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$A = \left(\begin{array}{ccc}0 & -1 & 0 \\0 & 0 & 1 \\-1 & -3 & 3\end{array}\right)$

I want to find eigenvalues of this matrix.

The answer given to me is: 1,1,1.

I cannot arrive at that answer. I get complex numbers for my eigenvalues.

Am I doing something wrong? Is there a way to get 1,1,1 as your eigenvalues?

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Calculate the characteristic polynomial, $\det|sI-A|$, $$\det\left(\begin{array}{ccc}s & 1 & 0 \\0 & s & -1 \\1 & 3 & s-3\end{array}\right)=s(s(s-3)+3)-1(1)=s^3-3s^2+3s-1=(s-1)^3$$ The three eigenvalues are the roots of this polynomial, and hence are all equal to $1$.

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The eigenvalues can be obtained by equating the determinant of $A-\lambda I$ to zero. $$ \det(A-\lambda I)= \left| \begin{array}{ccc} -\lambda & -1 & 0 \\ 0 & -\lambda & 1 \\ -1 & -3 & 3-\lambda \end{array} \right|= \lambda^2(3-\lambda)+1-3\lambda=(1-\lambda)^3 $$

Hence they are given by the solutions of $(1-\lambda)^3=0$

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    $\begingroup$ For future reference: $$ \begin{vmatrix} 1&2\\3&4 \end{vmatrix} $$ can be coded as \begin{vmatrix} 1&2\\3&4 \end{vmatrix}. $\endgroup$ Apr 10 '15 at 18:20
  • $\begingroup$ @Omnomnomnom Great thanks :) $\endgroup$
    – Math137
    Apr 10 '15 at 18:21

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