1. ## Linear Algebra/Differential equations

How can we show that the matrix differential equation dX/dt=AX+XB has the solution $\displaystyle X(t)=e^{At}X(0)e^{Bt}$ and prove that the solutions of dX/dt=AX-XA keep the same eigenvalues for all time.

2. You can check that that is indeed a solution by differentiating, now suppose we have a solution $\displaystyle X(t)$ we will prove it has to be of that form (uniqueness)

Remember that the product rule also holds for Matrixes, we have:

$\displaystyle \left( {e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} } \right)^\prime = \left( {e^{ - A \cdot t} \cdot X\left( t \right)} \right)^\prime \cdot e^{ - B \cdot t} + \left( {e^{ - A \cdot t} \cdot X\left( t \right)} \right) \cdot \left( { - B \cdot e^{ - B \cdot t} } \right)$

Hence: $\displaystyle \left( {e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} } \right)^\prime = \left[ {\left( { - A \cdot e^{ - A \cdot t} } \right) \cdot X\left( t \right) + e^{ - A \cdot t} \cdot X'\left( t \right)} \right] \cdot e^{ - B \cdot t} +$$\displaystyle \left( {e^{ - A \cdot t} \cdot X\left( t \right)} \right) \cdot \left( { - B \cdot e^{ - B \cdot t} } \right)$

$\displaystyle C$ and $\displaystyle e^C$ commute, for every matrix $\displaystyle C$.

Thus: $\displaystyle \left( {e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} } \right)^\prime = e^{ - A \cdot t} \left[ {\underbrace { - A \cdot X\left( t \right) + X'\left( t \right) - B \cdot X\left( t \right)}_{ = \bold{0}_{n \times n} }} \right]e^{ - B \cdot t} = \bold{0}_{n \times n}$

So it is a constant matrix ( because of the 0 derivative) : $\displaystyle e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} = M$

Let $\displaystyle t=0$ we get: $\displaystyle e^{ - A \cdot 0} \cdot X\left( 0 \right) \cdot e^{ - B \cdot 0} = {\text{Id}}_{n \times n} \cdot X\left( 0 \right) \cdot {\text{Id}}_{n \times n} = X\left( 0 \right) = M$

$\displaystyle e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} = X\left( 0 \right) \Rightarrow X\left( t \right) = e^{A \cdot t} \cdot X\left( 0 \right) \cdot e^{B \cdot t}$ And the uniqueness is proven.

For the second part of the question, let $\displaystyle B=-A$

The solutions is of the form: $\displaystyle e^{A \cdot t} \cdot X(0) \cdot e^{ - A \cdot t}$

Let's see the characteristic polynomial:

$\displaystyle \det \left( {e^{A \cdot t} \cdot X(0) \cdot e^{ - A \cdot t} - \lambda \cdot {\text{Id}}} \right) = \det \left( {e^{A \cdot t} \cdot X(0) \cdot e^{ - A \cdot t} - \lambda \cdot e^{A \cdot t} \cdot e^{ - A \cdot t} } \right)$ $\displaystyle = \det \left[ {e^{A \cdot t} \cdot \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right) \cdot e^{ - A \cdot t} } \right]$

$\displaystyle \det \left( {e^{A \cdot t} \cdot X\left( 0 \right) \cdot e^{ - A \cdot t} - \lambda \cdot {\text{Id}}} \right) = \det \left( {e^{A \cdot t} } \right) \cdot \det \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right) \cdot \det \left( {e^{ - A \cdot t} } \right)$ $\displaystyle = \det \left( {e^{A \cdot t} } \right) \cdot \det \left( {e^{ - A \cdot t} } \right) \cdot \det \left({X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right) = \det \left( {e^{A \cdot t} \cdot e^{ - A \cdot t} } \right) \cdot \det \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right)$

$\displaystyle = \det \left( {{\text{Id}}} \right) \cdot \det \left({X\left( 0 \right) - \lambda \cdot {\text{Id}}}\right) = \det \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right)$ !!!