Test 2 Practice

• Ax=0
• [1 2, 3 4]*[x1,x2]=[0,0]   (How many columns=How many x val.)
• x1 + 2x2= 0
• 3x2 + 4x2= 0
• (solve for x1 and x2)

ker(A)= [x1 value, x2 value]

(If A=[0 0, 0 0], then any vector x=[x1, x2] is a solution)

• Find the  RREF.
• [1 1 1 1, 0 1 2 3]
• The number of pivots is the span (what dimension it is) 
• RREF= [1 1 1 1, 0 1 2 3]
• The pivot columns are the first and second. 

Thus, Vectors the span the Im(A)= [1 1] and [1 2]

1. If it is a 2x2, find the det(A). If it is nonzero, then A is rank 2 and spans all of R2.

2. We can see clearly that all columns are a scalar multiple of c1=[1, -2]. Thus, T(x)=Ax is a line in R^2 along the direction of the vector [1, -2]

• [7   11 | 0]*11 [77 121 | 0]       [77 121 | 0]
• [11  7  | 0]*7  [77  49  | 0] -(1) [0   -72 | 0]
• = 77c1 + 121c2= 0
• = -72c2= 0

• So, c2=0
• --> 77c1+121(0)=0  --> c1=0

C1 and c2 both equal 0, therefore they are linearly independent.

• [1 2 3 | 0]          [1  2  3 | 0]
• [2 3 4 | 0] -2(1) [0 -1 -2 | 0]
• ⇒ c1+2c2+3c3= 0
• ⇒ -c2-2c3= 0

• c2= -2c3
• --> c1+2(-2c3)+3c3=0
• = c1-4c3+3c3=0
• = c1=c3

Therefore, all vectors all linearly dependent. [3, 4] is redundant because it can be written as a linear combination between the other two vectors.

• [1 1 1 | 0]        [1 1 1 | 0]         [1 1 1 | 0]
• [1 2 3 | 0] -(1) [0 1 2 | 0]         [0 1 2 | 0]
• [1 3 6 | 0] -(1) [0 2 5 | 0] -2(2) [0 0 1 | 0]
• --> c1+c2+c3=0
• --> c2+2c3=0
• --> c3=0

• c2+2(0)=0
• = c2=0

• c1+(0)+(0)=0
• c1=0

All columns equal 0, thus they are all linearly independent.

• [1 0 1]  [x1]
• [1 1 1]*[x2] = 0
• [1 0 1]  [x3]

• x1+0+x3=0
• x1+x2+x3=0
• x1+0+x3=0

• x1=-x3
• (-x3)+x2+x3=0
• x2=0
• Thus, x=[-x3, 0, x3] or x=[-1, 0, 1]

• Therefore, a redundant column vector is c3=[1,1,1] (because it is the same thing as c1)
• A non-zero vector in the kernel of A is: x=[-1,0,1]

• Check for linear independence:
• c1+c2=0
• c1+2c2=0
• c1+3c2=0

• c1=-c2
• --> (-c2)+2c2=0 = c2=0
• c1+(0)=0
• c1=0
• [Plug in to all equations to ensure it holds]

Thus, c1 and c2 are linearly independent (because they equal 0). Therefore, they form a basis for the image A= [1,1,1], [1,2,3]

• Check for linear independence:
• [1 2 3 | 0]        [1  2  3 | 0]          [1 2 3 | 0]-2(2) [1 0 -1 | 0]
• [4 5 6 | 0] -4(1)[0 -3 -6 | 0]*-1/3 [0 1 2 | 0]        [0  1 2  | 0]

• c1+0-c3=0
• 0+c2+2c3=0

• c1=c3
• c2=-2c3

• Thus, c1 and c2 are linearly independent because they do not depend on each other. c3 is a copy of c1 so it is dependent and a redundant vector.
• Therefore, the basis of the Im(A)= [1,4], [2,5]

We can clearly see that c2 is a scalar multiple of c1 (c2=3*c1)

• c2=k*c1
• [3,6]=k[1,2]
• 3=1k --> k=3
• 6=2k --> k=3

Therefore, c2 is a scalar multiple of c1, so c1 is independent and c2 is redundant.

• Find the Im(A):
• c1 is the only independent column, so Im(A)= [1, 2]

• Find the ker(A):
• Ax=0
• [1 3][x1]=0
• [2 6][x2]
• x1+3x2=0
• 2x1+6x2=0

• x1=-3x2
• Because we have a scalar multiple, we can say:
• x= [x1,x2]=[-3x2, x2]
• Thus, ker(A)= [-3, 1]

• c1= [1,3], c2=[2,4]
• 1c1+2c2=0
• 3c1+4c2=0

• c1= -2c2
• --> 3(-2c2)+4c2=0
• = -2c2=0
• = c2=0
• --> c1+2(0)=0 --> c1=0

Therefore, c1 and c2 are linearly independent so there are no redundant vectors.

• Basis of Im(A):
• Since c1 and c2 are linearly independent, they form a basis of Im(A).
• Im(A)= {[1,3], [2,4]}

• Basis of ker(A):
• Ax=0
• [1 2][x1]= 0
• [3 4][x2]
• = x1+2x2=0
• = 3x1+4x2=0

• x1=-2x2
• --> 3(-2x2)+4x2=0
• -2x2=0
• x2=0
• --> 3x1+4(0)=0
• = x1=0

• Thus the only solution to the system is x1=0 and x2=0, so the ker(A) contains only the zero vector.
• ker(A)= ø

• c1=[1,2], c2=[-2,4], c3=[3,6]
• We can see that there is a scalar multiple of:
• c3= 3*c1
• Therefore, c3 is redundant

• c2=k*c1
• [-2,4]=k[1,2]
• = -2=1k --> k=-2
• = 4= 2k --> k=2
• Since -2≠2, c2 is not redundant

• Basis of Im(A):
• c3 is redundant, so it is not linearly independent. Thus, c1 & c2 are independent. Im(A)= {[1,2], [-2,4]}

• Basis of ker(A):
• Ax=0     [x1]
• [1 -2 3]*[x2]= 0
• [2  4 6]  [x3]
• = x1-2x2+3x3=0
• = 2x1+4x2+6x3=0

• x1= 2x2-3x3
• --> 2(2x2-3x3)+4x2+6x3=0
• = 8x2=0 --> x2=0
• --> x1-2(0)+3x3=-
• = x1=-3x3

• x=[-3x3, 0, x3]
• = ker(A)= [-3, 0, 1]

• We can see that there is a scalar multiple of: c2=2*c1
• Thus, c2 is redundant.

• c3= kc1
• [1,2,3]=k[1,1,1]
• 1=k1 --> k=1
• 2=k1 --> k=2 ⇨ 1≠2≠3, so not redundant
• 3=k1 --> k=3

• Basis of Im(A):
• c1 and c3 are linearly independent. Therefore, Im(A)={[1,1,1],[1,2,3]}

• Basis of ker(A):
• Ax=0
• [1 2 1][x1]
• [1 2 2][x2]= 0
• [1 2 3][x3]
• =1x1+2x2+x3=0
• =1x1+2x2+2x3=0
• =1x1+2x2+3x3=0

• x1= -2x2-x3
• -->(-2x2-x3)+2x2+2x3=0
• = x3=0
• --> x1+2x2+(0)=0
• x1=-2x2
• --> (-2x2)+2x2+2(0)=0
• = 0=0

• [x1]   [-2x2]         [-2]                     [-2]
• [x2]= [  x2  ] = x2[1]   --> ker(A)= [1]
• [x3]   [  0  ]          [0]                       [0]

We see that there are multiple scalar multiples: c2=2*c1 and c3=3*c1. Thus, c2 & c3 are redundant.

• Basis of Im(A):
• c1 is the only linearly independent vector. Im(A)= [1]

• Basis of ker(A):
• Ax=0
• [1 2 3][x1,x2,x3]=0
• x1+2x2+3x3=0

• x1= -2x2-3x3
•                     [x1]    [-2x2-3x3]        [-2]       [-3]
• Therefore, x=[x2] = [     x2    ] -->x2[1] + x3[0]
•                     [x3]    [     x2    ]          [0]       [1]

•                        [-2]       [-3]
• Thus, ker(A)= [1 ] and [0 ]
•                        [0 ]        [1 ]

• c1 ([0,0]) is redundant because it is a zero vector.
• We also see that there is a scalar multiple of: c3=2c1

• c5=kc2+tc4
• [3,4]=k[1,0]+t[0,1]
• 3=k1+0 --> k=3
• 4=0+t --> t=4    ⇒ k≠t, so c5 is redundant

Thus only c2 and c4 is linearly independent. c1, c3, c5 is redundant.

• Basis of Im(A):
• c2 and c4 are independent, so Im(A)= [1,0] and [0,1]

• Basis of ker(A):
• Ax=0
• [0 1 2 0 3]*[x1,x2,x3,x4,x5]=0
• [0 0 0 1 4]
• = 0+x2+2x3+0+3x5=0
• = 0+0+0+x4+4x5=0

• x2=-2x3-3x5
• x4=-4x5
•   [      x1     ]
•   [ -2x3-3x5]
• =[     x3     ]
•   [    -4x5   ]
•   [      x5    ]
• 1) x3=1 and x5=0
• 2) x3=0 and x5=1

• 1) [0, -2(1)-3(0), 1, -4(0), 0]
• = [0, -2, 1, 0, 0]

• 2) [0, -2(0)-3(1), 0, -4(1), 1]
• = [0, -3, 0, -4, 1]

Therefore, ker(A)= [0,-2,1,0,0] & [0,-3,0,-4,1]

• [1 3 9]         [1   3    9 ]                 [1   3     9]         [1 3 9]
• [4 5 8]-4(1) [0  -7  -28]                 [0  -7  -28]*-1/7[0 1 4]
• [7 8 3]-7(1) [0  -15 -60]-2(-15/-7) [0   0     0]         [0 0 0]

• [1 3 9]-3(2) [1 0 -3]
• [0 1 4]        [0 1  4]
• [0 0 0]        [0 0  0]

• Basis of Im(A):
• Pivot columns in RREF in c1 and c2.
• Thus, Im(A)=[1,4,7], [3,5,8]

• Basis of ker(A):
• Ax=0 from RREF
• [1 0 -3]  [x1]
• [0 1  4]*[x2] = 0
• [0 0  0]  [x3]
• = x1+0-3x3=0
• = 0+x2+4x4=0

• x1=3x3
• x2=-4x3
• Thus, [x1]   [3x3]            [ 3]
•          [x2]= [-4x3] --> x3[-4]
•          [x3]  ��[  x3]            [ 1]
• Therefore, ker(A)= [3,-4,1]

c1v1+c2v2+c3v3+c4v4=0

• c1+c2+c3+c4=0          [1  1  1  1 | 0]
• c1-c2+2c3-2c4=0    → [1 -1  2 -2 |0]   --> Solve for RREF
• c1+c2+4c3+4c4=0  → [1  1  4  4 | 0]
• c1-c2+8c3-8c4=0        [1 -1  8 -8 | 0]

• [1  1    1    1   | 0]
• [0  1 -1/2  3/2 | 0]
• [0  0    1    1   | 0]
• [0  0    0    1   | 0]
• We see that there are pivots in all 4 columns. Therefore, the vectors are all linearly independent and form a basis of R^4

• 2x1+3x2+x3=0
• = 2x1=-3x2-x3
• --> x1= -3/2 x2 - 1/2 x3

Let x2=t and x3=s

• [x1]  [-3/2*t - 1/2*s]     [-3/2]      [-1/2]
• [x2]=[          t         ] = t[  1  ] + s[  0  ]
• [x3]  [          s         ]     [  0  ]      [  1  ]

Basis for the supspace: [-3/2, 1, 0] and [-1/2, 0, 1]

• x= c1v1+c2v2
• [2,3]= c1[1,0]+c2[0,1]
• = 2=c1(1)+c2(0) --> c1=2
• = 3=c1(0)+c2(1) --> c2=3

[x]β= [c1,c2]= [2,3]

• x=c1v1+c2v2
• [31,37]= c1[23,29]+c2[31,37]
• = 31=c1(23)+c2(31)
• = 37=c1(29)+c2(37)

• 23c1=31-31c2
• --> c1= (31-31c2) / 23
• = 37=((31-31c2) / 23)29 + c2(37)
• = 37*23=(31-31c2)29 + c2(37*23)
• = 851 = 29(31-31c2) + 851c2
• = 851 = 899-899c2 + 851c2
• = 851 = 899-48c2
• = -48 = -48c2
• = c2=1

• --> 31=23c1+31(1)
• = c1=0

Thus, [x]β= [0,1]

• x=c1v1+c2v2
• [3,3,4]=c1[1,1,0]+c2[0,-1,2]
• = 3=1c1+0c2
• = 3=1c1-1c2
• = 4=0c1+2c2

• c1=3
• --> 3= 1(3)-c2
• = c2=0
• Check: 4=0(3)+2(0) --> 4≠0

Thus, the vector x is not in the span V

• T(v1)=Av1--> A[1,1]= [0 1, 1 0][1,1] --> [0+1, 1+0]= [1,1]
• Tv1= [1,1]= 1*v1+0v2
•                = 1
• [T(v1)]β=[1,0]

• T(v2)=Av2--> A[1,-1]= [0 1, 1 0]*[1,-1]= [0-1, 1+0]=[-1,1]
• Tv2= [-1,1]= -1v1+1v2

• av1+bv2=[-1,1]
• a[1,1]+b[1,-1]=[-1,1]
• = a1+b1=-1
• = a1-b1=1

• a= -1-b
• --> (-1-b)-b1=1
• = -2b=2 --> b=-1

• a-(1)=-1
• = a=0
• Thus, [T(v2)]β=[0,-1]

so, B=[1 0, 0 -1]

• T(v1)= Av1=Av1--> [1 2, 3 6]*[1,3]=[7,21]
• T(v1)= a1v1+a2v2
• [7,21]=a1[1,3]+a2[-2,1]
• = 7=a1-2b
• = 21=3a1+a2 

• a1=7+2a2
• --> 21=3(7+2a2)+a2
• = 7a2=0 --> a2=0

• a1-2(0)=7
• = a1=7
• Thus, first column of B is: [7,0]

• T(v2)=Av2=A[-2,1] --> [1 2, 3 6]*[-2,1]= [-2+2,-6+6]=[0,0]
• Thus, second column of B is [0,0]

So, B= [7 0, 0 0]

• Find orthogonal basis:
• v=[1,2] is eigenvector of T
• Thus, v1=[1,2]

• Now find vector that is orthogonal (perpendicular) 
• = v*w=0

• This means v=[1,2] * w[a,b] must equal 0
• 1a+2b=0
• --> a=-2b

• We can choose b=1, which gives is a=-2
• Thus, v2=[-2,1]

Therefore, B=[1,2] and [-2,1]

• To be a subspace, a set must satisfy three conditions:
• 1. It contains the zero vector 0=[0,0,0]
• 2. It is closed under vector addition.
• 3. It is closed under scalar multiplication.
• =-=-=-=-=-=
• 1. Zero vector: W= (0+0+0) ≠ 1
• So, not a subspace

2. Zero vector: W= (0≤0≤0) is true.

• Vector addition: We would need, x1+x2 ≤ y1+y2 ≤ z1+z2 which does not guarantee the relationship holds.
• So, not a subspace

3. Zero vector: (0+0+0)= [0,0,0]

Vector addition: Adding, x1​+2y1​+3z1​,4x1​+5y1​+6z1​,7x1​+8y1​+9z1​ and x2​+2y2​+3z2​,4x2​+5y2​+6z2​,7x2​+8y2​+9z2 gives us the same form as the original vectors.

• Scalar multiplication: Let c be a scalar and consider c[x+2y+3z, 4x+5y+6z, 7x+8y+9z]
• =[c(x+2y+3z), c(4x+5y+6z), c(7x+8y+9z)]
• This is again of the same form, so the set is closed under scalar multiplication.
• Is in a subspace