As has been stated in this thread:
It's not hard to show that if the covariance matrix of the original data points $x_i$ was $\Sigma$, the variance of the new data points is just $u^T \Sigma u$.
However, below it has been proven that the variance isn't equal to $\Sigma$, because we haven't divided $A$ by the number of points. Then $A$ is obviously a different matrix than the covariance matrix, thus it has different eigenvectors and eigenvalues than covariance matrix.
Most sources say we should take eigenvectors of $\Sigma$ as the new basis, but here we're taking eigenvectors of $A$. I'm a bit confused...