Answer to Question #254068 in Linear Algebra for Sabelo Xulu

Question

Consider the matrix A=$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix}$

1.1. Show that the characteristic equation for the eigenvalues $\lambda$ of A is given by (λ²-1) (λ-3)=0.

1.2. Find an orthogonal matrix P which diagonalizes A.

1.3. Find Aⁿ (for n ∈ N) as a matrix

Answer 1

$|A-\lambda I|=\begin{vmatrix} 1-\lambda & 0&2 \\ 0& 1-\lambda&0\\ 2&0&1-\lambda \end{vmatrix}=0\\ (1-\lambda)^3-0-0-4(1-\lambda)-0-0=0\\ (1-\lambda)((1-\lambda)^2-4)=0\\ (1-\lambda )(1-\lambda-2)(1-\lambda+2)=0\\ (1-\lambda)(-1-\lambda)(3-\lambda)=0\\ (\lambda^2-1)(\lambda-3)=0\\ \lambda_1=1, \lambda_2=-1, \lambda_3=3$

$\lambda_1=1, \vec{a_1}=(x,y,z)\neq\vec{0}\\ (A-\lambda_1I)\vec{a_1}=\vec{0}\\ \begin{pmatrix} 0& 0&2 \\ 0 & 0&0\\ 2&0&0 \end{pmatrix}\cdot \begin{pmatrix} x \\ y\\z \end{pmatrix}=\begin{pmatrix} 0 \\ 0\\0 \end{pmatrix}\\ 2z=0,2x=0\\ x=z=0, y\in R\\ \vec{a_1}=(0,1,0)$

$\lambda_2=-1, \vec{a_2}=(x,y,z)\neq\vec{0}\\ (A-\lambda_2I)\vec{a_2}=\vec{0}\\ \begin{pmatrix} 2& 0&2 \\ 0 & 2&0\\ 2&0&2 \end{pmatrix}\cdot \begin{pmatrix} x \\ y\\z \end{pmatrix}=\begin{pmatrix} 0 \\ 0\\0 \end{pmatrix}\\ 2x+2z=0,2y=0\\ x=-z, y=0\\ \vec{a_2}=(1,0,-1)$

$\lambda_3=3, \vec{a_3}=(x,y,z)\neq\vec{0}\\ (A-\lambda_3I)\vec{a_3}=\vec{0}\\ \begin{pmatrix} -2& 0&2 \\ 0 & -2&0\\ 2&0&-2 \end{pmatrix}\cdot \begin{pmatrix} x \\ y\\z \end{pmatrix}=\begin{pmatrix} 0 \\ 0\\0 \end{pmatrix}\\ -2x+2z=0,-2y=0, 2x-2z=0\\ x=z=1, y=0\\ \vec{a_3}=(1,0,1)$

$\vec{a_1}=(0,1,0), |\vec{a_1}|=1,\\ \vec{a_2}=(1,0,-1), |\vec{a_2}|=\sqrt2,\\ \vec{a_3}=(1,0,1),|\vec{a_3}|=\sqrt2\\ P=\begin{pmatrix} 0 & 1&0 \\ \frac{1}{\sqrt2} & 0& -\frac{1}{\sqrt2}\\ \frac{1}{\sqrt2}&0 &\frac{1}{\sqrt2} \end{pmatrix}\\ B=P^{-1}AP\\ B=\begin{pmatrix} 1 & 0&0 \\ 0 & -1&0\\ 0&0&3 \end{pmatrix}$

$A^2=\begin{pmatrix} 1 & 0&2 \\ 0&1&0\\ 2&0&1 \end{pmatrix}\cdot\begin{pmatrix} 1 & 0&2 \\ 0&1&0\\ 2&0&1 \end{pmatrix}=\begin{pmatrix} 5 & 0&4 \\ 0&1&0\\ 4&0&5 \end{pmatrix}\\ A^3=A^2\cdot A=\begin{pmatrix} 5 & 0&4 \\ 0&1&0\\ 4&0&5 \end{pmatrix}\cdot \begin{pmatrix} 1 & 0&2 \\ 0&1&0\\ 2&0&1 \end{pmatrix}=\\ =\begin{pmatrix} 13 & 0&14 \\ 0&1&0\\ 14&0&13 \end{pmatrix}$

$A^4=A^3\cdot A=\begin{pmatrix} 13 & 0&14 \\ 0&1&0\\ 14&0&13 \end{pmatrix}\cdot \begin{pmatrix} 1 & 0&2 \\ 0&1&0\\ 2&0&1 \end{pmatrix}=\\ =\begin{pmatrix} 41 & 0&40 \\ 0&1&0\\ 40&0&41 \end{pmatrix}$

$A^n=\begin{pmatrix} a & 0&b \\ 0&1&0\\ b&0&a \end{pmatrix}$