snarble - 9 months ago 38

R Question

I have an example word by document matrix (from Landauer and Dumais, 1997):

`wxd <- matrix(c(1,1,1,0,0,0,0,0,0,0,0,0,`

0,0,1,1,1,1,1,0,1,0,0,0,

0,1,0,1,1,0,0,1,0,0,0,0,

1,0,0,0,2,0,0,1,0,0,0,0,

0,0,0,1,0,1,1,0,0,0,0,0,

0,0,0,0,0,0,0,0,0,1,0,0,

0,0,0,0,0,0,0,0,0,1,1,0,

0,0,0,0,0,0,0,0,0,1,1,1,

0,0,0,0,0,0,0,0,1,0,1,1)

,12, 9)

rownames(wxd) <- c("human", "interface", "computer", "user", "system",

"response", "time", "EPS", "survey", "trees", "graph", "minors")

colnames(wxd) <- c(paste0("c", 1:5), paste0("m", 1:4))

I can perform Singular Value Decomposition on this matrix using the

`svd()`

`U`

`S`

`V`

`SVD <- svd(wxd)`

U <- SVD$u

S <- diag(SVD$d)

V <- SVD$v

I can multiply these matrices and get my original matrix returned (within some small margin or error):

`U %*% S %*% t(V)`

I can also take the first two columns of the

`U`

`V`

`S`

`U[ , 1:2] %*% S[1:2, 1:2] %*% t(V[ , 1:2])`

I am wanting to make sure I understand what this function is doing (as best as I am able), and I have been able to generate the

`V`

`S`

`svd()`

`ATA <- t(wxd) %*% wxd`

V2 <- eigen(ATA)$vectors

S2 <- sqrt(diag(eigen(ATA)$values))

But, the

`U`

`U`

`U`

`svd()`

`AAT <- wxd %*% t(wxd)`

U2 <- eigen(AAT)$vectors

So my question is, why is the

`U`

Answer

`wxd`

has rank of `9`

. Therefore, your `AAT`

only has `9`

non-zero eigenvalues (the rest are very small `~1e-16`

). For those zero eigenvalues, the eigenvectors are arbitrary as long as they span the subspace orthogonal to that spanned by the other eigenvectors in R^12.

Now, by default `svd`

only computes `nu=min(n,p)`

left singular vectors (similarly for right eigenvectors) where `n`

is the number of rows and `p`

is the number of columns in the input (see `?svd`

). Therefore, you only get `9`

left singular vectors. To generate all `12`

, call `svd`

with:

```
svd(wxd,nu=nrow(wxd))
```

However, those extra `3`

left singular vectors will not correspond to those found with `eigen(AAT)$vectors`

again because these eigenvectors are determined somewhat arbitrarily to span that orthogonal subspace.

As for why some of the signs have changed, recall that eigenvectors are only determined up to a scale factor. Although these eigenvectors are normalized, they may differ by a factor of `-1`

. To check just divide one from `U`

with the corresponding one from `U2`

. You should get columns of all `1`

s or `-1`

s:

```
U[,1:9]/U2[,1:9]
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
## [1,] 1 -1 1 -1 1 -1 1 1 1
## [2,] 1 -1 1 -1 1 -1 1 1 1
## [3,] 1 -1 1 -1 1 -1 1 1 1
## [4,] 1 -1 1 -1 1 -1 1 1 1
## [5,] 1 -1 1 -1 1 -1 1 1 1
## [6,] 1 -1 1 -1 1 -1 1 1 1
## [7,] 1 -1 1 -1 1 -1 1 1 1
## [8,] 1 -1 1 -1 1 -1 1 1 1
## [9,] 1 -1 1 -1 1 -1 1 1 1
##[10,] 1 -1 1 -1 1 -1 1 1 1
##[11,] 1 -1 1 -1 1 -1 1 1 1
##[12,] 1 -1 1 -1 1 -1 1 1 1
```

This can be seen from the definition of the eigenvector. From Wikipedia,

In linear algebra, an eigenvector or characteristic vector of a linear transformation is a non-zero vector that does not change its direction when that linear transformation is applied to it.

In finite-dimensional vector space, the linear transformation is in terms of multiplying the vector with a square matrix `A`

, and therefore the definition is (This is where I wish SO supports LaTeX markdown as this is not an equation in code; that is `*`

is matrix-multiply here):

```
A * v = lambda * v
```

which is known as the *Eigenvalue Equation* for the matrix `A`

where `lambda`

is the eigenvalue associated with the eigenvector `v`

. From this equation, it is clear that if `v`

is an eigenvector of `A`

then any `k * v`

for some scalar `k`

is also an eigenvector of `A`

with associated eigenvalue `lambda`

.

Source (Stackoverflow)