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Math Help - find nondiagonal matrix whose eigenvalues and eigenvectors are given

  1. #1
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    [RESOLVED]find nondiagonal matrix whose eigenvalues and eigenvectors are given

    Find a 2x2 nondiagonal matrix whose eigenvalues are 2 and -3 and associated eigenvectors are
    \left[ {\begin{array}{c}<br />
-1 \\<br />
2 \\<br />
\end{array} } \right]<br />
and <br />
\left[ {\begin{array}{c}<br />
1 \\<br />
1 \\<br />
\end{array} } \right] respectivley.

    The answer key has:
    Let D=\left[ {\begin{array}{cc}<br />
2 & 0 \\<br />
0 & -3 \\<br />
\end{array} } \right]<br />
and P=\left[ {\begin{array}{cc}<br />
-1 & 1\\<br />
2 & 1 \\<br />
\end{array} } \right]<br />
    Then P^{-1}AP=D so A =PDP^{-1}=\frac{1}{3}<br />
\left[ {\begin{array}{cc}<br />
-4 & -5 \\<br />
-10 & 1 \\<br />
\end{array} } \right]<br />

    I don't see the intuition behind the first part, how do they know to come up with D and P by slamming together the eigenvectors and eigen values?
    Last edited by superdude; April 25th 2010 at 11:47 AM. Reason: resolved
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  2. #2
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    Quote Originally Posted by superdude View Post
    Find a 2x2 nondiagonal matrix whose eigenvalues are 2 and -3 and associated eigenvectors are
    \left[ {\begin{array}{c}<br />
-1 \\<br />
2 \\<br />
\end{array} } \right]<br />
and <br />
\left[ {\begin{array}{c}<br />
1 \\<br />
1 \\<br />
\end{array} } \right] respectivley.

    The answer key has:
    Let D=\left[ {\begin{array}{cc}<br />
2 & 0 \\<br />
0 & -3 \\<br />
\end{array} } \right]<br />
and P=\left[ {\begin{array}{cc}<br />
-1 & 1\\<br />
2 & 1 \\<br />
\end{array} } \right]<br />
    Then P^{-1}AP=D so A =PDP^{-1}=\frac{1}{3}<br />
\left[ {\begin{array}{cc}<br />
-4 & -5 \\<br />
-10 & 1 \\<br />
\end{array} } \right]<br />

    I don't see the behind the first part, how do they know to come up with D and P by slamming together the eigenvectors and eigen values?

    We know that if A is a square matrix which eigenvectors form a basis of the given vector space, then there exists an invertible matrix P , which its

    columns are eigenvectors of A , s.t. P^{-1}AP=D is a diagonal matrix whose diagonal elements are precisely the eigenvalues of A (all this is classic theory, not

    something new). Well, they just did all the process backwards...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    We know that if A is a square matrix which eigenvectors form a basis of the given vector space, then there exists an invertible matrix P , which its

    columns are eigenvectors of A

    Tonio
    How do you know which eigenvectors are which column? Does it matter? So in my example if I found the eigenvectors how would I know it wouldn't be <br />
D=\left[ {\begin{array}{cc}<br />
0 & 2 \\<br />
-3 & 0 \\<br />
\end{array} } \right]<br />
    Is it because we start with the smallest values of \lambda and work our way up and the eigenvectors found get put into the matrix D left from right? If this is true, then what happens if 2 eigenvectors are associated with one eigenvalue (is this even possible)?
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    Quote Originally Posted by superdude View Post
    How do you know which eigenvectors are which column? Does it matter? So in my example if I found the eigenvectors how would I know it wouldn't be <br />
D=\left[ {\begin{array}{cc}<br />
0 & 2 \\<br />
-3 & 0 \\<br />
\end{array} } \right]<br />


    This matrix isn't diagonal so it cannot be in our case , and if the first eigenvalue appears in D then the first column of the invertible matrix P will be an eigenvector corresponding to this eigenvalue...

    Tonio


    Is it because we start with the smallest values of \lambda and work our way up and the eigenvectors found get put into the matrix D left from right? If this is true, then what happens if 2 eigenvectors are associated with one eigenvalue (is this even possible)?
    .
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  5. #5
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    Quote Originally Posted by superdude View Post
    How do you know which eigenvectors are which column? Does it matter? So in my example if I found the eigenvectors how would I know it wouldn't be <br />
D=\left[ {\begin{array}{cc}<br />
0 & 2 \\<br />
-3 & 0 \\<br />
\end{array} } \right]<br />
    Is it because we start with the smallest values of \lambda and work our way up and the eigenvectors found get put into the matrix D left from right? If this is true, then what happens if 2 eigenvectors are associated with one eigenvalue (is this even possible)?
    You are confusing "eigenvalues" with "eigenvectors". D is the diagonal matrix having the eigenvalues on the diagonal. Swapping eigenvectors would give you P= \begin{bmatrix}1 & -1 \\ 1 & 2\end{bmatrix}. If you kept the same the same D matrix, you would get a matrix having the same eigenvalues and eigenvectors but with the eigenvector now corresponding to the othere eigenvalue.

    If you swap both, say P= \begin{bmatrix}1 & -1 \\ 1 & 2\end{bmatrix} and D= \begin{bmatrix}-3 & 0 \\ 0 & 2\end{bmatrix} then you would get another matrix having the same eigenvalues and the same eigenvectors corresponding to those eigenvalues as the first matrix.

    Choose D to be the diagonal matrix having the eigenvalues on the diagonal and P to be the matrix having the corresponding eigenvectors as columns- the first column is an eigenvector for the eigenvalue in the first column, the second column of P corresponds to the eigenvalue in the second column of D, etc.
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