Why eigenvectors of $\mathbf{A}^T\mathbf{A}$ are in row space of $\mathbf{A}$?

Question

I'm following SVD proof and I can't get why eigenvectors of $\mathbf{A}^T\mathbf{A}$ are in rowspace of $\mathbf{A}$. I can understand further why these eigenvectors are basis for the row space of $\mathbf{A}$, if they are in the row space of $\mathbf{A}$, but I miss this very step of proof about belonging to the rowspace of $\mathbf{A}$. I know it must be very simple, so the books I use don't specify it or specified it earlier in the book.

Answer 1

Posting the answer of @xbh given in the comments above:

If $\mathbf{A}^T\mathbf{A}\mathbf{v} = \mathbf{A}^T(\mathbf{A}\mathbf{v}) = c\mathbf{v}$ then $\mathbf{v}$ belongs to the row space of $\mathbf{A}$ and the column space of $\mathbf{A}^T$.

Answer 2

Let $\mathbf A$ be a matrix; then $\forall \boldsymbol x$, $\mathbf A \boldsymbol x$ is in the column space of $\mathbf A$. Thus $\mathbf A^T (\mathbf A \boldsymbol v)$ is in the column space of $\mathbf A^T$. But $\mathbf A^T \mathbf A \boldsymbol v = c \boldsymbol v$. Hence $c \boldsymbol v$ (or $\boldsymbol v$ if $c \ne 0$) is also in the column space of $\mathbf A^T$.