Finding the matrix exponential

Question

Find the matrix exponential of

1. $$\begin{bmatrix}1& 1\\ 0& 1\end{bmatrix}.$$ Since this matrix is not diagonalizable, you will have to use the definition of the matrix exponential.

2. $$\begin{bmatrix}-6& -2\\ 10&3\end{bmatrix}$$

Answer 1

For the first matrix:

The definition of exponential matrices is $$e^A=I+A+\frac12 A^2 + \frac1{3!}A^3 + ...$$

Try and calculate $A^i$ for some small values of $i$ to see if you can find a simple patern.

For the second matrix: is this matrix diagonalizable?

Answer 2

The first matrix can be factored as

$A=\begin{bmatrix}1& 1\\ 0& 1\end{bmatrix}=\begin{bmatrix}1& 0\\ 0& 1\end{bmatrix}+\begin{bmatrix}0& 1\\ 0& 0\end{bmatrix}=X+N$

So, $e^A=e^Xe^N=\begin{bmatrix}e& e\\ 0& e\end{bmatrix}$

The second is digonalizable:

$B=\begin{bmatrix}-6& -2\\ 3& 10\end{bmatrix}=\begin{bmatrix}-1/2& -5/2\\ 1& 1\end{bmatrix}\begin{bmatrix}-2& 0\\ 0& -1\end{bmatrix}\begin{bmatrix}-1/2& -5/2\\ 1& 1\end{bmatrix}^{-1}=PEP^{-1}$

So, $e^B=Pe^EP^{-1}=e^{-2}\begin{bmatrix}5-4e& 2(1-e)\\ 10(e-1)& 5e-4\end{bmatrix}$

Answer 3

For a Jordan Block, we have $A = I + N$:

$$e^A = e^{j_\lambda} = e^{\lambda I + N} = e^{\lambda}\sum_{j=0}^{n-1} \dfrac{1}{j!}N^j$$

This gives:

$$e^A = e\begin{bmatrix}1& 1\\ 0& 1\end{bmatrix} = \begin{bmatrix}e& e\\ 0& e\end{bmatrix}$$

For the second, we can diagonalize the matrix as:

$$A = PJP^{-1} = \begin{bmatrix}-1& -2\\ 2& 5\end{bmatrix}\begin{bmatrix}-2& 0\\ 0& -1\end{bmatrix}\begin{bmatrix}-5& 2\\ 2& 1\end{bmatrix}$$

Hence:

$$e^A = Pe^JP^{-1} = \begin{bmatrix}\frac{5-4 e}{e^2} &-\frac{2 (e-1)}{e^2} \\ \frac{10 (e-1)}{e^2}&\frac{5 e-4}{e^2}\end{bmatrix}$$