Answer to Question #235890 in Linear Algebra for lavanya

Question

Determine all eigenvalues and the corresponding eigenspaces for the matrix 𝐴 = [ −9 4 4 −8 3 4 −16 8 7 ]

Answer 1

$$A-\lambda I=\begin{pmatrix} -9-\lambda & 4 & 4 \\ -8& 3-\lambda & 4 \\ -16 & 8 & 7-\lambda \\ \end{pmatrix}$$

$$\det(A-\lambda I)=\begin{vmatrix} -9-\lambda & 4 & 4 \\ -8& 3-\lambda & 4 \\ -16 & 8 & 7-\lambda \\ \end{vmatrix}$$ $$=(-9-\lambda)\begin{vmatrix} 3-\lambda & 4 \\ 8 & 7-\lambda \end{vmatrix}-4\begin{vmatrix} -8 & 4 \\ -16 & 7-\lambda \end{vmatrix}$$

$$+4\begin{vmatrix} -8 & 3-\lambda \\ -16 & 8 \end{vmatrix}=(-9-\lambda)(21-10\lambda+\lambda^2-32)$$

$$-4(-56+8\lambda+64)+4(-64+48-16\lambda)$$

$$=99+11\lambda+90\lambda+10\lambda^2-9\lambda^2-\lambda^3$$

$$-32-32\lambda-64-64\lambda$$

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

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

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

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

$\lambda_1=-1, \lambda_2=-1, \lambda_3=3$

These are the eigenvalues: $-1, -1, 3.$

$\lambda=-1$

$$A-\lambda I=\begin{pmatrix} -9+1 & 4 & 4 \\ -8& 3+1 & 4 \\ -16 & 8 & 7+1 \\ \end{pmatrix}$$

$$=\begin{pmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \\ \end{pmatrix}$$

$R_2=R_2-R_1$

$$\begin{pmatrix} -8 & 4 & 4 \\ 0 & 0 & 0 \\ -16 & 8 & 8 \\ \end{pmatrix}$$

$R_3=R_3-2R_1$

$$\begin{pmatrix} -8 & 4 & 4 \\ 0 & 0 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}$$

$R_1=R_1/(-8)$

$$\begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 0 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}$$

$$\begin{pmatrix} 1 & -1/2 & -1/2 \\ 0 & 0 & 0 \\ 0 & 0 &0 \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}$$

If we take $x_2=t, x_3=s,$ then $x_1=\dfrac{1}{2}t+\dfrac{1}{2}s$

Thus

$$\vec x=\begin{pmatrix} s/2+t/2 \\ 1 \\ 1 \\ \end{pmatrix}s=\begin{pmatrix} 1/2 \\ 1 \\ 0 \\ \end{pmatrix}t+\begin{pmatrix} 1/2 \\ 0 \\ 1 \\ \end{pmatrix}s$$

The eigenvectors are

$$\begin{pmatrix} 1/2 \\ 1 \\ 0 \\ \end{pmatrix},\begin{pmatrix} 1/2 \\ 0 \\ 1 \\ \end{pmatrix}$$

$\lambda=3$

$$A-\lambda I=\begin{pmatrix} -9-3 & 4 & 4 \\ -8& 3-3 & 4 \\ -16 & 8 & 7-3 \\ \end{pmatrix}$$

$$=\begin{pmatrix} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 8 & 4 \\ \end{pmatrix}$$

$R_2=R_2-2R_1/3$

$$\begin{pmatrix} -12 & 4 & 4 \\ 0 & -8/3 & 4/3 \\ -16 & 8 & 4 \\ \end{pmatrix}$$

$R_3=R_3-4R_1/3$

$$\begin{pmatrix} -12 & 4 & 4 \\ 0 & -8/3 & 4/3 \\ 0 & 8/3 & -4/3 \\ \end{pmatrix}$$

$R_3=R_3+R_2$

$$\begin{pmatrix} -12 & 4 & 4 \\ 0 & -8/3 & 4/3 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_2=-3R_2/8$

$$\begin{pmatrix} -12 & 4 & 4 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_1=-R_1/12$

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

$R_1=R_1+R_2/3$

$$\begin{pmatrix} 1 &0 & -1/2 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$ $$\begin{pmatrix} 1 &0 & -1/2 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}$$

If we take $x_3=t,$ then $x_1=\dfrac{1}{2}t, x_2=\dfrac{1}{2}t$

Thus

$$\vec x=\begin{pmatrix} t/2 \\ t/2 \\ t \\ \end{pmatrix}=\begin{pmatrix} 1/2 \\ 1/2 \\ 1 \\ \end{pmatrix}t$$

The eigenvector is

$$\begin{pmatrix} 1/2 \\ 1/2 \\ 1 \\ \end{pmatrix}$$

Eigenvalue: −1, multiplicity: 2, eigenvectors: $\begin{pmatrix} 1/2 \\ 1 \\ 0 \\ \end{pmatrix},\begin{pmatrix} 1/2 \\ 0 \\ 1 \\ \end{pmatrix}$

Eigenvalue: 3, multiplicity: 1, eigenvector: $\begin{pmatrix} 1/2 \\ 1/2 \\ 1 \\ \end{pmatrix}$