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jcaruso

  • 2 years ago

A is a 2 × 2 matrix with eigenvectors v1=[1;-1] v2=[1;1] corresponding to eigenvalues λ1 = (1/2) and λ2 = 2,respectively, and X =[5;1]. Find A^10X.

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  1. jcaruso
    • 2 years ago
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    please help im lost

  2. jtvatsim
    • 2 years ago
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    hint: try to write X as a linear combination of v1 and v2

  3. jcaruso
    • 2 years ago
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    how do i determine that

  4. jtvatsim
    • 2 years ago
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    ok, first of all have you learned what a linear combination is?

  5. jcaruso
    • 2 years ago
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    not quite im very confused with this topic

  6. jtvatsim
    • 2 years ago
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    ok. A linear combination is just the addition and scalar multiplication of two vectors. Take for example \[2\left(\begin{matrix}1 \\ 0\end{matrix}\right) + -3\left(\begin{matrix}0 \\ 1\end{matrix}\right)\] this is a linear combination of (1 0) and (0 1) because we are multiplying by scalars and adding vectors... does that make sense?

  7. jcaruso
    • 2 years ago
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    ok i see that

  8. jcaruso
    • 2 years ago
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    so X=2v1+3v2

  9. jtvatsim
    • 2 years ago
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    excellent! that is right.

  10. jtvatsim
    • 2 years ago
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    OK, so now you are probably thinking... now what? :)

  11. jcaruso
    • 2 years ago
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    yea

  12. jtvatsim
    • 2 years ago
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    Well, let's write A^10 X using our new expression for X as a linear combination. We get \[A^{10}X = A^{10}(2v_1 + 3v_2) = A^{10}(2v_1) + A^{10}(3v_2)\] by distributing. Do you see that?

  13. jcaruso
    • 2 years ago
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    ok i get that

  14. jtvatsim
    • 2 years ago
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    Cool! Now, we get \[2A^{10}v_1 + 3A^{10}v_2\] since scalars can be moved around.

  15. jcaruso
    • 2 years ago
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    so then we can get 2(\[\lambda ^{10}\]v1+3(\[\lambda _{2^{10}}\]v2

  16. jtvatsim
    • 2 years ago
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    precisely!

  17. jcaruso
    • 2 years ago
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    i just dont know how to get the final answer

  18. jtvatsim
    • 2 years ago
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    OK, well I personally wouldn't actually calculate a number to the 10th power, just leave it with an exponent. I'll type it below.

  19. jtvatsim
    • 2 years ago
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    Maybe something like this \[\left(\begin{matrix}2^{-9} + 3 \times 2^{10} \\ -2^{-9} + 3 \times 2^{10}\end{matrix}\right)\]

  20. jtvatsim
    • 2 years ago
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    After I've just added to the two vector together after multiplying by the scalars.

  21. jtvatsim
    • 2 years ago
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    That 2^(-9) came from \[2(\frac{1}{2})^{10} = 2 \times 2^{-10} = 2^{-9}\]

  22. jtvatsim
    • 2 years ago
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    but however you want to write it :)

  23. jcaruso
    • 2 years ago
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    ok i tired that final answer and i got it incorrect

  24. jtvatsim
    • 2 years ago
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    could you take a screenshot maybe of what the input looks like?

  25. jtvatsim
    • 2 years ago
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    or else you could just input the decimal answer... maybe the computer would like that better...

  26. jcaruso
    • 2 years ago
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    i got it correct thanks so much for your help

  27. jtvatsim
    • 2 years ago
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    no problem, good luck on eigenvectors! :)

  28. helder_edwin
    • 2 years ago
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    since u have the eigenvectors and eigenvalues of the matrix A then A can be "factored" as \[\large A=S^{-1}\Lambda S \] where \[\large S=\begin{pmatrix} 1 & 1\\ -1 & 1 \end{pmatrix}\qquad\text{and}\qquad \Lambda=\begin{pmatrix} 1/2 & 0\\ 0 & 2 \end{pmatrix} \]

  29. helder_edwin
    • 2 years ago
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    therefore \[\large A^{10}x=(S^{-1}\Lambda S)^{10}x=(S^{-1}\Lambda^{10}S)x \]

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