You can use the decathlon dataset {FactoMineR} to reproduce this. The question is why the computed eigenvalues differ from those of the covariance matrix.

Here are the eigenvalues using princomp:

> library(FactoMineR);data(decathlon)
> pr <- princomp(decathlon[1:10], cor=F)
> pr$sd^2
      Comp.1       Comp.2       Comp.3       Comp.4       Comp.5       Comp.6 
1.348073e+02 2.293556e+01 9.747263e+00 1.117215e+00 3.477705e-01 1.326819e-01 
      Comp.7       Comp.8       Comp.9      Comp.10 
6.208630e-02 4.938498e-02 2.504308e-02 4.908785e-03 

And the same using PCA:

> res<-PCA(decathlon[1:10], scale.unit=FALSE, ncp=5, graph = FALSE)
> res$eig
          eigenvalue percentage of variance cumulative percentage of variance
comp 1  1.348073e+02           79.659589641                          79.65959
comp 2  2.293556e+01           13.552956464                          93.21255
comp 3  9.747263e+00            5.759799777                          98.97235
comp 4  1.117215e+00            0.660178830                          99.63252
comp 5  3.477705e-01            0.205502637                          99.83803
comp 6  1.326819e-01            0.078403653                          99.91643
comp 7  6.208630e-02            0.036687700                          99.95312
comp 8  4.938498e-02            0.029182305                          99.98230
comp 9  2.504308e-02            0.014798320                          99.99710
comp 10 4.908785e-03            0.002900673                         100.00000

Can you explain to me why the directly computed eigenvalues differ from those? (the eigenvectors are the same):

> eigen(cov(decathlon[1:10]))$values
 [1] 1.381775e+02 2.350895e+01 9.990945e+00 1.145146e+00 3.564647e-01
 [6] 1.359989e-01 6.363846e-02 5.061961e-02 2.566916e-02 5.031505e-03

Also, the alternative prcomp method gives the same eigenvalues as the direct computation:

> prc <- prcomp(decathlon[1:10])
> prc$sd^2
 [1] 1.381775e+02 2.350895e+01 9.990945e+00 1.145146e+00 3.564647e-01
 [6] 1.359989e-01 6.363846e-02 5.061961e-02 2.566916e-02 5.031505e-03

Why do PCA/princomp and prcomp give different eigenvalues?

  • $\begingroup$ PCA will give you different results depending on whether you use the covariance matrix or the correlation matrix. $\endgroup$ Apr 13, 2011 at 10:59
  • 7
    $\begingroup$ The differences seem relatively small, though probably too big to be simple numerical issues. Could it be the difference between normalizing by $n$ or $n-1$, for example, when calculating an estimate of the covariance prior to computing the SVD or eigenvalues decomposition? $\endgroup$
    – cardinal
    Apr 13, 2011 at 12:58
  • 7
    $\begingroup$ @cardinal Nice guess! Notice that the two different sequences of eigenvalues have identical successive ratios. Thus, one set is a constant multiple of the other. The multiple is 1.025 = 41/40 (exactly). It's unclear to me where this comes from. Maybe the dataset has 41 elements and the OP is revealing only the first 10? $\endgroup$
    – whuber
    Apr 13, 2011 at 14:24
  • 7
    $\begingroup$ @cardinal Indeed: Help page for princomp: "Note that the default calculation uses divisor N for the covariance matrix." Help page for prcomp: "Unlike princomp, variances are computed with the usual divisor N-1." $\endgroup$
    – caracal
    Apr 13, 2011 at 15:04
  • 2
    $\begingroup$ @caracal, you should copy your comment into an answer (and maybe make it CW) so that it can be accepted and the question can be marked as resolved. $\endgroup$
    – cardinal
    Apr 13, 2011 at 15:14

1 Answer 1


As pointed out in the comments, it's because princomp uses $N$ for the divisor, but prcomp and the direct calculation using cov both use $N-1$ instead of $N$.

This is mentioned in both the Details section of help(princomp):

Note that the default calculation uses divisor 'N' for the covariance matrix.

and the Details section of help(prcomp):

Unlike princomp, variances are computed with the usual divisor N - 1.

You can also see this in the source. For example, the snippet of princomp source below shows that $N$ (n.obs) is used as the denominator when calculating cv.

else if (is.null(covmat)) {
    dn <- dim(z)
    if (dn[1L] < dn[2L]) 
        stop("'princomp' can only be used with more units than variables")
    covmat <- cov.wt(z)
    n.obs <- covmat$n.obs
    cv <- covmat$cov * (1 - 1/n.obs)
    cen <- covmat$center

You can avoid this multiplication by specifying the covmat argument instead of the x argument.

princomp(covmat = cov(iris[,1:4]))$sd^2

Update regarding PCA scores:

You can set cor = TRUE in your call to princomp in order to perform PCA on the correlation matrix (instead of the covariance matrix). This will cause princomp to $z$-score the data, but it will still use $N$ for the denominator.

As as result, princomp(scale(data))$scores and princomp(data, cor = TRUE)$scores will differ by the factor $\sqrt{(N-1)/N}$.

  • 1
    $\begingroup$ You might consider replacing "guessed" with "already confirmed" (see comment stream above.) You might also consider editing your answer to make it CW. Cheers. $\endgroup$
    – cardinal
    Apr 13, 2011 at 15:35
  • $\begingroup$ @cardinal I didn't see those comments. I only saw those that had been up-voted. Thanks. Also, could you explain the rationale behind making the answer CW? What are the rules / guidelines for that? $\endgroup$ Apr 13, 2011 at 15:38
  • $\begingroup$ Can anybody guess why the code is not simply cv <- cov.wt(z, method="ML") making the 2 follwing lines unnecessary? $\endgroup$
    – caracal
    Apr 13, 2011 at 15:42
  • 2
    $\begingroup$ @Joshua: My suggestion regarding making the answer CW was due to the fact that the answer appeared via a stream of comments and was generated by a "community" discussion. Since it was resolved in the comments, my thoughts are that it makes the most sense to reformulate it as an answer, marked as CW to indicate this collaboration, and this allows for the answer to be accepted and the question to be marked as resolved. (Otherwise, it will automatically get bumped back up by the software after a certain amount of time.) $\endgroup$
    – cardinal
    Apr 13, 2011 at 17:01
  • 2
    $\begingroup$ @amoeba it would have been helpful to mention that in your edit comment. "Added 860 characters to body" to a ~450 character answer does not help anyone evaluate whether the edit is reasonable. $\endgroup$ Oct 30, 2016 at 14:51

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