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.


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

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

  • 1
    $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.