Eigenvalues: Spectral Decomposition

Setup

This vignette uses an example of a \(3 \times 3\) matrix to illustrate some properties of eigenvalues and eigenvectors. We could consider this to be the variance-covariance matrix of three variables, but the main thing is that the matrix is square and symmetric, which guarantees that the eigenvalues, \(\lambda_i\) are real numbers, and non-negative, \(\lambda_i \ge 0\).

A <- matrix(c(13, -4, 2, -4, 11, -2, 2, -2, 8), 3, 3, byrow=TRUE)
A

##      [,1] [,2] [,3]
## [1,]   13   -4    2
## [2,]   -4   11   -2
## [3,]    2   -2    8

Get the eigenvalues and eigenvectors using eigen(); this returns a named list, with eigenvalues named values and eigenvectors named vectors. We call these L and V here, but in formulas they correspond to a diagonal matrix, \(\mathbf{\Lambda} = diag(\lambda_1, \lambda_2, \lambda_3)\), and a (orthogonal) matrix \(\mathbf{V}\).

ev <- eigen(A)
# extract components
(L <- ev$values)

## [1] 17  8  7

(V <- ev$vectors)

##         [,1]    [,2]   [,3]
## [1,]  0.7454  0.6667 0.0000
## [2,] -0.5963  0.6667 0.4472
## [3,]  0.2981 -0.3333 0.8944

Matrix factorization

1. Factorization of A: A = V diag(L) V’. That is, the matrix \(\mathbf{A}\) can be represented as the product \(\mathbf{A}= \mathbf{V} \mathbf{\Lambda} \mathbf{V}'\).

V %*% diag(L) %*% t(V)

##      [,1] [,2] [,3]
## [1,]   13   -4    2
## [2,]   -4   11   -2
## [3,]    2   -2    8

2. V diagonalizes A: L = V’ A V. That is, the matrix \(\mathbf{V}\) transforms \(\mathbf{A}\) into the diagonal matrix \(\mathbf{\Lambda}\), corresponding to orthogonal (uncorrelated) variables.

diag(L)

##      [,1] [,2] [,3]
## [1,]   17    0    0
## [2,]    0    8    0
## [3,]    0    0    7

zapsmall(t(V) %*% A %*% V)

##      [,1] [,2] [,3]
## [1,]   17    0    0
## [2,]    0    8    0
## [3,]    0    0    7

Spectral decomposition

The basic idea here is that each eigenvalue–eigenvector pair generates a rank 1 matrix, \(\lambda_i \mathbf{v}_i \mathbf{v}_i '\), and these sum to the original matrix, \(\mathbf{A} = \sum_i \lambda_i \mathbf{v}_i \mathbf{v}_i '\).

A1 = L[1] * V[,1] %*% t(V[,1])
A1

##        [,1]   [,2]   [,3]
## [1,]  9.444 -7.556  3.778
## [2,] -7.556  6.044 -3.022
## [3,]  3.778 -3.022  1.511

A2 = L[2] * V[,2] %*% t(V[,2])
A2

##        [,1]   [,2]    [,3]
## [1,]  3.556  3.556 -1.7778
## [2,]  3.556  3.556 -1.7778
## [3,] -1.778 -1.778  0.8889

A3 = L[3] * V[,3] %*% t(V[,3])
A3

##      [,1] [,2] [,3]
## [1,]    0  0.0  0.0
## [2,]    0  1.4  2.8
## [3,]    0  2.8  5.6

Then, summing them gives A, so they do decompose A:

A1 + A2 + A3

##      [,1] [,2] [,3]
## [1,]   13   -4    2
## [2,]   -4   11   -2
## [3,]    2   -2    8

all.equal(A, A1+A2+A3)

## [1] TRUE

sum(A^2)

## [1] 402

c( sum(A1^2), sum(A2^2), sum(A3^2) )

## [1] 289  64  49

sum( sum(A1^2), sum(A2^2), sum(A3^2) )

## [1] 402

#' same as tr(A' A)
tr(crossprod(A))

## [1] 402

L^2

## [1] 289  64  49

cumsum(L^2)   # cumulative

## [1] 289 353 402

R(A1)

## [1] 1

R(A1 + A2)

## [1] 2

R(A1 + A2 + A3)

## [1] 3

# two dimensions
sum((A1+A2)^2)

## [1] 353

sum((A1+A2)^2) / sum(A^2)   # proportion

## [1] 0.8781