Exponentialish matrix

Question

"Explain why $e^{A(t+T)}=e^{At}e^{AT}$ using the formula $e^{At}=Se^{\Lambda t}S^{-1}$."

I really don't know where to start with this. For clarity $\Lambda$ is the diagonal eigenvalue matrix. $S$ is the eigenvector matrix. I can't tell from the context if $T$ is a matrix or means transpose.

Answer 1

It seems that $t$ and $T$ are scalars and that, since $\Lambda$ is diagonal, the matrix $e^{\Lambda t}$ denotes the diagonal matrix whose entries are obtained by replacing the diagonal entry $\Lambda_{ii}$ with $e^{\Lambda_{ii} t}$. Then

$$ e^{At}=Se^{\Lambda t}S^{-1},\qquad e^{AT}=Se^{\Lambda T}S^{-1},\qquad e^{A(t+T)}=Se^{\Lambda (t+T)}S^{-1}. $$

Now since $e^{\Lambda (t+T)}=e^{\Lambda t}e^{\Lambda T}$ (since this holds for scalar exponentials and therefore for diagonal matrices), it follows that $$ e^{\Lambda(t+T)}=Se^{\Lambda (t+T)}S^{-1}=Se^{\Lambda t}S^{-1}Se^{\Lambda T}S^{-1}=e^{\Lambda t}e^{\Lambda T}. $$

Answer 2

We are given that

$$e^{At}=Se^{\Lambda t}S^{-1}$$

Then, using $I=S^{-1}S$, we have

$$\begin{align} e^{A(t+T)}&=Se^{\Lambda (t+T)}S^{-1}\\\\ &=Se^{\Lambda t}e^{\Lambda T}S^{-1}\\\\ &=Se^{\Lambda t}(S^{-1}S)e^{\Lambda T}S^{-1}\\\\ &=e^{\Lambda t}e^{\Lambda T} \end{align}$$

as was to be shown!