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The matrix:

\begin{pmatrix} 2 & -1 & 2 \\ 5 & -3 & 3 \\ -1 & 0 & -2 \end{pmatrix}

so as per workflow I got:

$$\mathrm{det}(A-\lambda{E}) = \begin{pmatrix} 2-\lambda & -1 & -2 \\ 5 & -3-\lambda & 3 \\ -1 & 0 & -2-\lambda \end{pmatrix}$$

After all calculus I got some cubic equation:

$$\lambda^3 -7\lambda^2 +23\lambda -31 =0$$

which has really strange roots (as far as I remember they are $\lt 1)$

Is it normal, or have I done mistake in the proccess?

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  • $\begingroup$ What roots did you get? $\endgroup$ – AlgorithmsX May 18 '17 at 21:07
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    $\begingroup$ The top right entry of your matrix $A-\lambda E$ magically became $-2$. $\endgroup$ – Ken Duna May 18 '17 at 21:15
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    $\begingroup$ Your characteristic polynomial should be $-(\lambda+1)^3$ $\endgroup$ – Jean Marie May 18 '17 at 21:16
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    $\begingroup$ @JeanMarie It was a $2$ in the original matrix. There was a clerical error in which it became a $-2$. I just like using colorful language when talking about math. $\endgroup$ – Ken Duna May 18 '17 at 21:29
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    $\begingroup$ The trace is -3 which is a giveaway that you do not have the correct characteristic equation. $\endgroup$ – Doug M May 18 '17 at 21:49
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If we have: $$ A = \begin{bmatrix} 2 & -1 & 2 \\ 5 & -3 & 3 \\ -1 & 0 & -2 \end{bmatrix} $$ then we find the eigenvalues by solving the characteristic equation: $$ p(\lambda)=\det(A-\lambda I) = 0 $$ We see that \begin{align*} p(\lambda)=\det(A-\lambda I) &= \begin{vmatrix} 2-\lambda & -1 & 2 \\ 5 & -3-\lambda & 3 \\ -1 & 0 & -2-\lambda \end{vmatrix} \\ &= -1 \cdot \begin{vmatrix} -1 & 2 \\ -3-\lambda & 3 \end{vmatrix} + (-2-\lambda) \begin{vmatrix} 2-\lambda & -1 \\ 5 & -3-\lambda \end{vmatrix} \\ &= (-3-2\lambda)+(-2-\lambda)\left[(2-\lambda)(-3-\lambda)+5 \right] \\ &= -3-2\lambda - \left[(4-\lambda^2)(-3-\lambda)+5(2+\lambda) \right] \\ &= -3-2\lambda - \left[(\lambda^3-4\lambda + 3\lambda^2-12)+(10+5\lambda) \right] \\ &= -3-2\lambda - \left(\lambda^3+\lambda + 3\lambda^2-2 \right) \\ &= -\left(\lambda^3+3\lambda + 3\lambda^2+1 \right) \\ &= -(\lambda+1)^3 \end{align*} Therefore, $A$ has only one eigenvalue, $\lambda = -1$.

The set of eigenvectors of $A$ corresponding to the eigenvalue $-1$ is the nullspace of $A+I$, which we row reduce as follows: \begin{align*} \begin{bmatrix} 3 & -1 & 2 \\ 5 & -2 & 3 \\ -1 & 0 & -1 \end{bmatrix} \implies \begin{bmatrix} 0 & -1 & -1 \\ 0 & -2 & -2 \\ 1 & 0 & 1 \end{bmatrix} \implies \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \end{align*} So we see that the nullspace of $A+I$, the set of eigenvectors of $A$ for the eigenvalue $\lambda = -1$, is spanned by the following vector: $$ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} $$

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