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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.

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

2 Answers 2

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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 $\mathbf{A}^T$.

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

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  • $\begingroup$ Your answer would be a fine alternative to the accepted answer, except for the pointless opening remark. $\endgroup$
    – Lee Mosher
    Commented May 11, 2023 at 17:12

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