Can someone help me explain how to solve this problem below:
Let the matrix $\mathbf{A}$ be given as $$\mathbf{A}=\begin{bmatrix}0&-1&-1\\1&2&1\\1&1&2\end{bmatrix}$$ 1. Compute the characteristic equation.
Would something like this be correct?
$$(\lambda-2)(\lambda-2)(\lambda-0)-\ldots$$
But I'm not sure.
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$\begingroup$The easy and quick way to compute the characteristic equation of 3x3 matrix is to use the formulae $$x^3-tr(A)x^2+(A_{11}+A_{22}+A_{33})x-det(A)=0$$ For given matrix $$tr(A)=4, A_{11}(cofa_{11})=3, A_{22}(cofa_{22})=1, A_{33}(cofa_{33})=1, det(A)=2$$ so the char equation will be $x^3-4x^2+5x-2=0$
$\endgroup$ 6 $\begingroup$Quick way: Note that sum of each column of $A$ is $2$, hence $\lambda_1=2$ is an eigenvalue of $A$.Let $\lambda_1$ and $\lambda_2$ be other eigenvalues,using trace and determinant $\lambda_1 +\lambda_2=2$ and $\lambda_1 \lambda_2=1$,hence $\lambda_1=\lambda_2=1$ Hence Characteristic polynomial $p(x)$ is $$p(x)=(x-1)^2(x-2)$$
$\endgroup$ 4 $\begingroup$The characteristic equation is used to find the eigenvalues of a square matrix A.
First: Know that an eigenvector of some square matrix A is a non-zero vector x such that Ax = λx.
Second: Through standard mathematical operations we can go from this: Ax = λx, to this: (A - λI)x = 0
The solutions to the equation det(A - λI) = 0 will yield your eigenvalues. The previously mentioned equation is the characteristic equation.
-- SKIP TO THIS POINT FOR THE SHORT ANSWER --
Given some square matrix A, the characteristic equation is det(A - λI) = 0.
Example:
$$ A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$$$ λI = \begin{bmatrix} λ & 0 \\ 0 & λ \\ \end{bmatrix}$$
$$ A - λI = \begin{bmatrix} a-λ & b \\ c & d-λ \\ \end{bmatrix}$$
$$det\begin{bmatrix} a-λ & b \\ c & d-λ \\ \end{bmatrix} = 0$$ Characteristic Equation: $$(a-λ)(d-λ) - (c)(b) = 0$$
Source Cited: Lay, David C., et al. Linear Algebra and Its Applications, Fifth Edition. Pearson, 2016.
$\endgroup$ $\begingroup$That is incorrect. The characteristic equation comes from computing the $\lambda$ values necessary to give you a matrix determinant equal to zero.
$$\begin{vmatrix} -\lambda & -1 & -1 \\ 1 & 2 - \lambda & 1 \\ 1 & 1 & 2- \lambda \end{vmatrix} = 0 $$
Notice that for $\lambda = 1$ all three columns become identical and we can come up with two eigenvectors. Coming up with the last eigenvalue should be easy if we use the fact that the trace of the matrix is the sum of the eigenvalues.
The eigenvalues for this matrix are thus $\lambda_1 = 1, \lambda_2 = 1, \lambda_3 = 2$ as they make the matrix singular.
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