What is the relationship between the eigenvectors of $A A^T$ and $A^TA$? I tried using the fact that the left eigenvectors and right eigenvectors will form a diagonal matrix, but cannot proceed ahead.
 A: I will use the Hermitian conjugate, $A^H$, here instead of the transpose, $A^T$, so the answer is more general.
The eigenvectors of $AA^H$ are called the left-singular vectors of $A$ and the eigenvectors of $A^HA$ are the right-singular vectors of $A$. They are called this way because of their use in singular value decomposition. Say $A = U\Sigma V^H$, then the columns of $U$ are the left-singular vectors and the columns of $V$ are right-singular vectors.
We know that the non-zero eigenvalues of $AA^H$ and $A^HA$ are the same, they are the squares of the singular values of $A$. A singular of value of $A$ is a value $\sigma$ such that
$$Av = \sigma u ~~~~ A^Hu = \sigma v$$
where $u$ is a left-singular vector and $v$ is a right-singular vector. This can be taken as the definition of singular value, left-singular vector (which is an eigenvector of $AA^H$) and right-singular vector (which is an eigenvector of $A^HA$).
You can check how the left-singular and right-singular vector pair works with $AA^H$ and $A^HA$:
$$AA^H u = \sigma A v = \sigma^2 u$$
$$A^HA v = \sigma A^H u = \sigma^2 v$$ 
A: Let $v$ be an eigenvector of $AA^H$, relative to the eigenvalue $\lambda\ne0$ (which is real). Then, by definition,
$$
AA^Hv=\lambda v
$$
so
$$
(A^H\!A)(A^H\!v)=\lambda A^Hv
$$
which means that $A^H\!v$ is an eigenvector of $A^H\!A$ relative to $\lambda$ (here we use that $\lambda\ne0$). In particular, every nonzero eigenvalue of $AA^H$ is also an eigenvalue of $A^H\!A$. With the obvious changes, this also proves that every nonzero eigenvalue of $A^H\!A$ is an eigenvalue of $AA^H$.
If $E(X;\lambda)$ denotes the eigenspace of $X$ relative to $\lambda$, we see also that we have an injective linear map
$$
E(AA^H;\lambda)\to E(A^H\!A;\lambda),\qquad x\mapsto A^H\!x
$$
Therefore $\dim E(AA^H;\lambda)\le \dim E(A^H\!A;\lambda)$ and the converse map
$$
E(A^H\!A;\lambda)\to E(AA^H;\lambda),\qquad x\mapsto Ax
$$
shows equality. Hence both maps are isomorphisms.
Nothing can be said about the zero eigenvalue (if $0$ is an eigenvalue of one of the matrices in the first place). However, since both matrices are diagonalizable and they have the same rank as $A$, we already know the dimension of the null space of both.
Note: $A^H$ denotes the Hermitian transpose; for real matrices, just use the transpose.
A: I'm not sure about my solution,... but I'll post it anyway. maybe it helps you ;)
First of all I asumed A is a square matrix (Maybe I don't need this,... decide for yourselfe :) )
assume that $\lambda$ is a eigenvalue for $A^TA$ with the eigenvector x.
Therefor $\left(A^TA-\lambda I \right)x = 0$ now we multiply both sides with A and get $A\left(A^TA-\lambda I \right)x = A\cdot 0 = 0 \Leftrightarrow \left(AA^TA-A\lambda I \right)x = 0$ now $A$ commutates with $\lambda I$ therefor we get $\left(AA^TA-\lambda I A \right)x = 0 \Leftrightarrow \left(AA^T-\lambda I\right)Ax = 0$ Now I would interprete this as the Eigenvector of $AA^T$ is A times the Eigenvector of $A^TA$ 
If you find any mistakes, please report them ;)
