Matrix Exponential of Identity Matrix

Question

I was just wondering what would the sum be of $e^{I_n}$ where $I_n$ is the identity matrix. I know the maclaurin series for $e^x$ is $1+\frac x{1!}+\frac {x^2}{2!}+...$. I know that $e^0$ is 1 right? How should I calculate $e^{I_n}$

Answer 1

For a diagonal matrix $M$ the exponential is simply the exponential of each diagonal entry: $$e^M=\left(\begin{array}{cccc} e^{m_{11}} & 0 & \ldots & 0 \\ 0 & e^{m_{22}} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & e^{m_{nn}} \end{array}\right).$$ You can show this from the identity $$e^M = \sum_{k=0}^\infty \frac{1}{k!} M^k.$$ Since the product of two diagonal matrices $A$ and $B$ is the product of their diagonal elements, $$\left(\begin{array}{cccc} a_{11} & 0 & \ldots & 0 \\ 0 & a_{22} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & a_{nn} \end{array}\right) \left(\begin{array}{cccc} b_{11} & 0 & \ldots & 0 \\ 0 & b_{22} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & b_{nn} \end{array}\right) =\left(\begin{array}{cccc} a_{11}b_{11} & 0 & \ldots & 0 \\ 0 & a_{22}b_{22} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & a_{nn}b_{nn} \end{array}\right),$$ it easily follows that $e^M$ is as shown above.

Answer 2

You can use the Taylor expansion of the exponential,

$\displaystyle e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$.

All the powers of the identity matrix are equal to the identity matrix, so $\displaystyle I_n^k = I_n$.

Therefore,

$\displaystyle e^{I_n} = I_n \sum_{k=0}^\infty \frac{1}{k!} = e \, I_n$.

This is a diagonal matrix with all the terms in the diagonal being $e$.