# PCA, why variance of eigen values is measure of its utility?

Source - Murphy, 12.3

Heuristic for assessing applicability of PCA.

Let the empirical covariance matrix Σ have eigenvalues λ1≥λ2≥···≥λd>0, with mean λ. Explain why the variance of the eigen values, σ2=(1/d)∑(λi−λ)2 is a good measure of whether or not PCA would be useful for analysing the data (the higher the value of σ2 the more useful PCA)

In a more mathematical way, if we denote by $$\sigma_{tot}^2$$ the total variance over the dataset (not the variance of eigenvalues), we know that the mean of eigenvalues will be $$\frac{\sigma_{tot}^2}{n}$$. As all eigenvalues are non-negative (property of the covariance matrix), and based on the trace conservation property, we deduce that all eigenvalues are lower or equal to $$\sigma_{tot}^2$$. Then, the convexity of the square function implies that maximum variance among eigenvalues is reached when one of the eigenvalues is equal to $$\sigma_{tot}^2$$, and all other are null. Hence, eigenvalues maximum variance is reached when the first component wears the total dataset variance, which is obviously the optimal PCA case (all features reduced to a single dimension).