1.
⎣⎡11−1121141001⎦⎤ Find the reduced row echelon form
R2=R2−R1
⎣⎡10−112−1141−101⎦⎤ R3=R3+R1
⎣⎡10012−1341−111⎦⎤ R4=R4−R1
⎣⎡10002−1321−110⎦⎤ R2=−R2
⎣⎡100021321110⎦⎤ R1=R1−2R2
⎣⎡10000132−1110⎦⎤ R3=R3−3R2
⎣⎡10000102−11−20⎦⎤ R4=R4−2R2
⎣⎡10000100−11−2−2⎦⎤ R3=R3/(−2)
⎣⎡10000100−111−2⎦⎤ R1=R1+R3
⎣⎡10000100011−2⎦⎤ R2=R2−R3
⎣⎡10000100001−2⎦⎤ R4=R4+2R3
⎣⎡100001000010⎦⎤ To find the null space, solve the matrix equation
⎣⎡100001000010⎦⎤⎣⎡x1x2x3⎦⎤=⎣⎡0000⎦⎤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⎦⎤, it has no basis.
The nullity of the matrix is 0.
2.
A=[5−6−32]
Find the eigenvalues and eigenvectors
A−λI=[5−λ−6−32−λ]
det(A−λI)=∣∣5−λ−6−32−λ∣∣
=(5−λ)(2−λ)−18=10−7λ+λ2−18
=λ2−7λ−8 The characteristic equation
λ2−7λ−8=0
(λ+1)(λ−8)=0The roots are λ1=−1,λ2=8.
These are the eigenvalues.
Find the eigenvectors.
λ=−1
[5−λ−6−32−λ]=[6−6−33] R2=R2+R1
[60−30] R1=R1/6
[10−1/20] Solve the matrix equation
[10−1/20][x1x2]=[00] If we take x2=t, then x1=t/2.
Thus x=[t/2t]=[12]s
This is the eigenvector.
λ=8
[5−λ−6−32−λ]=[−3−6−3−6] R2=R2−2R1
[−30−30] R1=R1/(−3)
[1010] Solve the matrix equation
[1010][x1x2]=[00] If we take x2=t, then x1=−t.
Thus x=[−tt]=[1−1]s
This is the eigenvector.
Form the matrix P, whose column i is i -th eigenvector
P=[121−1] The matrix P that is diagonalizes A is
(i)
P=[121−1]
3.
1u+0v=00u+1v=01u+0v=02u+0v=0
⎣⎡10120100⎦⎤Find the reduced row echelon form
R3=R3−R1
⎣⎡10020100⎦⎤ R4=R4−2R1
⎣⎡10000100⎦⎤ To find the null space, solve the matrix equation
⎣⎡10000100⎦⎤[x1x2]=⎣⎡0000⎦⎤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], it has no basis.
The nullity of the matrix is 0. This is the dimension of the kernel of T.
Ker(T) is zero.
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