Calculating a matrix-exponential

Question

Let A be the following matrix. $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix} $$

I have to calculate $e^A$.

My idea was to diagonalize A because then $e^A = Pe^DP^-1$ if $A = PDP^-1$.

But A cannot be diagonalized since 1 is a double eigenvalue and therefore A does not have 2 linearly independent eigenvectors.

How else can I calculate $e^A$?

Thank you!

Answer 1

You can write : $A = I_2 + N$ with $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ nilpotent. Then, $N^2 = 0$ and $N$ commutes with $I_2$. So :

$$ A^n = \sum_{k = 0}^n \binom{n}{k} I_2^{n-k} N^k = I_2 + n N $$

Therefore :

$$ e^A = \sum_{k = 0}^{\infty} \frac{A^k}{k!} = I_2 \sum_{k = 0}^{\infty} \frac{1}{k!} + N \sum_{k = 0}^{\infty} \frac{k}{k!} = I_2 e + N e $$

This is a general strategy to compute the exponential even when the matrix is not diagonal.

Answer 2

Note that $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$.

So $e^A = \sum_{k=0}^{\infty} \frac{1}{k!}A^k = \sum_{k=0}^{\infty} \frac{1}{k!} \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \sum_{k=0}^{\infty} \frac{1}{k!} & \sum_{k=0}^{\infty} \frac{k}{k!} \\ 0 & \sum_{k=0}^{\infty} \frac{1}{k!} \end{pmatrix} = \begin{pmatrix} e & e \\ 0 & e \end{pmatrix}$

Answer 3

I offer a third method for finding matrix exponentials, which is a popular technique in Electrical Engineering.

Given matrix is $$ A = \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right] $$

We find $$ s I - A = \left[ \begin{array}{cc} s - 1 & -1 \\ 0 & s - 1 \\ \end{array} \right] $$ where $s$ is a complex variable.

We find that $$ (s I - A)^{-1} = \left[ \begin{array}{cc} {1 \over s - 1} & {1 \over (s - 1)^2} \\[2mm] 0 & {1 \over s - 1} \\[2mm] \end{array} \right] $$

The inverse Laplace transform of $(s I - A)^{-1}$ is the state transition matrix, which is also the matrix exponential, $e^{A t}$.

Hence, we get $$ e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right] = \left[ \begin{array}{cc} \mathcal{L}^{-1}\left( {1 \over s - 1} \right) & \mathcal{L}^{-1}\left( {1 \over (s - 1)^2} \right) \\[2mm] \mathcal{L}^{-1}(0) & \mathcal{L}^{-1}\left( {1 \over s - 1} \right) \\[2mm] \end{array} \right] = \left[ \begin{array}{cc} e^{t} & t e^t \\[2mm] 0 & e^t \\[2mm] \end{array} \right] $$

In particular, we get $$ e^{A} = \left[ \begin{array}{cc} e^{1} & 1 e^1 \\[2mm] 0 & e^1 \\[2mm] \end{array} \right] = \left[ \begin{array}{cc} e & e \\[2mm] 0 & e\\[2mm] \end{array} \right] $$