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Find an SVD of the indicated matrix. (Enter each matrix in the form [[row 1], [row 2], ...), where each row is a comma-separated list.) A= [12] (U, E, 1) = ( D. Need Help? Read It Watch It Talk to a Tutor 0/1 POINTS PREVIOUS ANSWERS POOLELINALG4 7.4.016. MY NOTES Find an SVD of the indicated matrix. (Enter each matrix in the form [[row 1], [row 2], ...), where each row is a comma-separated list.) A = [ 8 15] (U, E, D =( [[1]][[16.0]] | [16: 15) Need Help? Read It Talk to a Tutor 10. -/1 POINTS POOLELINALG4 7.4.018. MY NOTES Find an SVD of the indicated matrix. (Enter each matrix in the form [[row 1], [row 2], ...), where each row is a comma-separated list.) ſo 1 A = 1 0 [ 2-2 (U, E, 1) = D Need Help? Read It Talk to a Tutor

Answer 1

Find Singular Value Decomposition (SVD) of a Matrix ...
[ 5

12 ]

Solution:

A=

5 12

A⋅A′

A⋅A′=

25 60 60 144

Find Eigen vector for A⋅A′

|A⋅A′-λI|=0

(25-λ) 60 60 (144-λ)

= 0

∴(25-λ)×(144-λ)-60×60=0

∴(3600-169λ+λ2)-3600=0

∴(λ2-169λ)=0

∴λ(λ-169)=0

∴λ=0or(λ-169)=0

∴ The eigenvalues of the matrix A⋅A′ are given by λ=0,169,

1. Eigenvectors for λ=169

v1=

0.4167 1

2. Eigenvectors for λ=0

v2=

-2.4 1

For Eigenvector-1 (0.4167,1), Length L = √0.41672+12=1.0833

So, normalizing gives u1=(0.41671.0833,11.0833)=(0.3846,0.9231)

For Eigenvector-2 (-2.4,1), Length L = √(-2.4)2+12=2.6

So, normalizing gives u2=(-2.42.6,12.6)=(-0.9231,0.3846)

A′⋅A

A′⋅A=

[ 169 ]

Find Eigen vector for A′⋅A

|A′⋅A-λI|=0

(169-λ)

= 0

∴(169-λ)=0

∴(-λ+169)=0

∴-(λ-169)=0

∴(λ-169)=0

∴ The eigenvalues of the matrix A′⋅A are given by λ=169,

1. Eigenvectors for λ=169

v1=

[ 1 ]

For Eigenvector-1 (1), Length L = √12=1

So, normalizing gives v1=(11)=(1)

Solution

∴Σ=

√169 0

=

13 0

∴U=

[u1,u2]

=

0.3846 -0.9231 0.9231 0.3846

V is found using formula vi=1σiAT⋅ui

∴V=

[ 1 ]

----------------------------------------===============================--------------------------------------------------------

Solution:

A=

[ 8 15 ]

A⋅A′

A⋅A′=

[ 289 ]

Find Eigen vector for A⋅A′

|A⋅A′-λI|=0

(289-λ)

= 0

∴(289-λ)=0

∴(-λ+289)=0

∴-(λ-289)=0

∴(λ-289)=0

∴ The eigenvalues of the matrix A⋅A′ are given by λ=289,

1. Eigenvectors for λ=289

v1=

[ 1 ]

For Eigenvector-1 (1), Length L = √12=1

So, normalizing gives u1=(11)=(1)

A′⋅A

A′⋅A=

64 120 120 225

Find Eigen vector for A′⋅A

|A′⋅A-λI|=0

(64-λ) 120 120 (225-λ)

= 0

∴(64-λ)×(225-λ)-120×120=0

∴(14400-289λ+λ2)-14400=0

∴(λ2-289λ)=0

∴λ(λ-289)=0

∴λ=0or(λ-289)=0

∴ The eigenvalues of the matrix A′⋅A are given by λ=0,289,

1. Eigenvectors for λ=289

v1=

0.5333 1

2. Eigenvectors for λ=0

v2=

-1.875 1

For Eigenvector-1 (0.5333,1), Length L = √0.53332+12=1.1333

So, normalizing gives v1=(0.53331.1333,11.1333)=(0.4706,0.8824)

For Eigenvector-2 (-1.875,1), Length L = √(-1.875)2+12=2.125

So, normalizing gives v2=(-1.8752.125,12.125)=(-0.8824,0.4706)

Solution

∴Σ=

[ √289 0 ]

=

[ 17 0 ]

∴U=

[u1]

=

[ 1 ]

V is found using formula vi=1σiAT⋅ui

∴V=	0.4706 -0.8824 0.8824 0.4706

-----------------------------------------------==========================------------------------------------------------------------

Find Singular Value Decomposition (SVD) of a Matrix ...
[0 1

1 0

2- 2]

Solution:

A=	0 1 1 0 2 -2

A⋅A′

A⋅A′=

1 0 -2 0 1 2 -2 2 8

Find Eigen vector for A⋅A′

|A⋅A′-λI|=0

(1-λ) 0 -2 0 (1-λ) 2 -2 2 (8-λ)

= 0

∴(1-λ)((1-λ)×(8-λ)-2×2)-0(0×(8-λ)-2×(-2))+(-2)(0×2-(1-λ)×(-2))=0

∴(1-λ)((8-9λ+λ2)-4)-0(0-(-4))-2(0-(-2+2λ))=0

∴(1-λ)(4-9λ+λ2)-0(4)-2(2-2λ)=0

∴(4-13λ+10λ2-λ3)-0-(4-4λ)=0

∴(-λ3+10λ2-9λ)=0

∴-λ(λ-1)(λ-9)=0

∴λ=0or(λ-1)=0or(λ-9)=0

∴ The eigenvalues of the matrix A⋅A′ are given by λ=0,1,9,

1. Eigenvectors for λ=9

v1=	-0.25 0.25 1

2. Eigenvectors for λ=1

v2=

1 1 0

3. Eigenvectors for λ=0

v3=

2 -2 1

For Eigenvector-1 (-0.25,0.25,1), Length L = √(-0.25)2+0.252+12=1.0607

So, normalizing gives u1=(-0.251.0607,0.251.0607,11.0607)=(-0.2357,0.2357,0.9428)

For Eigenvector-2 (1,1,0), Length L = √12+12+02=1.4142

So, normalizing gives u2=(11.4142,11.4142,01.4142)=(0.7071,0.7071,0)

For Eigenvector-3 (2,-2,1), Length L = √22+(-2)2+12=3

So, normalizing gives u3=(23,-23,13)=(0.6667,-0.6667,0.3333)

A′⋅A

A′⋅A=

5 -4 -4 5

Find Eigen vector for A′⋅A

|A′⋅A-λI|=0

(5-λ) -4 -4 (5-λ)

= 0

∴(5-λ)×(5-λ)-(-4)×(-4)=0

∴(25-10λ+λ2)-16=0

∴(λ2-10λ+9)=0

∴(λ-1)(λ-9)=0

∴(λ-1)=0or(λ-9)=0

∴ The eigenvalues of the matrix A′⋅A are given by λ=1,9,

1. Eigenvectors for λ=9

v1=

-1 1

2. Eigenvectors for λ=1

v2=

1 1

For Eigenvector-1 (-1,1), Length L = √(-1)2+12=1.4142

So, normalizing gives v1=(-11.4142,11.4142)=(-0.7071,0.7071)

For Eigenvector-2 (1,1), Length L = √12+12=1.4142

So, normalizing gives v2=(11.4142,11.4142)=(0.7071,0.7071)

Solution

∴Σ=

√9 0 0 √1 0 0

=

3 0 0 1 0 0

∴U=

[u1,u2,u3]

=

-0.2357 0.7071 0.6667 0.2357 0.7071 -0.6667 0.9428 0 0.3333

V is found using formula vi=1σiAT⋅ui

∴V=	0.7071 0.7071 -0.7071 0.7071