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Math Help - Linear Algebra/Differential equations

  1. #1
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    Linear Algebra/Differential equations

    How can we show that the matrix differential equation dX/dt=AX+XB has the solution 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.
    Thanks in advance
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  2. #2
    Super Member PaulRS's Avatar
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    You can check that that is indeed a solution by differentiating, now suppose we have a solution X(t) we will prove it has to be of that form (uniqueness)

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

    <br />
\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)<br />

    Hence: <br />
\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}  + <br />
<br />
\left( {e^{ - A \cdot t}  \cdot X\left( t \right)} \right) \cdot \left( { - B \cdot e^{ - B \cdot t} } \right)<br />

    <br />
C<br />
and <br />
e^C <br />
commute, for every matrix C.

    Thus: <br />
\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} <br />

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

    Let t=0 we get: <br />
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<br />

    <br />
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} <br />
And the uniqueness is proven.


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

    The solutions is of the form: <br />
e^{A \cdot t}  \cdot X(0) \cdot e^{ - A \cdot t} <br />

    Let's see the characteristic polynomial:

    <br />
\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)<br />
=<br />
\det \left[ {e^{A \cdot t}  \cdot \left( {X\left( 0 \right) - \lambda  \cdot {\text{Id}}} \right) \cdot e^{ - A \cdot t} } \right]<br />

    <br />
\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)<br />
=<br />
\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)<br />

    <br />
 = \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)<br />
!!!
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