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.

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).
