What difference does centering (or de-meaning) your data make for PCA? I've heard that it makes the maths easier or that it prevents the first PC from being dominated by the variables' means, but I feel like I haven't been able to firmly grasp the concept yet.

For example, the top answer here How does centering the data get rid of the intercept in regression and PCA? describes how not centering would pull the first PCA through the origin, rather than the main axis of the point cloud. Based on my understanding of how the PC's are obtained from the covariance matrix's eigenvectors, I can't understand why this would happen.

Moreover, my own calculations with and without centering seem to make little sense.

Consider the setosa flowers in the iris dataset in R. I calculated the eigenvectors and eigenvalues of the sample covariance matrix as follows.

df <- iris[iris$Species=='setosa',1:4]
e <- eigen(cov(df))
> e
[1] 0.236455690 0.036918732 0.026796399 0.009033261

            [,1]       [,2]       [,3]        [,4]
[1,] -0.66907840  0.5978840  0.4399628 -0.03607712
[2,] -0.73414783 -0.6206734 -0.2746075 -0.01955027
[3,] -0.09654390  0.4900556 -0.8324495 -0.23990129
[4,] -0.06356359  0.1309379 -0.1950675  0.96992969

If I center the dataset first, I get exactly the same results. This seems quite obvious, since centering does not change the covariance matrix at all.

df.centered <- scale(df,scale=F,center=T)
e.centered<- eigen(cov(df.centered))

The prcomp function results in exactly this eigenvalue-eigenvector combination as well, for both the centered and uncentered dataset.

p.centered <- prcomp(df.centered)
Standard deviations:
[1] 0.48626710 0.19214248 0.16369606 0.09504347

                     PC1        PC2        PC3         PC4
Sepal.Length -0.66907840  0.5978840  0.4399628 -0.03607712
Sepal.Width  -0.73414783 -0.6206734 -0.2746075 -0.01955027
Petal.Length -0.09654390  0.4900556 -0.8324495 -0.23990129
Petal.Width  -0.06356359  0.1309379 -0.1950675  0.96992969

However, the prcomp function has the default option center = TRUE. Disabling this option results in the following PC's for the uncentered data (p.centered remains the same when center is set to false):

p.uncentered <- prcomp(df,center=F)
> p.uncentered
Standard deviations:
[1] 6.32674700 0.22455945 0.16369617 0.09766703

                    PC1         PC2        PC3         PC4
Sepal.Length -0.8010073  0.40303704  0.4410167  0.03811461
Sepal.Width  -0.5498408 -0.78739486 -0.2753323 -0.04331888
Petal.Length -0.2334487  0.46456598 -0.8317440 -0.19463332
Petal.Width  -0.0395488  0.04182015 -0.1946750  0.97917752

Why is this different from my own eigenvector calculations on the covariance matrix of the uncentered data? Does it have to do with the calculation? I've seen mentioned that prcomp uses something called the SVD method rather than the eigenvalue decomposition to calculate the PC's. The function princomp uses the latter, but its results are identical to prcomp. Does my issue relate to the answer I described at the top of this post?

EDIT: Issue was cleared up by the helpful @ttnphns. See his comment below, on this question: What does it mean to compute eigenvectors of a covariance matrix if the data were not centered first? and in this answer: https://stats.stackexchange.com/a/22520/3277. In short: a covariance matrix implicitly involves centering of the data already. PCA uses either SVD or eigendecomposition of the centered data $\bf X$, and the covariance matrix is then equal to ${\bf X'X}/(n-1)$.

  • 8
    $\begingroup$ Based on my understanding of how the PC's are obtained from the covariance matrix's eigenvectors... Please read comments in the answer you link to. Covariances imply centering of data, PCA "on covariances" = PCA on centered data. If you don't center the original variables X, PCA based on such data will be = PCA on X'X/n [or n-1] matrix. See also important overview: stats.stackexchange.com/a/22520/3277. $\endgroup$
    – ttnphns
    Jan 8, 2016 at 12:50
  • 3
    $\begingroup$ through the origin, rather than the main axis of the point cloud. PCA always pierces the origin. If data were centered, origin = the centroid. $\endgroup$
    – ttnphns
    Jan 8, 2016 at 12:54
  • $\begingroup$ Thank you, the overview especially helped clear things up for me. So it is simply an issue of naming conventions? I assumed PCA was a calculation based around the covariance matrix, whereas it is generally defined as a calculation performed on a centered dataset (where $\bf X'X$ is the covariance matrix). This seems like a confusing convention though. cont. $\endgroup$
    – Zenit
    Jan 8, 2016 at 18:48
  • 3
    $\begingroup$ Many sources emphasise the importance of centering your data and then go on to explain how the eigenvectors of the covariance matrix are of interest, seemingly ignoring the fact that $\bf S$ is implicitly centered already. The PCA entry on wikipedia does not make this distinction either: To find the axes of the ellipse, we must first subtract the mean of each variable from the dataset to center the data around the origin. Then, we compute the covariance matrix of the data, and calculate the eigenvalues and corresponding eigenvectors of this covariance matrix. $\endgroup$
    – Zenit
    Jan 8, 2016 at 18:50
  • 1
    $\begingroup$ See Principal component analysis: a review and recent developments for a discussion of centered vs. un-centered PCA (which are generally not the same). $\endgroup$
    – Yibo Yang
    Mar 24, 2019 at 22:34

3 Answers 3


As you remarked yourself and as explained by @ttnphns in the comments, computing covariance matrix implicitly performs centering: variance, by definition, is the average squared deviation from the mean. Centered and non-centered data will have identical covariance matrices. So if by PCA we understand the following procedure: $$\mathrm{Data}\to\text{Covariance matrix}\to\text{Eigen-decomposition},$$ then centering does not make any difference.

[Wikipedia:] To find the axes of the ellipse, we must first subtract the mean of each variable from the dataset to center the data around the origin. Then, we compute the covariance matrix of the data...

And so you are right to observe that this is not a very accurate formulation.

When people talk about "PCA on non-centered data", they mean that instead of covariance matrix the eigen-decomposition is performed on the $\mathbf X^\top \mathbf X/(n-1)$ matrix. If $\mathbf X$ is centered then this will be exactly the covariance matrix. If not then not. So if by PCA we understand the following procedure:

$$\text{Data } \mathbf X\to\text{Matrix } \mathbf X^\top \mathbf X/(n-1)\to\text{Eigen-decomposition},$$

then centering matters a lot and has the effect described and illustrated by @ttnphns in How does centering the data get rid of the intercept in regression and PCA?

It might seem weird to even mention this "strange" procedure, however consider that PCA can be very conveniently performed via singular value decomposition (SVD) of the data matrix $\mathbf X$ itself. I describe this in detail here: Relationship between SVD and PCA. How to use SVD to perform PCA? In this case the procedure is as follows:

$$\text{Data } \mathbf X \to \text{Singular value decomposition}.$$

If $\mathbf X$ is centered then this is equivalent to standard PCA done via covariance matrix. But if not, then it's equivalent to the "non-centered" PCA as described above. Since SVD is a very common and very convenient way to perform PCA, in practice it can be quite important to remember to center the data before calling svd function. I certainly had my share of bugs because of forgetting to do it.

  • 2
    $\begingroup$ I've got an advice for your answer to make a cleaner logic. Since in your two first (eigen) examples you are speaking of the decomposition MSCP matrix X'X/(n-1) and not of SSCP matrix X'X - it would be in the svd example, showing the equivalence, to decompose X/sqrt(n-1) instead of X (as you currently speak of). [Of course the difference is only in the scale part (eigenvalues) not eigenvectors, but it is better didactically, I would think.] My second point would be to remind that as soon as centering of X takes no place the Bessel correction n-1 becomes unnecessary, n is welcome. $\endgroup$
    – ttnphns
    Jan 8, 2016 at 23:02
  • 2
    $\begingroup$ Both good points, @ttnphns. I will think about how I can incorporate them into my answer: I wanted this answer to be understandable by a broader audience and so tried to avoid going into unnecessary mathematical details. $\endgroup$
    – amoeba
    Jan 8, 2016 at 23:10

I'll try to provide a mathematical justification.
By centering, I assume you mean applying eigendecomposition on $XX^T$ instead of $(X-\mu)(X-\mu)^T$
Here $\mu$ denotes the sample mean and not the true mean, thus $\mu = \sum_i x_i / n$
Let $B = \sum_i x_ix_i^T$
Let $A = \sum_i (x_i - \mu)(x_i - \mu)^T$ $A= \sum_i (x_ix_i^T -\mu x_i^T - x_i \mu^T + \mu\mu^T) = B + \sum_i (-\mu x_i^T - x_i \mu^T + \mu\mu^T)$
$A = B - \mu(\sum_i x_i)^T - (\sum_i x_i)\mu^T + \sum_i \mu\mu^T$
$A = B - \mu(n\mu)^T - (n\mu)\mu^T + n\mu\mu^T$
$A = B - n\mu\mu^T - n\mu\mu^T + n\mu\mu^T$
$A = B - n\mu\mu^T$
Usually, we take the eigendecomposition of $C = A/(n-1)$ (This is definition of the sample covariance matrix)
Here, say if you take the eigendecomposition of $D = B/(n-1)$, thus,
$$\boxed{D = C + \frac{n\mu\mu^T}{n-1}}$$
Then it is clear that the eigenvalues and eigenvectors corresponding to $D$ won't be the same as those for $C$ (unless $\mu$ = 0)
Thus, you would obtain wrong principal components (The correct principal components correspond to the eigendecompostion of $C$).


Carefully read through this thread which helps me greatly on understanding the whole PCA process, here is my summary (not sure if I get the true idea or not):

PCA <-> Eigen-decomposition on cov matrix -> it will find axis on whose direction the data has max spread. it doesn't matter whether we center the data data before hand, the cov matrix is the same and thus we will always get axis maximize the data spread.

I saw some text book when introducing PCA, they assume that the data matrix is centered. I think this is the reason that triggers my confusion, it induced me that the data needs to be centered before doing PCA analysis. Now my feeling is that I need to look these in a revers way: centering the data helps to justify that eigen vectors of covariance matrix is the right axis that we are looking for.

  • 3
    $\begingroup$ I think you missed the main point: PCA can be conducted on the sum-of-squares-and-products matrix. This is not the covariance matrix at all. It matters hugely whether the data are centered. $\endgroup$
    – whuber
    Jul 12, 2020 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.