I have a matrix with dimension nxp. No:of observations are not known. But no: of variables is 3. So p=3. Population mean is defined as µ1=1,µ2=-1,µ3=2.


\begin{bmatrix} 1 & k & 0\\ k & 1 & k\\ 0 & k & 1\end{bmatrix} How to find the value of k such that the principal components PC1 and PC2 account for more than some percentage (say 75%) of total variation of X?

  • $\begingroup$ It looks like Sigma is a correlation matrix not a covariance matrix because of the ones on the diagonal. Could you precise what is X ? $\endgroup$ – manu190466 Apr 10 '20 at 7:19
  • $\begingroup$ X = (X1, X2, X3) distributed as N3(µ, Σ). I did determinant( Σ - lamda * I ) = 0. To get value of k in terms of lamda. $\endgroup$ – StatsMonkey Apr 10 '20 at 11:52
  • $\begingroup$ When you say "they account ..." what would you mean exactly ? The two first eigenvectors ? Please edit your question to give more details. $\endgroup$ – manu190466 Apr 10 '20 at 12:28
  • $\begingroup$ Yes..There was mistake in question. I edited $\endgroup$ – StatsMonkey Apr 10 '20 at 13:11

Given that the cumulative ratio of variance explained is equal to the sum of the reduced eigenvalues (see 1 for details), you have to

  • calculate the three eigenvalues of your matrix in function of k
  • reduce and sum the two first eigenvalues
  • find the value of k that gives you the needed percentage (ie : 0.75)
  • 1
    $\begingroup$ I think we have to find all the eigenvalues of the matrix to find sum of the reduced eigenvalues. (λ1+λ2) / (λ1+λ2+λ3) $\endgroup$ – StatsMonkey Apr 10 '20 at 15:07
  • $\begingroup$ I didn't mentionned it but you are right ! Answer edited. $\endgroup$ – manu190466 Apr 10 '20 at 15:14

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