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$A$ is an $n\times m$ matrix and $AA^{T}$ is a symmetric real matrix. Also, we have: $\operatorname{rank}(AA^{T})=r\stackrel{?}{=}\operatorname{rank}(A)$. Let $Q= \begin{Bmatrix} q_1,...,q_{n-r} \end{Bmatrix}$ be a basis for the Null space of $AA^{T}$. i.e. $AA^{T}q_i=0$, show that $A^{T}q_i=0$. I guess one proof can be that the Null space for $A^{T}$ is a subspace for Null space for $AA^{T}$, then the question would be why $\operatorname{rank}(A)=\operatorname{rank}(AA^{T})$?

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up vote 12 down vote accepted

Let $q_i$ be a null vector of $A A^\top$, i.e. $ A A^\top q_i =0 $, then $ 0 = q_i^\top A A^\top q_i = \vert\vert A^\top q_i \vert\vert_2$, and thus $q_i$ is also a null vector of $A^\top$. Thus $\mathrm{rank}(A A^\top) = \mathrm{rank}(A^\top) = \mathrm{rank}(A)$.

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I like above answer; it says only what it needed. However, I tried the following. It takes a slightly different approach as it does not rely on $||A^T q_i$$||$ properties.

Let $X = Null(A)$, then $\forall x \in X, Ax = 0$. Assume that $Y = Null(A^TA)$. $\forall y \in Y, A^TAy = 0$. This implies,

  1. $Ay= 0$; or
  2. $A^TAy=0$ and $Ay \not= 0$

If case (1) is true than we are done: $y$ is in $X$. Now

$A^TAy = 0 \implies y^TA^TA = 0^T \implies y^TA^T = 0^T$ ($A$ is non-zero) $\implies Ay = 0$. And this is same as case (1). Thus $X = Y$.

Now one can argue about their ranks.

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?? I don't think that you can conclude from $y^TA^TA=O^T$ that you also have $y^TA^T=0$ based on $A\neq0$ alone. After all, $y^TA^T$ could be in the (left-sided) null space of $A$. – Jyrki Lahtonen Mar 8 '12 at 8:23
@JyrkiLahtonen Thanks for pointing it out. Seems like there is no way without bringing in $y^TA^TAy = 0^T$ and then arguing about $|| . ||$? – Dilawar Mar 8 '12 at 12:09

I have found this proof on the Web. I find it easy to read, and I have not found anything clearer on the waters.

We wish to prove that $null(A^TA) \subset null(A)$ Take $x\in R^n$ such that $A^T Ax=0$. Then, as $x$ is orthogonal to every row vector of $A^T A$, and since $A^T A$ is symmetric, then x is orthogonal to every column vectors of $A^TA$.

Thus, $x^T A^T Ax=0$ $\implies$ $(Ax)^T Ax=0$. This implies $Ax.Ax=0$ and $Ax=0$. Which finishes the proof.

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