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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.

Please suggest.

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    In English, we use "eigenvalue" and "eigenvector" (single word).2012-05-14
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    $\text{tr}(A) = \sum_i \lambda_i,$ so $ 20 = \lambda + l + 0.$2012-05-14
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    how it is clear that determinant is 0?by hand calculation?2013-05-11
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    @Tsotsi I suppose it is clear by hand calculation.2013-05-11

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Hint: The trace of the matrix is $40+30+(-50)$. As you observed, $0$ is an eigenvalue.

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    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.2012-05-14
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    @srijan i do not understand why 0 is an eigen value, explain kor2013-05-11
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    @Tsotsi As the determinant of the matrix is zero which shows that matrix $A$ is singular hence it must have atleast one eigenvalue 02013-05-11
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    @Tsotsi: Each row has sum $0$.2013-05-11
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    @AndréNicolas Thank you!2013-05-11
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    You are welcome. It is clear when you see it, but if one doesn't notice, it looks as if one needs to do the full determinant calculation.2013-05-11