Matrixexponential

Description

The matrixexponention generalised the Exponentialfunktion to matrices. The idea is simple. We take the Taylorentwicklung of the exponential function and plug in a matrix.

Definition

Let $A\in M^{n\times n}$ be square matrix. The matrix exponential is defined as: $e^A=\sum _{k=0}^{\infty }\frac{A^k}{k!}$

Properties

Respects conjugation (in a weird way)

If $B=TA{T}^{-1}$, then $e^{tB}=T{e}^{tA}{T}^{-1}$.

Little thought: What still confuses me is the fact that there are two ways of solving linear equations $x^{\prime }=Ax$. The first is quite easy to understand. It’s $e^{tA}$ and is calculated by taking the Eigenvektor of $A$ to write $A$ in Jordannormalform form. The second way is to calculate the eigenvalues ${\lambda }_i$ and eigenvectors $v_i$ and then just say that the $v_i{e}^{{\lambda }_{i}t}$ generate all solutions. At least in the diagonal case. Otherwise it is slightly more complicated. How are these two connected? I think I got it. Diagonalise $T^{1}AT=D$. Then the latter solution set is just $T{e}^{tD}$. Or more exactly, the products, when we multiply the basis vectors to $T{e}^{tD}$. The right term gives us $e^{t{\lambda }_i}$. Then the second term makes $v_i{e}^{t{\lambda }_i}$ out of it. But the issue here is that if we want to find out what the solution is for the initial value $(2,0)$ then multiplying this does’t make any sense since the input is in terms of the eigenvectors and the output in terms of the problem. We must use the full $T{e}^{tD}{T}^{-1}$ which we can do, by adding the inverse of zero: $T{e}^{tD}{T}^{-1}=T{e}^{tD}{e}^{0D}{T}^{-1}=\Phi (t){\Phi }^{-1}(0)$. I don’t really like this but I guess, it is useful because you are given $\Phi$ immediately. Though you could also just do $e^{tA}=T{e}^{tD}{T}^{-1}=\Phi (t)T^{-1}$. Yeah, whatever. I don’t like it.
Nevermind, it turns out that this is actually useful. It turns out that the “transformation matrix way” is generally not as complex, since the matrices are not as complex.

Examples

Diagnonal matrix and Jordan blocks

Let $A=\left(\begin{array}{ccc}{\lambda }_1& ...& 0\\ ⋮& \ddots & ⋮\\ 0& ..& {\lambda }_n\end{array}\right)$. Then the exponential matrix is given as $e^{tA}=\left(\begin{array}{ccc}{e}^{t{\lambda }_1}& ...& 0\\ ⋮& \ddots & ⋮\\ 0& ...& e^{t{\lambda }_n}\end{array}\right)$ Let $B=\left(\begin{array}{cccc}{\lambda }_1& 1& & 0\\ & \ddots & \ddots & \\ & & \ddots & 1\\ 0& & & {\lambda }_n\end{array}\right)$. Then the exponential matrix is given as $e^{tB}=\left(\begin{array}{cccc}{e}^{\lambda t}& t{e}^{\lambda t}& ...& \frac{t^{n-1}}{(v-1)!}{e}^{\lambda t}\\ & \ddots & & ⋮\\ & & \ddots & t{e}^{\lambda t}\\ 0& & & e^{\lambda t}\end{array}\right)$

I would love to understand the deeper connection here but I don’t have time for that Well, easy to explain is the following fact: If $t=0$, then nothing happened and indeed, the matrix is the identity. Note: The matrix exponential is not the same thing as the Fundamentalmatrix. The fundamental matrix are just the linearly independent solutions written next to another. But both seem to solve the system of equations.