# Complex eigenvalues of real matrices

Given a matrix $$A = \begin{pmatrix} 40 & -29 & -11\\ \ -18 & 30\ & -12 \\\ \ 26 &24 & -50 \end{pmatrix}$$ has a certain complex number $l\neq0$ as an eigenvalue. Which of the following must also be an eigenvalue of $A$: $$l+20, l-20, 20-l, -20-l?$$

It seems that complex eigenvalues occur in conjugate pairs. It is clear that the determinant of the matrix is zero, then $0$ seems to be one of the eigenvalues.

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In English, we use "eigenvalue" and "eigenvector" (single word). –  Arturo Magidin May 14 '12 at 18:57
$\text{tr}(A) = \sum_i \lambda_i,$ so $20 = \lambda + l + 0.$ –  user2468 May 14 '12 at 19:00
how it is clear that determinant is 0?by hand calculation? –  La Belle Noiseuse May 11 '13 at 12:53
@Tsotsi I suppose it is clear by hand calculation. –  srijan May 11 '13 at 15:06

Hint: The trace of the matrix is $40+30+(-50)$. As you observed, $0$ is an eigenvalue.
Given matrix is a $3\times 3$ matrix(odd order) with one complex eigen value.Complex eigen value will ocur in a conjugate pair. You must have noticed that 0 is its real eigen value while $l$ is given complex eigen value. Find remaining eigen value by using above hints and that must be complex one. –  srijan May 14 '12 at 20:39
@Tsotsi As the determinant of the matrix is zero which shows that matrix $A$ is singular hence it must have atleast one eigenvalue 0 –  srijan May 11 '13 at 15:07
@Tsotsi: Each row has sum $0$. –  André Nicolas May 11 '13 at 15:40