Question #256065
  1. What is the basis for the null space of the set 121110110141\begin{vmatrix} 1 &2& 1 \\ 1 &1 & 0\\ -1 & 1 & 0\\ 1 & 4 & 1 \end{vmatrix}
  2. For a given matrix A = [5362]\begin{bmatrix} 5 & -3 \\ -6 & 2 \end{bmatrix} , the matrix P that is diagonalizes A is

(i) P = [1121]\begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}

(ii) P = [1121]\begin{bmatrix} 1 & -1 \\ -2 & 1 \end{bmatrix}

(iii) P = [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

3.. Suppose T:R2 -> M22 is a linear defined by T(u,v) = [uvu2u]\begin{bmatrix} u & v \\ u & 2u \end{bmatrix} then Ker (T) is ...


1
Expert's answer
2021-10-26T17:43:46-0400

1.


[121110110141]\begin{bmatrix} 1 & 2 & 1 \\ 1 & 1 & 0 \\ - 1 & 1 & 0 \\ 1 & 4 & 1 \\ \end{bmatrix}

Find the reduced row echelon form

R2=R2R1R_2=R_2-R_1


[121011110141]\begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & -1 \\ - 1 & 1 & 0 \\ 1 & 4 & 1 \\ \end{bmatrix}

R3=R3+R1R_3=R_3+R_1


[121011031141]\begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & 3 & 1 \\ 1 & 4 & 1 \\ \end{bmatrix}

R4=R4R1R_4=R_4-R_1


[121011031020]\begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & 3 & 1 \\ 0 & 2 & 0 \\ \end{bmatrix}

R2=R2R_2=-R_2


[121011031020]\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 3 & 1 \\ 0 & 2 & 0 \\ \end{bmatrix}

R1=R12R2R_1=R_1-2R_2


[101011031020]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 3 & 1 \\ 0 & 2 & 0 \\ \end{bmatrix}

R3=R33R2R_3=R_3-3R_2


[101011002020]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \\ 0 & 2 & 0 \\ \end{bmatrix}

R4=R42R2R_4=R_4-2R_2


[101011002002]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \\ 0 & 0 & -2 \\ \end{bmatrix}

R3=R3/(2)R_3=R_3/(-2)


[101011001002]\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 &1 \\ 0 & 0 & -2 \\ \end{bmatrix}

R1=R1+R3R_1=R_1+R_3


[100011001002]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 &1 \\ 0 & 0 & -2 \\ \end{bmatrix}

R2=R2R3R_2=R_2-R_3


[100010001002]\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &1 \\ 0 & 0 & -2 \\ \end{bmatrix}

R4=R4+2R3R_4=R_4+2R_3


[100010001000]\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


[100010001000][x1x2x3]=[0000]\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 [000],\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}, it has no basis.

The nullity of the matrix is 0.0.


2.


A=[5362]A=\begin{bmatrix} 5 &-3 \\ -6 & 2 \end{bmatrix}


Find the eigenvalues and eigenvectors


AλI=[5λ362λ]A-\lambda I=\begin{bmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{bmatrix}

det(AλI)=5λ362λ\det (A-\lambda I)=\begin{vmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{vmatrix}

=(5λ)(2λ)18=107λ+λ218=(5-\lambda)(2-\lambda)-18=10-7\lambda+\lambda^2-18

=λ27λ8=\lambda^2-7\lambda-8

The characteristic equation


λ27λ8=0\lambda^2-7\lambda-8=0

(λ+1)(λ8)=0(\lambda+1)(\lambda-8)=0

The roots are λ1=1,λ2=8.\lambda_1=-1, \lambda_2=8.

These are the eigenvalues.

Find the eigenvectors.


λ=1\lambda=-1

[5λ362λ]=[6363]\begin{bmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{bmatrix}=\begin{bmatrix} 6 &-3 \\ -6 & 3 \end{bmatrix}

R2=R2+R1R_2=R_2+R_1

[6300]\begin{bmatrix} 6 &-3 \\ 0 & 0 \end{bmatrix}

R1=R1/6R_1=R_1/6

[11/200]\begin{bmatrix} 1 &-1/2 \\ 0 & 0 \end{bmatrix}

Solve the matrix equation 


[11/200][x1x2]=[00]\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 x2=t,x_2=t, then x1=t/2.x_1=t/2.


Thus x=[t/2t]=[12]s\vec x=\begin{bmatrix} t/2 \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}s

This is the eigenvector.


λ=8\lambda=8


[5λ362λ]=[3366]\begin{bmatrix} 5-\lambda &-3 \\ -6 & 2-\lambda \end{bmatrix}=\begin{bmatrix} -3 &-3 \\ -6 & -6 \end{bmatrix}

R2=R22R1R_2=R_2-2R_1


[3300]\begin{bmatrix} -3 &-3 \\ 0 & 0 \end{bmatrix}

R1=R1/(3)R_1=R_1/(-3)


[1100]\begin{bmatrix} 1 &1 \\ 0 & 0 \end{bmatrix}

Solve the matrix equation 


[1100][x1x2]=[00]\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 x2=t,x_2=t, then x1=t.x_1=-t.


Thus x=[tt]=[11]s\vec x=\begin{bmatrix} -t \\ t \end{bmatrix}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}s

This is the eigenvector.


Form the matrix P,P, whose column ii is ii -th eigenvector


P=[1121]P=\begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}

The matrix PP that is diagonalizes AA is

(i)


P=[1121]P=\begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix}

3.


1u+0v=00u+1v=01u+0v=02u+0v=0\begin{matrix} 1u+0v=0 \\ 0u+1v=0\\ 1u+0v=0 \\ 2u+0v=0\\ \end{matrix}


[10011020]\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ 2 & 0 \\ \end{bmatrix}

Find the reduced row echelon form

R3=R3R1R_3=R_3-R_1


[10010020]\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 2 & 0 \\ \end{bmatrix}

R4=R42R1R_4=R_4-2R_1


[10010000]\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix}

To find the null space, solve the matrix equation


[10010000][x1x2]=[0000]\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 [00],\begin{bmatrix} 0 \\ 0 \\ \end{bmatrix}, it has no basis.

The nullity of the matrix is 0.0. This is the dimension of the kernel of T.T.

Ker(T)(T) is zero.




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