ \begin{equation} 1 + \frac{1}{m} A = \begin{pmatrix} 1 + \frac{a_{11}}{m} & \frac{a_{12}}{m} \\ \frac{a_{21}}{m} & 1 + \frac{a_{22}}{m} \\ \end{pmatrix} \end{equation} $$

If, in the calculation of the determinant, we ignore the contributions of order $1/m^2$ and higher, the off-diagonal elements can be ignored (in a determinant, there are no terms containing only one off-diagonal element as a factor). In this approximation, the determinant is simply the product of the diagonal elements.

$$ \begin{align*} \text{Det}(1 + \frac{1}{m}A) & \approx (1 + \frac{a_{11}}{m}) (1 + \frac{a_{22}}{m})\\ & \approx 1 + \frac{1}{m} (a_{11} + a_{22}) + \mathcal{O}\left(\frac{1}{m^{2}}\right). \end{align*} $$

In general, one finds= $$ \begin{align*} \text{Det}(1 + \frac{1}{m}A) = 1 + \frac{1}{m} \text{Tr}(A) + \mathcal{O}\left(\frac{1}{m^{2}}\right). \end{align*} $$ where $\text{Tr}(A)$ is the trace of the matrix $A$, this means the sum of its diagonal elements, $$ \begin{align*} \text{Tr}(A) = \sum_{k} a_{kk}. \end{align*} $$

Substituting this in eq. (3) gives

$$ \begin{align*} \text{Det}(e^{A}) & = \lim_{m \to \infty} \left( 1 + \frac{1}{m} \text{Tr}(A) + \mathcal{O} \left( \frac{1}{m^{2}} \right)\right)^{m}\\ & = e^{\text{Tr}(A)} \end{align*} $$

and thus we arrive at $$ \begin{equation} \text{Det}(e^{A})= e^{\text{Tr}(A)}. \end{equation} $$

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Published on May 6, 2021

Last revised on Aug 6, 2021

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