Introducing Latent Semantic Analysis

Introducing Latent Semantic Analysis

Tomas K Landauer et al., “An introduction to latent semantic analysis,” Discourse Processes, Vol. 25 (2-3), pp. 259-284, 1998.Scott Deerwester et al., “Indexing by latent semantic analysis,” Journal of the American Society for Information Science, Vol. 41 (6), pp. 391-407, 1990.Kirk Baker, “Singular Value Decomposition Tutorial,” Electronic document, 2005.

Aug 22, 2014Hee-Gook Jun

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Outline

SVD SVD to LSA Conclusion

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Eigendecomposition vs. Singular Value Decomposition

Eigendecomposition– Must be a diagonalizable matrix– Must be a square matrix– Matrix (n x n size) must have n linearly independent eigenvector

e.g. symmetric matrix ..

Singular Value Decomposition– Computable for any size (M x n) of matrix

A U ∑ VT

A P Ʌ P-1

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U: Left Singular Vectors of A

Unitary matrix– Columns of U are orthonormal (orthogonal + normal)– orthonormal eigenvectors of AAT

A U ∑ VT

and is orthogonal

= [0,0,0,1] = [0,1,0,0]

= (0x0) + (0x1) + (0x0) + (1x0)

= 0

is normal vector

= [0,0,0,1]

|| =

= 1

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V: Right Singular Vectors of A

Unitary matrix– Columns of V are orthonormal (orthogonal + normal)– orthonormal eigenvectors of ATA

A U ∑ VT

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∑ (or S)

Diagonal Matrix– Diagonal entries are the singular values of A

Singular values– Non-zero singular values– Square roots of eigenvalues from U (or V) in descending order

A U ∑ VT

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Calculation Procedure

1. U is a list of eigenvectors of AAT

1. Compute AAT

2. Compute eigenvectors of AAT

3. Matrix Orthonormalization

2. V is a list of eigenvectors of ATA1. Compute ATA2. Compute eigenvalues of ATA3. Orthonormalize and transpose

3. ∑ is a list of eigenvalues of U or V1. (eigenvalues of U = eigenvalues of V)

A U ∑ VT

① ② ③

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1.1 Matrix U – Compute AAT

Start with the matrix

Transpose of A

Then

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1.2 Matrix U – Eigenvectors and Eigenvalues [1/2]

Eigenvector– Nonzero vector that satisfies the equation– A is a square matrix, is an eigenvalue (scalar), is the eigenvector

≡rearrange

set determinent of the coefficient matrix to zero

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1.2 Matrix U – Eigenvectors and Eigenvalues [2/2]

Thus, set of eigenvectors [𝟏 𝟏𝟏 −𝟏]

② For

Calculated eigenvalues

① For

eigenvector

eigenvector

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1.3 Matrix U – Orthonormalization

Gram-Schmidt orthonormalization

𝑤𝑘=𝑣𝑘−∑𝑖=1

𝑘−1

(𝑢𝑖 ∙𝑣𝑘 )×𝑢𝑖

set of eigenvectors orthonormal matrix

𝑣1𝑣2 𝑢1𝑢2

normalize v1

normalize w2

find w2 (orthogonal to u1)

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2.1 Matrix VT – Compute ATA

Start with the matrix

Transpose of A

Then

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2.2 Matrix VT – Eigenvectors and Eigenvalues [1/2]

Eigenvector– Nonzero vector that satisfies the equation– A is a square matrix, is an eigenvalue (scalar), is the eigenvector

≡rearrange

set determinent of the coefficient matrix to zeroby cofactor expansion ( 여인수 전개 )

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2.2 Matrix VT – Eigenvectors and Eigenvalues [2/2]

Thus, set of eigenvectors

② For

① For eigenvector

[𝟏 𝟏𝟏 −𝟏]

③ For

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2.3 Matrix VT – Orthonormalization and Transformation

Gram-Schmidtorthonormalization

𝑤𝑘=𝑣𝑘−∑𝑖=1

𝑘−1

(𝑢𝑖 ∙𝑣𝑘 )×𝑢𝑖

set of eigenvectors orthonormal matrix

𝑣1𝑣2𝑣3 𝑢1𝑢2𝑢3

normalize v1

normalize w2find w2 (orthogonal to u1)

normalize w3

find w3 (orthogonal to u2)

Transpose

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3.1 Matrix ∑ (= S)

Square roots of the non-zero eigenvalues– Populate the diagonal with the values– Diagonal entries in ∑ are the singular values of A

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Outline

SVD SVD to LSA

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Latent Semantic Analysis

Use SVD (Singular Value Decomposition)– to simulate human learning of word and passage meaning

Represent word and passage meaning– as high-dimensional vectors in the semantic space

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LSA Example

doc 1 " modem the steering linux. modem, linux the modem. steering the modem. linux "

doc 2 " linux; the linux. the linux modem linux. the modem, clutch the modem. petrol "

doc 3 " petrol! clutch the steering, steering, linux. the steering clutch petrol. clutch the petrol; the clutch "

doc 4 " the the the. clutch clutch clutch! steering petrol; steering petrol petrol; steering petrol "

First analysis – Document Similarity

Second analysis – Term Similarity

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LSA Example: Build a Term Frequency Matrix

d1 d2 d3 d4

linux 3 4 1 0

modem 4 3 0 1

the 3 4 4 3

clutch 0 1 4 3

steer-ing

2 0 3 3

petrol 0 1 3 4

Let Matrix A =

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LSA Example: Compute SVD of Matrix A

dim1

dim2

dim3

dim4

t1-

0.33

-0.53

0.36

-0.14

t2-

0.32

-0.53

-0.48

0.35

t3-

0.61

-0.09

0.26

-0.14

t4-

0.37

0.42

0.60

-0.23

t5-

0.35

0.25

-0.68

-0.46

t6-

0.37

0.42

0.01

0.74

dim1

dim2

dim3

dim4

dim1

11.4 0 0 0

dim2

0 6.27 0 0

dim3

0 0 2.21 0

dim4

0 0 0 1.28

d1 d2 d3 d4

dim1

-0.42

-0.56

-0.64

-0.29

dim2

-0.48

-0.52

0.61 0.33

dim3

-0.56

0.44 0.27 -0.63

dim4

-0.51

0.46 -0.35

0.63

d1 d2 d3 d4

t1 (linux) 3 4 1 0

t2 (mo-dem)

4 3 0 1

t3 (the) 3 4 4 3

t4 (clutch) 0 1 4 3

t5 (steer-ing)

2 0 3 3

t6 (petrol) 0 1 3 4

A

x x

6 x 4 4 x 4 4 x 4

U S VT

=

- R code -result ← svd(A)

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LSA Example: Reduced SVD

dim1

dim2

dim3

dim4

t1-

0.33

-0.53

0.36

-0.14

t2-

0.32

-0.53

-0.48

0.35

t3-

0.61

-0.09

0.26

-0.14

t4-

0.37

0.42

0.60

-0.23

t5-

0.35

0.25

-0.68

-0.46

t6-

0.37

0.42

0.01

0.74

dim1

dim2

dim3

dim4

dim1

11.4 0 0 0

dim2

0 6.27 0 0

dim3

0 0 2.21 0

dim4

0 0 0 1.28

d1 d2 d3 d4

dim1

-0.42

-0.56

-0.64

-0.29

dim2

-0.48

-0.52

0.61 0.33

dim3

-0.56

0.44 0.27 -0.63

dim4

-0.51

0.46 -0.35

0.63

x x

6 x 4 4 x 4 4 x 4

dim1

dim2

dim3

dim4

t1-

0.33

-0.53

0.36

-0.14

t2-

0.32

-0.53

-0.48

0.35

t3-

0.61

-0.09

0.26

-0.14

t4-

0.37

0.42

0.60

-0.23

t5-

0.35

0.25

-0.68

-0.46

t6-

0.37

0.42

0.01

0.74

dim1

dim2

dim3

dim4

dim1

11.4 0 0 0

dim2

0 6.27 0 0

dim3

0 0 2.21 0

dim4

0 0 0 1.28

d1 d2 d3 d4

dim1

-0.42

-0.56

-0.64

-0.29

dim2

-0.48

-0.52

0.61 0.33

dim3

-0.56

0.44 0.27 -0.63

dim4

-0.51

0.46 -0.35

0.63

x x

6 x 2 2 x 2 2 x 4

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LSA Example: Document Similarity

dim1

dim2

dim3

dim4

dim1

11.4 0 0 0

dim2

0 6.27 0 0

dim3

0 0 2.21 0

dim4

0 0 0 1.28

d1 d2 d3 d4

dim1

-0.42

-0.56

-0.64

-0.29

dim2

-0.48

-0.52

0.61 0.33

dim3

-0.56

0.44 0.27 -0.63

dim4

-0.51

0.46 -0.35

0.63

x

2 x 2 2 x 4

S V

=

d1 d2 d3 d4

dim1

-4.83

5.49 -6.49

-5.86

dim2

-3.52

-3.28

2.79 2.88

d1 d2 d3 d4

d1 1 0.99 0.51 0.46

d2 0.99 1 0.58 0.54

d3 0.51 0.58 1 0.99

d4 0.46 0.54 0.99 1

𝑆𝑖𝑚 ( 𝐴 ,𝐵 )=𝑐𝑜𝑠𝑖𝑛𝑒𝜃=𝐴 ∙𝐵

|𝐴||𝐵|=

(−4.83×5.49 )+(−3.52×−3.28)

√ (−4.83 )2+(−3.52 )2×√(5.49 )2+ (−3.28 )2doc 1"modem the steering linux. modem, linux the modem. steering the modem. linux "

doc 2"linux; the linux. the linux modem linux. the modem, clutch the modem. petrol "

doc 3"petrol! clutch the steering, steering, linux. the steering clutch petrol. clutch the petrol; the clutch "

doc 4 "the the the. clutch clutch clutch! steering petrol; steering petrol petrol; steering petrol "

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LSA Example: Term Similarity

dim1

dim2

dim3

dim4

t1-

0.33

-0.53

0.36

-0.14

t2-

0.32

-0.53

-0.48

0.35

t3-

0.61

-0.09

0.26

-0.14

t4-

0.37

0.42

0.60

-0.23

t5-

0.35

0.25

-0.68

-0.46

t6-

0.37

0.42

0.01

0.74

dim1

dim2

dim3

dim4

dim1

11.4 0 0 0

dim2

0 6.27 0 0

dim3

0 0 2.21 0

dim4

0 0 0 1.28

x =

S V

𝑆𝑖𝑚 ( 𝐴 ,𝐵 )=𝑐𝑜𝑠𝑖𝑛𝑒𝜃=𝐴 ∙𝐵

|𝐴||𝐵|

dim1

dim2

t1-

3.76

-3.33

t2-

3.65

-3.35

t3-

7.01

-0.61

t4-

4.30

2.63

t5-

4.09

1.59

t6-

4.24

2.65

t1 t2 t3 t4 t5 t6

t1 10.99

0.80

0.29

0.45

0.28

t20.99

1.00

0.79

0.27

0.44

0.26

t30.80

0.79

10.80

0.89

0.79

t40.29

0.27

0.80

10.98

0.99

t50.45

0.44

0.89

0.98

10.98

t60.28

0.26

0.79

0.99

0.98

1

linux modem the clutch steering petrol

linux modem the

modem linux the

the linux modem clutch steering petrol

clutch the steering petrol

steering the clutch petrol

petrol the clutch steering

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Conclusion

Pros– Compute document similarity– even if they do not have common words

Cons– Statistical foundation missing → PLSA

dim1

dim2

dim3

dim4

t1-

0.33

-0.53

0.36

-0.14

t2-

0.32

-0.53

-0.48

0.35

t3-

0.61

-0.09

0.26

-0.14

t4-

0.37

0.42

0.60

-0.23

t5-

0.35

0.25

-0.68

-0.46

t6-

0.37

0.42

0.01

0.74

dim1

dim2

dim3

dim4

dim1

11.4 0 0 0

dim2

0 6.27 0 0

dim3

0 0 2.21 0

dim4

0 0 0 1.28

d1 d2 d3 d4

dim1

-0.42

-0.56

-0.64

-0.29

dim2

-0.48

-0.52

0.61 0.33

dim3

-0.56

0.44 0.27 -0.63

dim4

-0.51

0.46 -0.35

0.63

x x

Which one is to be chosen to reduce?