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Suppose A is a $m \times n$ matrix and the vectors $x$ and $y$ are such that $Az=y$ for some vector $z$ and $A^T x=0$. Which one is correct?

  1. $x^Ty=0$
  2. $||x||_2=||y||_2$
  3. $||x||_2 < ||y||_2$
  4. $x=ay$ for some real values of $a$
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I couldn't find a title for the question! –  Gigili Feb 15 '12 at 19:33
2  
Write out $x^Ty$ using $Az=y$ and then remember that $(X^T)^T=X$ and $(XY)^T=Y^TX^T$ –  Bill Cook Feb 15 '12 at 19:34
1  
How about, for the title, "What does $Az=y$ and $A^Tx=0$ imply about the relationship between $x$ and $y$?" –  David Mitra Feb 15 '12 at 19:40
    
@BillCook: $x^Ty=x^T(Az)=((Az)^T x)^T=(z^T A^T x)^T$, what should I do next? –  Gigili Feb 15 '12 at 19:41
3  
Observe $(z^T A^Tx)^T=(z^T 0)^T=0$ :) –  David Mitra Feb 15 '12 at 19:46

1 Answer 1

up vote 2 down vote accepted

So there's an answer...

(1) is the correct choice.

$x^Ty=x^TAz=x^T(A^T)^Tz=(A^Tx)^Tz=0z=0$

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I'd like to know how (2), (3) and (4) are wrong, if it's possible for you. Thank you for your answer anyway. –  Gigili Feb 15 '12 at 19:56
    
Take an Identity matrix as A and x becomes a zero vector. Y could be anything you want. Thus, option 2 gets kicked out. Similarly, you can take an invertible matrix A with x,y,z all as zero vectors and still satisfy everything. Thus option 3 gets kicked out. Kicking out option 4 is a little taxing on my brain right now. –  Inquest Feb 15 '12 at 20:03
    
@Nunoxic: That's enough, thank you again. –  Gigili Feb 15 '12 at 20:08
    
All the rest of the parts can hold by picking the right choice of $A$, $z$, and $x$. They can also fail by picking the "wrong" choice. (1) is the only part which always holds. –  Bill Cook Feb 15 '12 at 20:40

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