For a square matrix $M$, its matrix exponential is defined by
\[e^M = \sum_{i=0}^\infty \frac{M^k}{k!}.\]

Suppose that $M$ is a diagonal matrix
\[ M = \begin{bmatrix} m_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n n} \end{bmatrix}.\]

Find the matrix exponential $e^M$.

Solution.

First, we find $M^k$ for each integer $k \geq 0$. The first couple powers can be calculated directly,
\[M^0 = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} , \quad M = \begin{bmatrix} m_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n n} \end{bmatrix},\] \[M^2 = \begin{bmatrix} m^2_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^2_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^2_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^2_{n n} \end{bmatrix} , \quad M^3 = \begin{bmatrix} m^3_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^3_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^3_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^3_{n n} \end{bmatrix}.\]

The general pattern can now be seen:
\[M^k = \begin{bmatrix} m^k_{1 1} & 0 & 0 & \cdots & 0 \\ 0 & m^k_{2 2} & 0 & \cdots & 0 \\ 0 & 0 & m^k_{3 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^k_{n n} \end{bmatrix}.\]

Now, we can calculate the infinite series $e^M$:
\begin{align*}
e^M &= \sum_{k=0}^{\infty} \frac{ M^k }{k!} \\
&= \sum_{k=0}^\infty \frac{1}{k!} \begin{bmatrix} m^k_{1, 1} & 0 & 0 & \cdots & 0 \\ 0 & m^k_{2, 2} & 0 & \cdots & 0 \\ 0 & 0 & m^k_{3, 3} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m^k_{n, n} \end{bmatrix} \\
&= \begin{bmatrix} \sum_{k=0}^\infty \frac{ m^k_{1 1} }{k!} & 0 & 0 & \cdots & 0 \\ 0 & \sum_{k=0}^\infty \frac{ m^k_{2 2} }{k!} & 0 & \cdots & 0 \\ 0 & 0 & \sum_{k=0}^\infty \frac{ m^k_{3 3} }{k!} & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \sum_{k=0}^\infty \frac{ m^k_{n n} }{k!} \end{bmatrix} . \end{align*}

Now, for any real number $c$ we can write $e^c$ as the series
\[e^c = \sum_{k=0}^\infty \frac{ c^k }{k!}.\]

Thus, the matrix exponential $e^M$ is
\[e^M = \begin{bmatrix} e^{ m_{1 1} } & 0 & 0 & \cdots & 0 \\ 0 & e^{ m_{2 2} } & 0 & \cdots & 0 \\ 0 & 0 & e^{ m_{3 3} } & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & e^{ m_{n n} } \end{bmatrix}.\] Click here if solved 35

The post The Matrix Exponential of a Diagonal Matrix first appeared on Problems in Mathematics.