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Loser66

  • one year ago

If \(A=\left[\begin{matrix}1&0\\0&2\end{matrix}\right]\), then \(e^{tA}=\left[\begin{matrix}e^t&0\\0&e^{2t}\end{matrix}\right]\) Now, if \(B=\left[\begin{matrix}0&1\\0&0\end{matrix}\right]\), then \(e^{tB}=\left[\begin{matrix}1&t\\0&1\end{matrix}\right]\) I don't get why t there. Please, help

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  1. Loser66
    • one year ago
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    @oldrin.bataku

  2. jtvatsim
    • one year ago
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    Are you specifically referring to the \[e^{tB}\] matrix?

  3. Loser66
    • one year ago
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    Yes, sir

  4. jtvatsim
    • one year ago
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    OK, lemme check. I believe exponentiation of matrices is defined by its infinite series, but let me see if there is a more elegant approach.

  5. jtvatsim
    • one year ago
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    OK, I got it. I'll let @oldrin.bataku reply first and see if they have any additional insights.

  6. jtvatsim
    • one year ago
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    We first need that \[e^x = 1 + x + x^2/2! + x^3/3! + \cdots \]Are you familiar and/or comfortable with that?

  7. Loser66
    • one year ago
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    Yes, I do

  8. jtvatsim
    • one year ago
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    OK, great. This definition can be extended to matrices so that \[e^{tB} = I + tB + (tB)^2/2! + (tB)^3/3! + \cdots\] In general, \[e^A = \sum_{n=0}^\infty \frac{A^n}{n!}\]

  9. jtvatsim
    • one year ago
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    Expanded, we have \[e^{tB} = I + tB + t^2B^2/2! + t^3B^3/3! + \cdots\]

  10. jtvatsim
    • one year ago
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    OK, so far?

  11. Loser66
    • one year ago
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    Yes

  12. jtvatsim
    • one year ago
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    Alright, awesome! Now, the annoying thing is the powers of matrices. We really hope that we have a "nice" matrix, otherwise this can get messy. Fortunately, B isn't that bad.

  13. jtvatsim
    • one year ago
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    \[B = \left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right], \ \ B^2 = \left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]\]

  14. jtvatsim
    • one year ago
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    The zero matrix times any matrix is still the zero matrix, so all powers of B above 1 will be zeroed out.

  15. Loser66
    • one year ago
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    I know it

  16. jtvatsim
    • one year ago
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    Hence, our problem reduces to this: \[e^{tB} = I + tB + 0 + 0 + \cdots = I + tB\] and the result follows immediately.

  17. Loser66
    • one year ago
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    yyyyyyyyyyyyyes!

  18. jtvatsim
    • one year ago
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    \[I + tB = \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] + \left[\begin{matrix}0 & t \\ 0 & 0\end{matrix}\right] = \left[\begin{matrix}1 & t \\ 0 & 1\end{matrix}\right]\]

  19. jtvatsim
    • one year ago
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    Yay!! Indeed, any questions?

  20. Loser66
    • one year ago
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