Answer to Question #256065 in Linear Algebra for Three

Question

1. What is the basis for the null space of the set $\begin{vmatrix} 1 &2& 1 \\ 1 &1 & 0\\ -1 & 1 & 0\\ 1 & 4 & 1 \end{vmatrix}$
2. For a given matrix A = $\begin{bmatrix} 5 & -3 \\ -6 & 2 \end{bmatrix}$ , the matrix P that is diagonalizes A is

(i) P = $\begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}$

(ii) P = $\begin{bmatrix} 1 & -1 \\ -2 & 1 \end{bmatrix}$

(iii) P = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

3.. Suppose T:R² -> M₂₂ is a linear defined by T(u,v) = $\begin{bmatrix} u & v \\ u & 2u \end{bmatrix}$ then Ker (T) is ...

Answer 1

1.

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

Find the reduced row echelon form

$R_2=R_2-R_1$

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

$R_3=R_3+R_1$

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

$R_4=R_4-R_1$

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

$R_2=-R_2$

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

$R_1=R_1-2R_2$

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

$R_3=R_3-3R_2$

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

$R_4=R_4-2R_2$

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

$R_3=R_3/(-2)$

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

$R_1=R_1+R_3$

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

$R_2=R_2-R_3$

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

$R_4=R_4+2R_3$

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

To find the null space, solve the matrix equation

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

Since this system has a unique solution, the null space contains only a zero vector.

The nullity of a matrix is the dimension of the basis for the null space.

Thus, the nullity of the matrix is 0.

The null space is $\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix},$ it has no basis.

The nullity of the matrix is $0.$

2.

$$A=\begin{bmatrix} 5 &-3 \\ -6 & 2 \end{bmatrix}$$

Find the eigenvalues and eigenvectors

$$A-\lambda I=\begin{bmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{bmatrix}$$

$$\det (A-\lambda I)=\begin{vmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{vmatrix}$$

$$=(5-\lambda)(2-\lambda)-18=10-7\lambda+\lambda^2-18$$

$$=\lambda^2-7\lambda-8$$

The characteristic equation

$$\lambda^2-7\lambda-8=0$$

$$(\lambda+1)(\lambda-8)=0$$

The roots are $\lambda_1=-1, \lambda_2=8.$

These are the eigenvalues.

Find the eigenvectors.

$\lambda=-1$

$$\begin{bmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{bmatrix}=\begin{bmatrix} 6 &-3 \\ -6 & 3 \end{bmatrix}$$

$R_2=R_2+R_1$

$$\begin{bmatrix} 6 &-3 \\ 0 & 0 \end{bmatrix}$$

$R_1=R_1/6$

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

Solve the matrix equation 

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

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

Thus $\vec x=\begin{bmatrix} t/2 \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}s$

This is the eigenvector.

$\lambda=8$

$$\begin{bmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{bmatrix}=\begin{bmatrix} -3 &-3 \\ -6 & -6 \end{bmatrix}$$

$R_2=R_2-2R_1$

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

$R_1=R_1/(-3)$

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

Solve the matrix equation 

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

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

Thus $\vec x=\begin{bmatrix} -t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}s$

This is the eigenvector.

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

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

The matrix $P$ that is diagonalizes $A$ is

(i)

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

3.

$$\begin{matrix} 1u+0v=0 \\ 0u+1v=0\\ 1u+0v=0 \\ 2u+0v=0\\ \end{matrix}$$

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

Find the reduced row echelon form

$R_3=R_3-R_1$

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

$R_4=R_4-2R_1$

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

To find the null space, solve the matrix equation

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

Since this system has a unique solution, the null space contains only a zero vector.

The nullity of a matrix is the dimension of the basis for the null space.

Thus, the nullity of the matrix is 0.

The null space is $\begin{bmatrix} 0 \\ 0 \\ \end{bmatrix},$ it has no basis.

The nullity of the matrix is $0.$ This is the dimension of the kernel of $T.$

Ker$(T)$ is zero.