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rsst123

  • one year ago

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  1. rsst123
    • one year ago
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  2. anonymous
    • one year ago
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    use your calculator

  3. rsst123
    • one year ago
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    without calculator

  4. zzr0ck3r
    • one year ago
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    do you know how to find eigen values and eigen vectors? Do you know what diagonalization is ?

  5. zzr0ck3r
    • one year ago
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    there are other ways, so I need to know what you are doing in class if not.

  6. rsst123
    • one year ago
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    yes isn't it A=SDS^-1?

  7. zzr0ck3r
    • one year ago
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    Correct. But you first need to find the Eigen values and vectors so that you can construct \(S\)

  8. rsst123
    • one year ago
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    ok so after finding eigen values and vectors what would i do to find the matrix?

  9. zzr0ck3r
    • one year ago
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    Ok if we call your 2x2 matrix given above A and the matrix of eigen vectors S. Then you will find D by D = \(SAS^{-1}\). Now once you have D you just take the square root of the entries (they will only be non 0 on the diagonal). You with me?

  10. rsst123
    • one year ago
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    oh ok i see, I would get a diagonal matrix and then I can just square root the diagonal is that correct?

  11. zzr0ck3r
    • one year ago
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    then \(\sqrt{A} = S^{-1}\sqrt{D}S\)

  12. zzr0ck3r
    • one year ago
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    yes

  13. rsst123
    • one year ago
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    ok perfect thank you!

  14. zzr0ck3r
    • one year ago
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    np

  15. amoodarya
    • one year ago
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    \[\left[\begin{matrix}4 & 3 \\ 2 & 3\end{matrix}\right]=\left[\begin{matrix}-1 & \frac{3}{2} \\ 1 & 1\end{matrix}\right]\times \left[\begin{matrix}1 & 0 \\ 0 & 6 \end{matrix}\right] \times \left[\begin{matrix}\frac{-2}{5} & \frac{3}{5} \\ \frac{2}{5} & \frac{2}{5} \end{matrix}\right]=\\=S D S^{-1}\\\sqrt{A}==S \left[\begin{matrix}\sqrt{1} & 0 \\ 0 & \sqrt{6}\end{matrix}\right] S^{-1}\]

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