Let $A$ be a $m*n$ martrix and suppose $v$ is an eigenvector of $AA^T$ with eigenvalue $\lambda\ne0$.
a) show that $u=A^{T}v$ cannot be the zero vector.
b) show that $\lambda$ is also an eigenvalue of $A^{T}A$. (Hint: consider $u$ given in (a).)
This is what i have for part (a): $$(AA^T)v=\lambda v$$ $$A^{-1}AA^Tv=\lambda A^{-1}v$$ $$A^Tv=\lambda A^{-1}v $$
Now, $\lambda\ne0$ is given, and $A^{-1}$ and $v$ cannot be zero.
Then, $A^Tv$ is not a zero vector.
How do i prove part (b)?
You are given that $\mathbf{v}$ is an eigenvector of $AA^T$ with eigenvalue $\lambda\ne 0$. Rephrase using the definition of what it means to be an eigenvector. $$AA^T\mathbf{v}=\lambda\mathbf{v}$$ for some $\mathbf{v}\ne\mathbf{0}$.
For part (a) ask yourself what would happen if $A^T\mathbf{v}=\mathbf{0}$ and why this is impossible.
For part (b) again use the definition. If $\lambda$ is an eigenvalue of $A^TA$ then there must exist a nonzero $\mathbf{x}$ such that $$A^TA\mathbf{x}=\lambda\mathbf{x}.$$ The hint suggests what $\mathbf{x}$ you might try.