SVD says $X = U \Sigma V^\top$ and it is used for PCA (Principal Component Analysis).
In my understanding, Principal Components are eigenvectors of $X$. If so, where are eigenvectors? Is it in $U$ or $V$?
Suppose there is a matrix $X$ of shape $(N, D) = (1000, 784)$ having $1000$ images and $D$ is the image size $(28 \times 28)$, from where can I get the eigenvectors of $X$?
Alternatively, if the matrix is shaped as $(D, N) = (784, 1000)$ as in Singular Value Decomposition (SVD): Matrix Approximation to hold the same $1000$ images, from where can I get the eigenvectors of $X$? In the video, $X$ is sequence of face images where $X_i$ is a face image.
Principal components are not eigenvectors of $X$, but of $X^T X$ (it doesn't even make sense to talk of eigenvalues of non-square matrices).
From your SVD you can compute $$ X^TX = (U\Sigma V^T)^TU \Sigma V^T = V \Sigma U^T U \Sigma V^T = V \Sigma^2 V^T,$$ so the principal compontents are the columns of $V$.