enter image description here

I have this question and I think I can just use maths for (ii) X'Xa=lambda a XX'Xa=XX'(Xa)=lambda(Xa) And so if lambda and a are eigenvalues/eigenvectors of X'X then lambda and Xa will be eigenvalues/eigenvectors of XX'

Not sure if this is enough though, but my main problem is that I can't figure out why it is important that X is a centered data matrix. How does that change the eigenvalues? For the first part of the question I would just divide lambda by (n-1) but that just seems too simplistic

  • $\begingroup$ Welcome to CV! You may want to add the self-study tag and read its wiki. (To the CV veterans, I'm voting leave open: While clearly textbook, the OP's specificity about X being centered is in keeping with the spirit and letter of the self-study tag.) $\endgroup$ Apr 25, 2017 at 13:43
  • $\begingroup$ I'm so sorry, did I break the rules? I'll read the wiki asap thanks! $\endgroup$
    – Jyn
    Apr 25, 2017 at 13:45
  • $\begingroup$ I don't believe you broke them. For self-study questions, we ask that posts outline a good faith attempt to solve the problem, and that they be specific about where the confusion lies. In my view you've done both of those things. $\endgroup$ Apr 25, 2017 at 13:47
  • 6
    $\begingroup$ It's just that if X is not centered then $X^\top X/(n-1)$ is not a covariance matrix (look up definition of covariance matrix - it includes "centering"). $\endgroup$
    – amoeba
    Apr 25, 2017 at 13:49
  • $\begingroup$ How does the matrix being centered change the eigenvalues/eigenvectors? The question title does not correspond to the question asked. Centering of columns of data X (n cases by p variables) does affect values of the eigenvalues and eigenvectors; but the book excerpt is not about the values, it is about some basic properties or rules of PCA. It tells the story that (first p) eigenvalues of X'X and of XX' are same. And that one can arrive from eigenvectors V of X'X to eigenvectors U of XX'. Which follows from the property of svd(X)=USV'. $\endgroup$
    – ttnphns
    Apr 25, 2017 at 15:18

1 Answer 1


In case you are still interested in some answer to your question title, a paper by Paul Henoeine (2014) (link: http://arxiv.org/abs/1407.2904v1) may be of some relevance. In particular, Lemma 1 and Theorem 3 of the paper give relationships in eigenvalues of $X'X$ between centered and non-centered $X$ matrix.

Let $K = X'X$ and $K_c$ = $X_c'X_c$ where $X$ is the nxp matrix and $X_c$ = $(I - \frac {1}{n}11')X$ (i.e., the centered counterpart). Note that $\frac {1}{n}1'X$ = ($\bar x_1$, $\bar x_2$, ..., $\bar x_p$), a row vector of sample means of the p variables, which is denoted as $\mu'$. Conventionally, the eigenvalues are ordered as a decreasing sequence.

Then, the eigen decompositions of $K$ and $K_c$ are $K = A\Lambda A'$ and $K_c = B\Lambda_c B'$, respectively. Note that $\Lambda$ = Diag{$\lambda_i$, i = 1, 2, ..., p }, the diagonal matrix of the eigenvalues of $K$ and $\Lambda_c$ = Diag{$\lambda_{ci}$, i = 1, 2, ..., p }, the diagonal matrix of the eigenvalues of $K_c$. Also, the columns of matrices A and B are the eigenvectors associated with the corresponding eigenvalues.

Applying Lemma 1 of Henoeine (2014), one has the following: $$\sum_{i=1}^p \lambda_{ci} = \sum_{i=1}^p \lambda_{i} - n\mu'\mu$$

Applying Theorem 3, one has the following interlacing property among the eigenvalues: $$\lambda_{cp} \le \lambda_p \le ... \le\lambda_{i+1} \le \lambda_{ci} \le \lambda_{i} \le ...\le \lambda_{2} \le \lambda_{c1} \le \lambda_{1}$$

  • $\begingroup$ This is interesting and may be very useful, so it would be nice to know at least a little about what those relationships are! Could you add a sentence or two about the nature of these relationships so that readers can easily decide whether to read the paper itself? $\endgroup$
    – whuber
    Jan 28, 2018 at 0:18
  • 1
    $\begingroup$ @whuber - appreciate your suggestion. More details are offered now. Hopefully, they are useful. $\endgroup$
    – T Lin
    Jan 29, 2018 at 2:35

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.