Chapter 6, Problem 1RE

To determine

To find: The eigenvalues and corresponding eigenvectors of the matrix $A=\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)$.

Answer to Problem 1RE

The eigenvalues and corresponding eigenvectors of the given matrix are ${\text{λ}}_1=-4,v_1=\left(\begin{array}{c}-2\\ 1\end{array}\right)$ and ${\text{λ}}_2=11,v_2=\left(\begin{array}{c}1\\ 2\end{array}\right)$.

Explanation of Solution

Given:

The given matrix is,

$A=\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)$

Approach:

A nonzero vector v is an eigenvector of the square matrix A if there is a number $\text{λ}$, called an eigenvalue of A. So that,

$Av=\text{λ}v$

If A is $n\times n$ equation ${\left(-1\right)}^n\det \left(A-\text{λI}\right)=0$ is called the characteristic equation of A; ${\left(-1\right)}^n\det \left(A-\text{λI}\right)$ is called the characteristic polynomial of A.

Calculation:

Consider the matrix.

$A=\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)$

The characteristic polynomial is given by,

$\begin{array}{c}\left|A-\text{λI}\right|=\left|\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)-\text{λ}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right|\\ =\left|\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)-\left(\begin{array}{cc}\text{λ}& 0\\ 0& \text{λ}\end{array}\right)\right|\\ =\left|\left(\begin{array}{cc}-1-\text{λ}& 6\\ 6& 8-\text{λ}\end{array}\right)\right|\\ =\left(-1-\text{λ}\right)\cdot \left(8-\text{λ}\right)-36\end{array}$

Simplify further.

$\begin{array}{c}\left(-1-\text{λ}\right)\cdot \left(8-\text{λ}\right)-36=-8+\text{λ}-8\text{λ}+{\text{λ}}^2-36\\ ={\text{λ}}^2-7\text{λ}-44\\ ={\text{λ}}^2-11\text{λ}+4\text{λ}-44\\ =\left(\text{λ}-11\right)\left(\text{λ}+4\right)\end{array}$

Equate to zero and find the value of $\text{λ}$.

$\begin{array}{c}\left(\text{λ}-11\right)\left(\text{λ}+4\right)=0\\ \text{λ}=-4,11\end{array}$

The eigenvalues are ${\text{λ}}_1=-4,{\text{λ}}_2=11$.

The eigenvector corresponding to eigenvalue ${\text{λ}}_1=-4$.

$\begin{array}{c}\left(A-\text{λI}\right)=\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)-\left(-4\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ =\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)+\left(\begin{array}{cc}4& 0\\ 0& 4\end{array}\right)\\ =\left(\begin{array}{cc}3& 6\\ 6& 12\end{array}\right)\end{array}$

Form the augmented matrix by reducing the column to zero.

$\left(\begin{array}{cc}3& 6\\ 6& 12\end{array}\right)\stackrel{R_2-2R_2}{\to }\left(\begin{array}{cc}3& 6\\ 0& 0\end{array}\right)$

It show the equation $3x_1+6y_1=0$, suppose $y_1=t$, then $x_1=-2t$. The eigenvector is $v_1=\left(\begin{array}{c}-2t\\ t\end{array}\right)$.

So, $v_1=\left(\begin{array}{c}-2\\ 1\end{array}\right)$ is the eigenvector corresponding to eigenvalue ${\text{λ}}_1=-4$.

The eigenvector corresponding to eigenvalue ${\text{λ}}_2=11$.

$\begin{array}{c}\left(A-\text{λI}\right)=\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)-\left(11\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ =\left(\begin{array}{cc}-1& 6\\ 6& 8\end{array}\right)-\left(\begin{array}{cc}11& 0\\ 0& 11\end{array}\right)\\ =\left(\begin{array}{cc}-12& 6\\ 6& -3\end{array}\right)\end{array}$

Form the augmented matrix by reducing the column to zero.

$\left(\begin{array}{cc}-12& 6\\ 6& -3\end{array}\right)\stackrel{R_2+\frac{1}{2}{R}_2}{\to }\left(\begin{array}{cc}-12& 6\\ 0& 0\end{array}\right)$

It show the equation $-12x_2+6y_2=0$, suppose $x_2=t$, then $y_2=2t$. The eigenvector is $v_1=\left(\begin{array}{c}1\\ 2\end{array}\right)t$, that is, $v_1=\left(\begin{array}{c}1\\ 2\end{array}\right)$ is the eigenvector corresponding to eigenvalue ${\text{λ}}_2=11$.

Therefore, the eigenvalues and corresponding eigenvectors of the given matrix are ${\text{λ}}_1=-4,v_1=\left(\begin{array}{c}-2\\ 1\end{array}\right)$ and ${\text{λ}}_2=11,v_2=\left(\begin{array}{c}1\\ 2\end{array}\right)$.