Show that if 0 < θ < π then the matrix has no real eigenvalues and consequently no eigenvectors.
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Originally Posted by chuckienz Show that if 0 < θ < π then the matrix has no real eigenvalues and consequently no eigenvectors. So we take the determinate of the above matrix to get This will only have real solutions if the discriminate is posative So the discriminate of this quadratic is This is negative for So the equation have no Real solutions
FYI, its eigenvalues are
Originally Posted by Moo FYI, its eigenvalues are
Originally Posted by Gamma I meant theta
ahhh! Okay, I should have realized that is what you meant. Sorry
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