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$$ \begin{bmatrix} 4 & \frac {69}{28} \\ \frac {69}{28} & \frac {15}{7} \\ \end{bmatrix} $$ This is my initial matrix. So, I calculated my eigenvalues with no problem but got stuck with eigenvectors. My eigenvalues are 5.7049 and 0.4380 (they are right - I checked with Matlab).

Then I performed row manipulation and got

$$ \begin{bmatrix} -1.70631 & 2.464 \\ 0 & 0 \\ \end{bmatrix} $$

And now I don't know what to do to get the correct eigenvectors. There's one free variable so I set $x_1$ as 't' and tried solving for $x_2$ but that gives me the correct answer. According to Matlab, the correct eigenvector should be $$ \begin{bmatrix} -0.8224 \\ 0.5689\\ \end{bmatrix} $$

What do I have to do to get that??

Thanks in advance:)

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    $\begingroup$ An eigenvector isn't unique. Did you get something that is a scalar multiple of what Matlab got? $\endgroup$ – G Tony Jacobs Oct 14 '17 at 17:13
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    $\begingroup$ Since the eigenvector is only unique upto multiplication with a scalar, you can choose one coordinate free, but not $0$ (for example $x_2=1$). This gives an eigenvector colinear with the one given by Matlab. $\endgroup$ – Peter Oct 14 '17 at 17:15
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    $\begingroup$ Matlab seems to have normalised its vector so that it is a unit vector $\endgroup$ – Mark Bennet Oct 14 '17 at 17:15
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    $\begingroup$ To get a unit vector (See Mark's comment), just didvide the vector you have calculated by its length. Upto possibly a sign, this gives you the vector given by Matlab. $\endgroup$ – Peter Oct 14 '17 at 17:17
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    $\begingroup$ I did not know that Matlab automatically normalizes the eigenvector. That solves it! Thanks a lot!! $\endgroup$ – Shelly_Cooper Oct 14 '17 at 17:20

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