I am trying to decompose a covariance matrix based on a sparse / gappy data set. I'm noticing that the sum of lambda (explained variance), as calculated with svd, is being amplified with increasingly gappy data. Without gaps, svd and eigen yeild the same results.

This does not seem to happen with an eigen decomposition. I had been leaning towards using svd because the lambda values are always positive, but this tendency is worrying. Is there some sort of correction that needs to be applied, or should I avoid svd altogether for such a problem.

###Make complete and gappy data set
x <- 1:100
y <- 1:100
grd <- expand.grid(x=x, y=y)

#complete data
z <- matrix(runif(dim(grd)[1]), length(x), length(y))
image(x,y,z, col=rainbow(100))

#gappy data
zg <- replace(z, sample(seq(z), length(z)*0.5), NaN)
image(x,y,zg, col=rainbow(100))

###Covariance matrix decomposition
#complete data
C <- cov(z, use="pair")
E <- eigen(C)
S <- svd(C)


#gappy data (50%)
Cg <- cov(zg, use="pair")
Eg <- eigen(Cg)
Sg <- svd(Cg)


###Illustration of amplification of Lambda
frac <- seq(0,0.5,0.1)
E.lambda <- list()
S.lambda <- list()
for(i in seq(frac)){
    zi <- z
    NA.pos <- sample(seq(z), length(z)*frac[i])
    if(length(NA.pos) > 0){
        zi <- replace(z, NA.pos, NaN)
    Ci <- cov(zi, use="pair")
    E.lambda[[i]] <- eigen(Ci)$values
	S.lambda[[i]] <- svd(Ci)$d

x11(width=10, height=5)
YLIM <- range(c(sapply(E.lambda, range), sapply(S.lambda, range)))

for(i in seq(E.lambda)){
    if(i == 1) plot(E.lambda[[i]], t="n", ylim=YLIM, ylab="lambda", xlab="", main="Eigen Decomposition")
    lines(E.lambda[[i]], col=i, lty=1)
abline(h=0, col=8, lty=2)
legend("topright", legend=frac, lty=1, col=1:length(frac), title="fraction gaps")

for(i in seq(S.lambda)){
    if(i == 1) plot(S.lambda[[i]], t="n", ylim=YLIM, ylab="lambda", xlab="", main="Singular Value Decomposition")
    lines(S.lambda[[i]], col=i, lty=1)
abline(h=0, col=8, lty=2)
legend("topright", legend=frac, lty=1, col=1:length(frac), title="fraction gaps")

enter image description here

  • $\begingroup$ I'm sorry for being not able to follow your code (don't know R), but here's one or two notions. Negative eigenvalues can appear in eigen-decomposition of a cov. matrix if the raw data had many missing values and those were deleted pairwisely when computing the cov. SVD of such a matrix will report (misleadingly) those negative eigenvalues as positive. Your pictures show that both eigen and svd decompositions behave similarly (if not exactly same) besides that only difference regarding negative values. $\endgroup$
    – ttnphns
    Commented Aug 20, 2012 at 15:13
  • 1
    $\begingroup$ P.S. Hope you understood me: sum of eigenvalues must be equal to the trace (diagonal sum) of the cov. matrix. However, SVD is "blind" to the fact that some eigenvalues might be negative. SVD is rarely used to decompose non-gramian cov. matrix, it is typically used either with knowingly gramian (positive semidefinite) matrix or with raw data $\endgroup$
    – ttnphns
    Commented Aug 20, 2012 at 15:31
  • 1
    $\begingroup$ @ttnphns - Thanks for your insight. I guess I wouldn't be so worried about the result given by svd if it weren't for the different shape of the eigenvalues. The result is obviously giving more importance to the trailing eigenvalues than it should. $\endgroup$ Commented Aug 20, 2012 at 19:47

1 Answer 1


You need to do the sum of the absolute value of eigen values i.e, sum(abs(Eg$values)) and compare it with the sum of the singular values. They would be equal.

The reason is that if you multiply the rows or columns that correspond to the negative eigenvalues with $-1$, then the eigen-value of the new matrix becomes positive and the orthogonality of the eigen-vectors is not disturbed.

The proof of the converse of this beautiful theorem appeared in The algebra of hyperboloids of revolution, Javier F. Cabrera, Linear Algebra and its Applications, Princeton University (now at Rutgers).

Another way to reason this is by the fact that sqrt(eigen(t(Cg) %*% Cg)) are equal to the singular values of Cg. But when the eigenvalues are negative, the data must be represented in a hermitian form with the complex plane taken into account, which is what was missed in the original formulation, i.e the data formed by the symmetric square-root of the matrix with negative eigen values would have complex entries.

  • 1
    $\begingroup$ many thanks for this explanation. I was aware of the relationship between eigen and svd, but was unaware that they would still be related given the above-mentioned differences in decomposing the original matrix. Just curious - as far as I can tell, an eigen decomposition will give negative values when a matrix is not "positive definite". Is this the case with all covariance matrices based on gappy data? $\endgroup$ Commented Aug 21, 2012 at 10:57
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    $\begingroup$ Marc, positive definiteness of a real symmetric matrix is equivalent to having all positive eigenvalues. There is no close relationship with "gappy data," understanding that to mean sparse matrices with many zero entries. After all, the very sparsest among the nonsingular matrices (symmetric or not) are the diagonal ones, which exhibit their eigenvalues as their entries. $\endgroup$
    – whuber
    Commented Aug 21, 2012 at 16:11
  • $\begingroup$ @whuber - thanks for your comment. In this case, I interpret "gappy" differently to "sparse" in that the non-values are NaNs and not 0 (zero). Thus the covariance values are scaled by the number of common values (i.e. divided by n-1). In that respect, I don't believe that the covariance matrix actually contained any zeros. $\endgroup$ Commented Aug 21, 2012 at 18:52
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    $\begingroup$ Finding a way even to estimate the covariance matrix from such missing data is a challenge: I asked a similar question long ago and received some stimulating answers. $\endgroup$
    – whuber
    Commented Aug 21, 2012 at 19:19
  • $\begingroup$ I have asked an additional question elaborating on my interest in the subject for the use in Empirical Orthogonal Function (EOF) analysis here: stats.stackexchange.com/questions/34832/… $\endgroup$ Commented Aug 22, 2012 at 9:49

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