0
$\begingroup$

Let the least square problem be $Ax=b$. If the vector $v$ satisfies $A^*v=0$, why does adding a multiple of $v$ to the right hand side $b$ does not change the solution, e.g., $Ax=b+tv$, for $t$ a constant?

Thanks.

  • 1
    I'm not sure the question makes sense. If $A$ is an $m\times n$ matrix, then $x$ and $v$ are $n\times 1$, $b$ is $m\times1$, and $b+tv$ doesn't make sense (unless $m=n$ --- did you mean to assume $A$ is square?).2012-11-02
  • 0
    Ah, I see, $A^*$ is being used as notation for the transpose of $A$. Please ignore previous comment.2012-11-02
  • 0
    Caught me for a while too...2012-11-02

1 Answers 1