number and size of eigenvectors in PCA As I understand, the size of eigenvector produced in PCA should be min{n,N}, where N=number of samples and n=dimension of each sample (Right?). However, I have seen in couple of cases that this size is max{n,N}, not min. Why?
And the next relevant question is,
Are number of produced eigenvectors equal to their size (I know we pick the first couple of them in PCA, but I am talking about the produced ones)? It should be, as the number of eigenvectors is always equal to size of each eigenvector.
 A: One of the ways to compute PCA is by using eigen-decomposition of the covariance matrix of the data matrix $\underset{N \times n}{X}$, which is $\underset{n\times n}{X^TX}$ if $X$ is mean-centralized. Since we generally have $n \ll N$, it's computationally much faster to compute the eigenvectors of the covariance matrix $X^TX$. It outputs $n$ eigenvalues and an $n \times n$ matrix with $n$ orthonormal eigenvectors as columns where the dimension of each eigenvector is $n \times 1$.
But, sometimes we have $N \ll n$ (e.g., for patient-genes data), then it's computationally very expensive to do the eigen-decomposition of $X^TX$, instead it's easier to do the eigen-decomposition of $\underset{N \times N}{XX^T}$, since both have same eigenvalues and the eigenvectors of $X^TX$ can be computed easily from the eigenvectors of $XX^T$.
For example, let $XX^T$ has eigenvalue $λ$ and the corresponding eigenvector $v$, then we have,
$XX^Tv=λv⟹(X^TX)(X^Tv)=λ(X^Tv)$, i.e, $\underset{n\times N}{X^T}.\underset{N \times 1}{v}$ is an eigenvector of $X^TX$ (upto normalization).
