Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$A$ 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 $\operatorname{rank}(A)=\operatorname{rank}(AA^{T})$?

share|improve this question
add comment

2 Answers

up vote 11 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 $\mathrm{rank}(A A^\top) = \mathrm{rank}(A^\top) = \mathrm{rank}(A)$.

share|improve this answer
add comment

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.

share|improve this answer
    
?? 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.