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A principal component analysis is carried out using the correlation matrix R of a data set with n = 25 observations and p = 4 variables. The ordered eigenvalues of R are given by

$$\lambda_1 = 2.25, \lambda_2 = 1.20, \lambda_3 = 0.35, \,\text{and} \, \lambda_4 = 0.2$$

with the associated eigenvectors

$$\gamma_1 = \pmatrix{0.57 \\ 0.51 \\ -0.25 \\ 0.59}, \, \gamma_2 = \pmatrix{-0.26 \\ 0.51 \\ 0.81 \\ 0.14}, \, \gamma_3 = \pmatrix{0.73 \\ 0.06 \\ *_1 \\ -0.62}, \, \gamma_4 = \pmatrix{0.30 \\ *_2 \\ *_3 \\ 0.50}$$

Where the $*$ denote the missing values.

Now I know a way to solve this would be to use that

$$R \gamma = \lambda \gamma$$

where $R$ is the correlation matrix. But this is only a small part of the total question I am looking it. So surely I am missing a much quicker way to find these values.

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    $\begingroup$ Eigenvectors form an orthonormal (orthogonal matrix). Therefore their column and row sums of squares... Therefore in your particular example..... $\endgroup$ – ttnphns May 14 '18 at 17:14
  • $\begingroup$ The sum of squares of the row/columns are equal to 1? $\endgroup$ – user162934 May 14 '18 at 17:20
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    $\begingroup$ Yes. In your example it is easy to find the missing entries. Given that column and row SS of the eigenvector matrix $V$ are all 1, find first absolute values of *1, and then of *2 and *3. Finally, set the signs for the values so that $VV'=V'V=I$ (identity matrix). See Wikipedia article on Orthogonal matrix. $\endgroup$ – ttnphns May 14 '18 at 17:56

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