Answer to Question #272373 in Linear Algebra for Pan

Question

Known Matrix:

$$A=\begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{bmatrix}$$ $$B=\begin{bmatrix} 3 & 0 & 2 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}$$

1. Determine the characteristic equation det(A − λI) = 0 of the matrix above
2. Determine the eigenvalues of the matrix and the basis of the eigenspace

Note: In matrix B, let the value of a be so that the eigenvalues and the basis of the eigenspace are dependent on a.

Answer 1

a)

$$A=\begin{bmatrix} 2& 1 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{bmatrix}$$

$$A-\lambda I=\begin{bmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}$$

The characteristic equation

$$\begin{vmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{vmatrix}=0$$

$$(2 -\lambda)\begin{vmatrix} 2 -\lambda & 2 \\ 1 & 3 -\lambda \end{vmatrix}-(1)\begin{vmatrix} 1 & 2 \\ 1 & 3 -\lambda \end{vmatrix}$$

$$+(2)\begin{vmatrix} 1 & 2-\lambda \\ 1 & 1 \end{vmatrix}$$

$$=(2-\lambda)((2-\lambda)(3-\lambda)-2)-(3-\lambda-2)$$

$$+2(1-(2-\lambda))$$

$$=(2-\lambda)(4-5\lambda+\lambda^2)-1+\lambda+2\lambda-2$$

$$=8-10\lambda+2\lambda^2-4\lambda+5\lambda^2-\lambda^3+3\lambda-3$$

$$=-\lambda^3+7\lambda^2-11\lambda+5=0$$

$$-\lambda^2(\lambda-1)+6\lambda(\lambda-1)-5(\lambda-1)=0$$

$$-(\lambda-1)(\lambda^2-6\lambda+5)=0$$

Matrix $A.$ The characteristic equation is

$$-(\lambda-1)(\lambda-1)(\lambda-5)=0$$

$$B=\begin{bmatrix} 3& 0 & 2 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}$$

$$B-\lambda I=\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}$$

The characteristic equation

$$\begin{vmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{vmatrix}=0$$

$$(3 -\lambda)\begin{vmatrix} 1 -\lambda & a \\ 2 & 2a -\lambda \end{vmatrix}-(0)\begin{vmatrix} 0 & 2 \\ 2 & 2a -\lambda \end{vmatrix}$$

$$+(0)\begin{vmatrix} 0 & 2 \\ 1-\lambda & a \end{vmatrix}$$

$$=(3-\lambda)((1-\lambda)(2a-\lambda)-2a)$$

$$+2(1-(2-\lambda))$$

$$=(3-\lambda)(2a-\lambda-2a\lambda+\lambda^2-2a)$$

$$=(3-\lambda)\lambda(\lambda-(1+2a))=0$$

Matrix $B.$ The characteristic equation is

$$(3-\lambda)\lambda(\lambda-(1+2a))=0$$

2.

Matrix $A.$ The characteristic equation is

$$-(\lambda-1)(\lambda-1)(\lambda-5)=0$$

The eigen values are $\lambda_1=5, \lambda_2=1, \lambda_3=1.$

$\lambda_1=5$

$$\begin{bmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} -3 & 1 & 2 \\ 1 & -3 & 2 \\ 1 & 1 & -2 \end{bmatrix}$$

$R_1=R_1/(-3)$

$$\begin{bmatrix} 1 & -1/3 & -2/3 \\ 1 & -3 & 2 \\ 1 & 1 & -2 \end{bmatrix}$$

$R_2=R_2-R_1$

$$\begin{bmatrix} 1 & -1/3 & -2/3 \\ 0 & -8/3 & 8/3 \\ 1 & 1 & -2 \end{bmatrix}$$

$R_3=R_3-R_1$

$$\begin{bmatrix} 1 & -1/3 & -2/3 \\ 0 & -8/3 & 8/3 \\ 0 & 4/3 & -4/3 \end{bmatrix}$$

$R_2=-3R_2/8$

$$\begin{bmatrix} 1 & -1/3 & -2/3 \\ 0 & 1 & -1 \\ 0 & 4/3 & -4/3 \end{bmatrix}$$

$R_1=R_1+R_2/3$

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 4/3 & -4/3 \end{bmatrix}$$

$R_3=R_3-4R_2/3$

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$

Solve the matrix equation

$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $x_3=t,$ then $x_1=t, x_2=t, x_3=t.$

The eigen vector is $\begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix}.$

$\lambda_2=1$

$$\begin{bmatrix} 2 -\lambda & 1 & 2 \\ 1 & 2-\lambda & 2 \\ 1 & 1 & 3-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 1 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{bmatrix}$$

$R_2=R_2-R_1$

$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 1 & 1 & 2 \end{bmatrix}$$

$R_3=R_3-R_1$

$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

Solve the matrix equation

$$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $x_2=t,x_3=s,$ then $x_1=-t-2s, x_2=t, x_3=s.$

The eigen vectors are $\begin{bmatrix} -1 \\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0\\ 1 \end{bmatrix}$

Matrix $B.$ The characteristic equation is

$$(3-\lambda)\lambda(\lambda-(1+2a))=0$$

The eigen values are $\lambda_1=3, \lambda_2=1+2a, \lambda_3=0.$

$\lambda_1=3$

$$\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}=\begin{bmatrix} 0 & 0 & 2 \\ 0 & -2 & a \\ 0 & 2 & 2a-3 \end{bmatrix}$$

Swap the rows 1and 2

$$\begin{bmatrix} 0 & -2 & a \\ 0 & 0 & 2 \\ 0 & 2 & 2a-3 \end{bmatrix}$$

$R_1=R_1/(-2)$

$$\begin{bmatrix} 0 & 1 & -a/2 \\ 0 & 0 & 2 \\ 0 & 2 & 2a-3 \end{bmatrix}$$

$R_3=R_3-2R_1$

$$\begin{bmatrix} 0 & 1 & -a/2 \\ 0 & 0 & 2 \\ 0 & 0& 3a-3 \end{bmatrix}$$

$R_2=R_2/2$

$$\begin{bmatrix} 0 & 1 & -a/2 \\ 0 & 0 & 1 \\ 0 & 0& 3a-3 \end{bmatrix}$$

$R_1=R_1+aR_2/2$

$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0& 3a-3 \end{bmatrix}$$

$R_3=R_3-3(a-1)R_2$

$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0& 0 \end{bmatrix}$$

Solve the matrix equation

$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $x_1=t,$ then $x_1=t, x_2=0, x_3=0.$

The eigen vector is $\begin{bmatrix} 1 \\ 0\\ 0 \end{bmatrix}.$

$\lambda_2=2a+1$

$$\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}=\begin{bmatrix} 2-2a & 0 & 2 \\ 0 & -2a & a \\ 0 & 2 & -1 \end{bmatrix}$$

$R_1=R_1/(2-2a), a\not=1$

$$\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & -2a & a \\ 0 & 2 & -1 \end{bmatrix}$$

$R_2=R_2/(-2a)$

$$\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & 1 & -1/2 \\ 0 & 2 & -1 \end{bmatrix}$$

$R_3=R_3-2R_2$

$$\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \end{bmatrix}$$

Solve the matrix equation

$$\begin{bmatrix} 1 & 0 & 1/(1-a) \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $x_3=t,$ then $x_1=1/(a-1), x_2=1/2, x_3=1.$

The eigen vector is $\begin{bmatrix} 1 /(a-1)\\ 1/2\\ 1 \end{bmatrix}.$

$\lambda_3=0$

$$\begin{bmatrix} 3 -\lambda & 0 & 2 \\ 0 & 1-\lambda & a \\ 0 & 2 & 2a-\lambda \end{bmatrix}=\begin{bmatrix} 3 & 0 & 2 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}$$

$R_1=R_1/3$

$$\begin{bmatrix} 1 & 0 & 2/3 \\ 0 & 1 & a \\ 0 & 2 & 2a \end{bmatrix}$$

$R_3=R_3-2R_2$

$$\begin{bmatrix} 1 & 0 & 2/3 \\ 0 & 1 & a \\ 0 & 0 & 0 \end{bmatrix}$$

Solve the matrix equation

$$\begin{bmatrix} 1 & 0 & 2/3 \\ 0 & 1 & a \\ 0 & 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

If we take $x_3=t,$ then $x_1=-2/3, x_2=-a, x_3=1.$

The eigen vector is $\begin{bmatrix} -2/3\\ -a\\ 1 \end{bmatrix}.$