TITLE: Relation between the eigenvectors of $A$ and $A^TA$
QUESTION [8 upvotes]: Is there a relation between the eigenvectors of matrix $A$ and of matrix $A^TA$? 
This question is related to eigenvectors, not eigenvalues. Further, the sizes of matrices $A$ and $A^TA$ are different if $A$ is not a square matrix. Thus, my question has no relation to the question about eigenvalues of $A$ ans $A^T$.

REPLY [3 votes]: There is no simple relation between eigenvectors of $A$ and of $A^*A$ unless $A$ is normal. For example, if 
$$
A=\pmatrix{0&1\\0&0},
$$
$A$ has only one dimensional eigenvector subspace spanned by $[1,0]^T$, but $A^*A=0$ so any nonzero 2-vector is its eigenvector. Even if $A$ is diagonalizable, the eigenspaces of $A$ and $A^*A$ can be completely different.
If $A$ is normal, then $A=UDU^*$ for a unitary $U$ and diagonal $D$, so $A^*A=U|D|^2U^*$ is the spectral decomposition of $A^*A$. 
However, there is a simple relation between left singular vectors of $A$ and the eigenvectors of $A^*A$ as indicated in another answer.