Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $a$
share|cite|improve this question
I couldn't find a title for the question! – Gigili Feb 15 '12 at 19:33
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
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
Observe $(z^T A^Tx)^T=(z^T 0)^T=0$ :) – David Mitra Feb 15 '12 at 19:46
up vote 2 down vote accepted

So there's an answer...

(1) is the correct choice.


share|cite|improve this answer
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

Your Answer


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.