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  1. #1
    Super Member craig's Avatar
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    Another matrix transformation

    Hi.

    A transformation T : R^2 \rightarrow R^2 is represented by the matrix A = \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right).

    There is a line through the origin for which every point is mapped onto itself under T.

    Find the cartesian equation of this line.

    I have already worked out the two eigenvalues for this matrix, 1 and -6.

    Forming the equation:

    \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = \lambda \left(\begin{array}{cc}x\\y\end{array}\right)

    Where \lambda are my respective eigenvalues.

    The question I have is which on do I use, using 1 I get 5y = 3x and using -6 I get y = 2x.

    My answers tell me that;

    \lambda = 1 is associated with point invariant line
    So 1 is the value I should use, but what exactly does the above expression mean?

    Thanks in advance
    Last edited by craig; June 1st 2009 at 09:00 AM.
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  2. #2
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    Quote Originally Posted by craig View Post
    Hi.

    A transformation T : R^2 \rightarrow R^2 is represented by the matrix A = \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right).

    There is a line through the origin for which every point is mapped onto itself under T.

    Find the cartesian equation of this line.

    I have already worked out the two eigenvalues for this matrix, 1 and -6.

    Forming the equation:

    \left(\begin{array}{cc}4&4\\4&-2\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = \lambda \left(\begin{array}{cc}x\\y\end{array}\right)

    Where \lambda are my respective eigenvalues.

    The question I have is which on do I use, using 1 I get 5y = 3x and using -6 I get y = 2x.

    My answers tell me that;



    So 1 is the value I should use, but what exactly does the above expression mean?

    Thanks in advance
    ????
    How did A = \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right) mysteriously change into \left(\begin{array}{cc}4&4\\4&-2\end{array}\right)?
    Is one of those a typo?

    Okay, the first has eigenvalues 1 and -6 as you say while the second has eigenvalues 6 and -4 so I assume the second was probably accidently copied from a different problem.

    Notice an important difference between this and a similar problem that was posted (perhaps by you?) earlier on this board. Before the problem asked for "invariant lines" which meant, of course, lines that are mapped into themselves. Those turn out to be the lines in the "characteristic directions" or in the directions given by the eigenvectors. But saying that a line is "invariant" doesn't mean that points on that line are mapped into themselves. It means points on that line are mapped into other points on that same line.

    Here, you are asked for a line on which "every point is mapped into itself". That is, you are asked for "invariant points" which happen to lie on the line corresponding to eigenvalue 1.

    For example, the point (1, 2) is on the line y= 2x. If you apply the matrix to that, \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin{array}{c}1 \\ 2\end{array}\right)= \left(begin{array}{c}-6 \\ -12\end{array}\right), a different point because each component is multiplied by -6, but on the same line. If we take the point (3, 5), on the line 3y= 5x, and do the same, \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}3 \\ 5\end{array}\right)= \left(begin{array}{c}3 \\ 5\end{array}\right), exactly the same point: sure, it has been "multiplied" by its eigenvalue, 1!

    Any eigenvector gives an "invariant line" but only eigenvectors corresponding to eigenvalue 1 give "invariant points".
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  3. #3
    Super Member craig's Avatar
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    Hi thanks for the reply.

    Quote Originally Posted by HallsofIvy View Post
    Okay, the first has eigenvalues 1 and -6 as you say while the second has eigenvalues 6 and -4 so I assume the second was probably accidently copied from a different problem.
    Yes that was my mistake sorry about that.

    Quote Originally Posted by HallsofIvy View Post
    For example, the point (1, 2) is on the line y= 2x. If you apply the matrix to that, \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right) \left(\begin{array}{c}1\\2\end{array}\right)= \left(\begin{array}{c}-6\\-12\end{array}\right), a different point because each component is multiplied by -6, but on the same line. If we take the point (3, 5), on the line 3y= 5x, and do the same, \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}3 \\ 5\end{array}\right)= \left(\begin{array}{c}3 \\ 5\end{array}\right), exactly the same point: sure, it has been "multiplied" by its eigenvalue, 1!

    Any eigenvector gives an "invariant line" but only eigenvectors corresponding to eigenvalue 1 give "invariant points".
    I get what your saying, that's the best anyone's explained it! So if you want invariant points you must have an eigenvalue of 1?

    Also, the example you gave above is wrong :P, it should be:

    <br /> <br />
\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}5 \\3\end{array}\right)= \left(\begin{array}{c}5\\3\end{array}\right)<br />

    As

    <br /> <br />
\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}3 \\ 5\end{array}\right)= \left(\begin{array}{c}-13 \\ -27\end{array}\right)<br />

    Thanks again for the help, really appreciate it
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  4. #4
    Super Member craig's Avatar
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    Oh and you have a couple of Latex errors, instead of:

    Quote Originally Posted by HallsofIvy View Post
    \left(begin{array}{c}3 \\
    It should be:

    \left(\begin{array}{c}3 \\
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