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Thread: eigen value and vector using properties help

  1. #1
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    eigen value and vector using properties help

    i know how to find eigen values and vectors but i'm having difficulties in questions 2 3 and 4 (part 2 and 3) eigen value and vector using properties help-agen.pngeigen value and vector using properties help-agen2.png
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  2. #2
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    Re: eigen value and vector using properties help

    ok you say that you were able to find the eigenvalues and associated eigenvectors of $A$ so you know that they are

    $2, ~(0,1,1), \\ \\1, ~(-1,1,1), \\ \\0, ~(-1,1,0)$

    $Q$ is simply the normalized eigenvectors set as columns, i.e.

    $\large Q = \begin{pmatrix}
    0 & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} \\
    \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} \\
    \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & 0 \\
    \end{pmatrix}$

    $\large Q^{-1} = \begin{pmatrix}
    \sqrt{2} & \sqrt{2} & 0 \\
    -\sqrt{3} & -\sqrt{3} & \sqrt{3} \\
    0 & \sqrt{2} & -\sqrt{2} \\
    \end{pmatrix}$

    and

    $D=Q^{-1}AQ = \begin{pmatrix}2 &0 &0 \\0 &1 &0 \\0 &0 &0 \end{pmatrix}$

    i.e. a diagonal matrix of the eigenvalues. Note the eigenvalue appears in the same column that it's associated vector appears in $Q$.

    Now,

    $A= QDQ^{-1}$ so

    $A^5 = \left(QDQ^{-1}\right)^5 = QD^5Q^{-1} = Q\begin{pmatrix}2^5 &0 &0\\0 &1 &0 \\0 &0 &0 \end{pmatrix}Q^{-1} =Q\begin{pmatrix}32 &0 &0\\0 &1 &0 \\0 &0 &0 \end{pmatrix}Q^{-1} $

    and I leave you to do the matrix algebra
    Last edited by romsek; Mar 1st 2018 at 11:49 PM.
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  3. #3
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    Re: eigen value and vector using properties help

    For part (c)

    $A^3 = QD^3Q^{-1}$ as shown in part (b)

    but this corresponds to a matrix that the same eigenvectors as $A$ but with the associated eigenvalues cubed.

    To determine the eigenvalues/vectors of $A+6I$ we first note that $I=QIQ^{-1}$ so

    $(A+6I) = QDQ^{-1} + Q(6I)Q^{-1} = Q(D+6I)Q^{-1}$

    by examination this is a matrix with the same eigenvectors as $A$ but with the eigenvalues increased by 6.

    Apply similar logic to above to determine the eigensystem of $(A-5I)^{-1}$

    I leave that to you.
    Last edited by romsek; Mar 2nd 2018 at 12:06 AM.
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  4. #4
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    Re: eigen value and vector using properties help

    for (4)

    i) you say you know how to compute the eigensystem so do it.

    ii) we essentially did this in problem 2. In that case $k=6$

    iii) this just uses both the properties learned in problem 2. It should be clear that $C$ has the same eigenvectors as $A$ but with associated eigenvalues $\lambda \to (\lambda+2)^3$
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  5. #5
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    Re: eigen value and vector using properties help

    for (3)

    It's trivial to show that $|A| =-1 \neq 0 \Rightarrow \exists~A^{-1}$

    I can't see any particularly clever way to show that

    $A^3 - A = A^2 - I$

    worst comes to worst you can always just do the matrix algebra and show both sides simplify to the same quantity.

    once that is done though it's simple to multiply both sides by $A^{-1}$ on the right

    $A^3 A^{-1} - A A^{-1} = A^2 A^{-1} - I A^{-1}$

    $A^2 - I = A - A^{-1}$

    $A^{-1} = A + I - A^2$
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    Thumbs up Re: eigen value and vector using properties help

    thanks man with your help and YouTube videos (3b1b) i was able to understand the tutorial on eigen you seem to be very well educated in maths so i guess you know how it is in engineering i.e covering the whole eigen part in 2 lectures and not understanding anything apart from the basics coupled with a test on Tuesday and again thank you i appreciate the help
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