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

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.

• Hint: what is the relation of the row space of $\mathbf A$ and the column space of $\mathbf A^{\mathrm T}$?
– xbh
Oct 23, 2018 at 10:04
• They are the same Oct 23, 2018 at 10:08
• If $\mathbf A ^{\mathrm T} \mathbf A \mathbf v = \mathbf A^{\mathrm T}(\mathbf {Av}) = c \mathbf v$, then where does $\mathbf v$ belong now?
– xbh
Oct 23, 2018 at 10:12
• Oh, I see, $\mathbf{v}$ belong to row space of $\mathbf{A}$ and column space of $\mathbf{A}^T$. Thank you! Oct 23, 2018 at 10:26
• You are welcome. Glad to help!
– xbh
Oct 23, 2018 at 10:37

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 $$\mathbf{A}^T$$.
• This assumes $c \ne 0$. Though I guess the question is only for non zero eigenvalued eigenvectors Oct 5, 2020 at 8:25
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 $$\mathbf A^T$$.