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When I carry out Principal Component Analysis, the outputs are the Eigen-values and Eigen-vectors for each PC.

Question: are the Eigen-values directly proportional to the Variance explained by the PC? For example, for a three-dimensional data-set, my Eigen-values came out as: 0.007, 0.002, 0.001. Does that mean that the first PC captures 0.007 / (0.007 + 0.002 + 0.001) ~ 77% of the variance?

Jan Stuller
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    Does [https://stats.stackexchange.com/questions/22569/pca-and-proportion-of-variance-explained?rq=1](this) answer your question? – tchainzzz Jul 12 '21 at 18:46
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    Does this answer your question? [PCA and proportion of variance explained](https://stats.stackexchange.com/questions/22569/pca-and-proportion-of-variance-explained) – tchainzzz Jul 12 '21 at 18:47
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    the answer is always YES – Aksakal Jul 12 '21 at 19:10

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There are 2 concepts to look out for and it can get confusing at first. Let me put it the best way you can understand:

  1. Variance of one PC:

    The variance for the $i^{th}$ principal component is equal to the $i^{th}$ Eigenvalue. In your case, the variance of the $1^{st}$ PC is equal to the $1^{st}$ Eigenvalue which is 0.007.

  2. Proportion of Variance by one PC:

    The proportion of variation explained by the $i^{th}$ principal component is then defined to be the eigenvalue for that component divided by the sum of the eigenvalues. In other words, the $i^{th}$ principal component explains the proportion of the total variation for total number of factors let's say $n$ as:

    $x_i$/$\sum$$x_{i\to{n}}$

Your interpretation is the 2nd one which explains the proportion of variance captured.

If you divide the $1^{st}$ PC by the sum of all EVs, you will get the proportion of variance captured.

Hope this helps!

Vishva Shah
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