Question

2. Let A=$\begin{pmatrix} - 3 & 1 & 0\\ - 6 & 2 & 0\\ - 3 & 1 & 0 \end{pmatrix}$

a) Find the characteristic polynomial of A and show that the eigenvalues are 0 and - 1

b) Find a basis for each eigenspace of A

c) Explain why is A diagonalisable

d) Find an invertible matrix P and a diagonal matrix D such that A=PDP-¹

e) Hence, or otherwise, calculate A²⁰¹⁸

Answer 1

a)

$$A=\begin{pmatrix} - 3 & 1 & 0\\ - 6 & 2 & 0\\ - 3 & 1 & 0 \end{pmatrix}$$

$$A-\lambda I=\begin{pmatrix} - 3-\lambda & 1 & 0\\ - 6 & 2-\lambda & 0\\ - 3 & 1& 0-\lambda \end{pmatrix}$$

Characteristic polynomial

$$\det(A-\lambda I)=|A-\lambda I|=\begin{vmatrix} - 3-\lambda & 1 & 0\\ - 6 & 2-\lambda & 0\\ - 3 & 1 & 0 -\lambda \end{vmatrix}$$

$$=(-3-\lambda)\begin{vmatrix} 2-\lambda& 0 \\ 1 & -\lambda \end{vmatrix}-1\begin{vmatrix} -6 & 0 \\ -3& -\lambda \end{vmatrix}+0\begin{vmatrix} -6 & 2-\lambda \\ -3 & 1 \end{vmatrix}$$

$$=(-3-\lambda)(-2\lambda+\lambda^2-0)-(6\lambda+0)+0$$

$$=6\lambda-3\lambda^2+2\lambda^2-\lambda^3-6\lambda=-\lambda^3-\lambda^2$$

The characteristic equation of $A$ is

$$|A-\lambda I|=0$$

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

The roots are $\lambda_1=\lambda_2=0, \lambda_3=-1.$

These are the eigenvalues.

b) Find the eigenvectors.

$\lambda=-1$

$$A-\lambda I=\begin{pmatrix} - 2 & 1 & 0\\ - 6 & 3 & 0\\ - 3 & 1& 1 \end{pmatrix}$$

$R_2=R_2-3R_1$

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

$R_3=R_3-3R_1/2$

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

Swap the rows 2 and 3

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

$R_2=-2R_2$

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

$R_1=R_1-R_2$

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

$R_1=R_1/(-2)$

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

Solve the matrix equation

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

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

Thus

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

The null space of this matrix is 

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

$\lambda=0$

$$A-\lambda I=\begin{pmatrix} - 3 & 1 & 0\\ - 6 & 2 & 0\\ - 3 & 1 & 0 \end{pmatrix}$$

$R_2=R_2-2R_1$

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

$R_3=R_3-R_1$

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

$R_1=R_1/(-3)$

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

Solve the matrix equation

$$\begin{pmatrix} 1 & -1/3 & 0\\ 0 & 0 & 0\\ 0 & 0& 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$$

If we take $x_2=t,x_3=s,$ then $x_1=(1/3)t.$

Thus

$$\vec x=\begin{pmatrix} (1/3)t\\ t\\ s \end{pmatrix}=\begin{pmatrix} 1/3\\ 1\\ 0 \end{pmatrix}t+\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}s$$

The null space of this matrix is 

$$\bigg\{\begin{pmatrix} 1/3\\ 1\\ 0 \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}\bigg\}$$

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

Eigenvalue: $0,$ multiplicity: $2,$ eigenvectors: $\begin{pmatrix} 1/3\\ 1\\ 0 \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$

c) A matrix is diagonalizable if and only of for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue.

For the eigenvalue $-1$ this is trivially true as its multiplicity is only one and we find one nonzero eigenvector associated to it.

For the eigenvector  we find two linearly indepedent eigenvectors.

Therefore the matrix $A$ is diagonalizable.

d)

Form the matrix $P,$ whose column $i$  is $i$-th eigenvector

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

Form the diagonal matrix $D$ whose element at row $i,$ column $i$ is $i$-th eigenvalue

$$D=\begin{pmatrix} -1 & 0& 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$$

$$\det P=\begin{vmatrix} 1 & 1/3 & 0\\ 2 & 1 & 0\\ 1 & 0& 1 \end{vmatrix}=1\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}-(1/3)\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix}+0$$

$$=1-2/3=1/3\not=0$$

The cofactor matrix is

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

The adjugate matrix is

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

The inverse matrix is the adjugate matrix divided by the determinant.

$$P^{-1}=\begin{pmatrix} 3 & -1 & 0\\ -6 & 3 & 0\\ -3 & 1 & 1 \end{pmatrix}$$

$$A=PDP^{-1}$$

$$A^{2018}=PD^{2018}P^{-1}$$

$$=\begin{pmatrix} 1 & 1/3 & 0\\ 2 & 1 & 0\\ 1 & 0& 1 \end{pmatrix}\begin{pmatrix} (-1)^{2018} & 0& 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} 3 & -1 & 0\\ -6 & 3 & 0\\ -3 & 1 & 1 \end{pmatrix}$$

$$=\begin{pmatrix} 1 & 0 & 0\\ 2 & 0 & 0\\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 3 & -1 & 0\\ -6 & 3 & 0\\ -3 & 1 & 1 \end{pmatrix}$$

$$=\begin{pmatrix} 3 & -1 & 0\\ 6 & -2 & 0\\ 3 & -1 & 0 \end{pmatrix}$$