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# The following list of matrices and their respective characteristic polynomials is referred to in Exercise.$A=\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right]$, $p(t) = (t − 3)(t − 1),$$B=\left[\begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array}\right]$, $p(t)=(t-2)^{2}$,$C=\left[\begin{array}{rrr} -6 & -1 & 2 \\ 3 & 2 & 0 \\ -14 & -2 & 5 \end{array}\right]$, $p(t)=-(t-1)^{2}(t+1)$,$D=\left[\begin{array}{rrr} -7 & 4 & -3 \\ 8 & -3 & 3 \\ 32 & -16 & 13 \end{array}\right]$, $p(t)=-(t-1)^{3}$,$E=\left[\begin{array}{llll} 6 & 4 & 4 & 1 \\ 4 & 6 & 1 & 4 \\ 4 & 1 & 6 & 4 \\ 1 & 4 & 4 & 6 \end{array}\right]$, $p(t) =(t + 1)(t + 5)^2(t − 15),$,$F=\left[\begin{array}{rrr} 1 & -1 & -1 & -1 \\ -1 & 1 & -1 & -1 \\ -1 & -1 & 1 & -1 \\ -1 & -1 & -1 & 1 \end{array}\right]$, $p(t)=(t+2)(t-2)^{3}$. Find a basis for the eigenspace $E_{\lambda}$ for the given matrix and the value of $\lambda$. Determine the algebraic and geometric multiplicities of $\lambda$. $E, \lambda=-1$

Solution

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We have

\begin{align*} E&=\begin{bmatrix} 6&4&4&1\\ 4&6&1&4\\ 4&1&6&4\\ 1&4&4&6 \end{bmatrix}, p(t)=(t+1)(t+5)^{2}(t-15), \lambda=-1 \end{align*}

Find the eigenvectors. According to $\textbf{\color{#c34632}Formula $(1)$ Section 4.5 }$the eigenvectors corresponding to an eigenvalue $\lambda$ of $A$ are the nonzero solutions of

$\begin{equation} (A-\lambda I)x=\theta \end{equation}$

or

$Ax=0.$

For $\lambda=-1$ a non-trivial system is

$\begin{equation} \begin{bmatrix} 7&4&4&1\\ 4&7&1&4\\ 4&1&7&4\\ 1&4&4&7 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ x_{4} \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix} \end{equation}$

Using Gaussian elimination we solve the equation $(2)$. Reduce matrix to row echelon form

\begin{align*} \begin{bmatrix} 7&4&4&1\\ 4&7&1&4\\ 4&1&7&4\\ 1&4&4&7 \end{bmatrix}{\small\begin{matrix} R_{2}-\dfrac{4}{7}R_{1}\rightarrow R_{2}\\[7pt] R_{3}-\dfrac{4}{7}R_{1}\rightarrow R_{3} \end{matrix}}&\sim \begin{bmatrix} 7&4&4&1\\ 0&\dfrac{33}{7}&-\dfrac{9}{7}&\dfrac{24}{7}\\[7pt] 0&-\dfrac{9}{7}&\dfrac{33}{7}&\dfrac{24}{7}\\[7pt] 1&4&4&7 \end{bmatrix}\\ \begin{bmatrix} 7&4&4&1\\ 0&\dfrac{33}{7}&-\dfrac{9}{7}&\dfrac{24}{7}\\[7pt] 0&-\dfrac{9}{7}&\dfrac{33}{7}&\dfrac{24}{7}\\[7pt] 1&4&4&7 \end{bmatrix}{\small\begin{matrix} R_{4}-\dfrac{1}{7}R_{1}\rightarrow R_{4}\\ R_{3}+\dfrac{3}{11}R_{2}\rightarrow R_{3} \end{matrix}}&\sim \begin{bmatrix} 7&4&4&1\\ 0&\dfrac{33}{7}&-\dfrac{9}{7}&\dfrac{24}{7}\\[7pt] 0&0&\dfrac{48}{11}&\dfrac{48}{11}\\[7pt] 0&\dfrac{24}{7}&\dfrac{24}{7}&\dfrac{48}{7} \end{bmatrix}\\ \begin{bmatrix} 7&4&4&1\\ 0&\dfrac{33}{7}&-\dfrac{9}{7}&\dfrac{24}{7}\\[7pt] 0&0&\dfrac{48}{11}&\dfrac{48}{11}\\[7pt] 0&\dfrac{24}{7}&\dfrac{24}{7}&\dfrac{48}{7} \end{bmatrix}{\small\begin{matrix} R_{4}-\dfrac{8}{11}R_{2}\rightarrow R_{4}\\ R_{4}-R_{3}\rightarrow R_{4} \end{matrix}}&\sim \begin{bmatrix} 7&4&4&1\\ 0&\dfrac{33}{7}&-\dfrac{9}{7}&\dfrac{24}{7}\\[7pt] 0&0&\dfrac{48}{11}&\dfrac{48}{11}\\[7pt] 0&0&0&0 \end{bmatrix} \end{align*}

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