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if given that the determinant of a 3 by 3 matrix with only one real entries is 82 and an eigenvalue is given to be 4+5i, what are the other eigenvalues? Since the determinant is a non zero number, I am having a hard time figuring out how to do this.

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closed as too broad by GNUSupporter 8964民主女神 地下教會, Don Thousand, max_zorn, user10354138, Leucippus Oct 31 '18 at 4:05

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  • $\begingroup$ Have you made any attempt to solve this problem? If so, please explain what you have tried in the body of your question. $\endgroup$ – Carl Schildkraut Oct 31 '18 at 1:08
  • $\begingroup$ I am confused since the determinant is not zero. I have not done these kinds of questions. $\endgroup$ – Madison Oct 31 '18 at 1:13
  • $\begingroup$ Do you know any properties of the determinant as it relates to eigenvalues? Or properties of the eigenvalues when the matrix has all real entries? $\endgroup$ – Carl Schildkraut Oct 31 '18 at 1:17
  • $\begingroup$ Yes, I am currently in linear algebra land just learned about that. $\endgroup$ – Madison Oct 31 '18 at 1:20
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    $\begingroup$ The characteristic polynomial has real coefficients. If one of its roots is complex, what can you say about its other roots? $\endgroup$ – amd Oct 31 '18 at 1:21
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a 3x3 real matrix has up to 3 eigenvalues, since one of them is known to be complex, its conjugate is also an eigenvalue the third eigenvalue x must be real the determinant equals the product of eigenvalues Therefore (4+5i)*(4-5i)*x = 82 Solve for x, you're done

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    $\begingroup$ A $3\times3$ real matrix has exactly three eigenvalues (counting with multiplicity). $\endgroup$ – Gerry Myerson Oct 31 '18 at 2:54
  • $\begingroup$ So I got lambda=78/25-4/5i. Is that correct? $\endgroup$ – Madison Oct 31 '18 at 14:39
  • $\begingroup$ No, you should expand (4+5i)*(4-5i) and you will get a real number... Recall $(a + b)(a - b) = a^2 - b^2$ $\endgroup$ – phaedo Oct 31 '18 at 16:28

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