Calculating eigenvalues of a matrix

Question

I have this quadratic matrix: \begin{align*} A = \begin{bmatrix} 0 & -1 & -2\\ -1 & 0 & -2\\ -2 & -2 & -3 \end{bmatrix} \implies A - \lambda I = \begin{bmatrix} 0 - \lambda & -1 & -2\\ -1 & 0 - \lambda & -2\\ -2 & -2 & -3 - \lambda \end{bmatrix} \end{align*} After forming the characteristic polynomial and using the Rule of Sarrus, I get this equation: $$-\lambda^3 - 3\lambda^2 + 9\lambda - 5$$ However the correct form of this term would be: $$-(5 + \lambda)(1 - \lambda)^2$$ So the eigenvalues are $1$ and $-5$. The thing I do not understand, is how to form this equation to get the values. Which approach and rules can solve this?

Thank you very much.

Answer 1

$$p(\lambda)=-\lambda^3 - 3\lambda^2 + 9\lambda - 5$$

Proceed from your work,

$$\begin{align} p(\lambda)&=-\lambda^3-5\lambda^2+2\lambda^2+9\lambda-5\\ \\ &=-\lambda^2(\lambda+5)+(2\lambda-1)(\lambda+5)\\ \\ &=(\lambda+5)(-\lambda^2+2\lambda-1)\\ \\ &=-(\lambda+5)(\lambda^2-2\lambda+1)\\ \\ &=-(\lambda+5)(\lambda-1)^2\end{align} $$

Answer 2

You can use the rational roots theorem.

In your case, the equation reads

$$-x^3 - 3x^2 + 9x - 5$$

so you may expect that every rational root must be among the numbers $\{\pm 1, \pm 5\}$ as indeed is.

Answer 3

For matrices of $\mathbb{R}^{3 \times 3}$ and above, the method is to try to apply row operations and factorize. Avoid computing the determinant directly, or expanding the terms first.

1. Apply row operations to $ |A - \lambda I|$. $$ \begin{matrix} \\ -R_1+R_2 \rightarrow\\ \\ \end{matrix} \begin{vmatrix} -\lambda & -1 & -2 \\ -1 & -\lambda & -2 \\ -2 & -2 & -3 -\lambda \\ \end{vmatrix} \implies \begin{vmatrix} -\lambda & -1 & -2 \\ -(1-\lambda) & 1-\lambda & 0 \\ -2 & -2 & -3-\lambda \\ \end{vmatrix} $$
2. Find the characteristic polynomial from the reduced expression. We will exploit column $3$. $$ \begin{align} p(\lambda) = & -2 \bigl( 2(1-\lambda) + 2(1-\lambda) \bigl) + (-3-\lambda)\bigl( -\lambda(1-\lambda) -(1-\lambda) \bigl)\\ = & -8(1-\lambda) + (-3-\lambda)(1-\lambda)(-\lambda-1)\\ = & -8(1-\lambda) + (3+\lambda)(1-\lambda)(1+\lambda)\\ = & (1-\lambda)\bigl(-8 + (3+\lambda)(1+\lambda) \bigl) \\ = & (1-\lambda)(-8+\lambda^2+4\lambda+3)\\ = & (1-\lambda)(\lambda^2+4\lambda-5)\\ = & (1-\lambda)(\lambda+5)(\lambda-1)\\ = & -(1-\lambda)^2(\lambda+5) \end{align} $$

You may refer to a similar question I have answered.

Link: Finding Eigen Vectors and Respective Eigen Values for 3x3 Matrice