2
$\begingroup$

We also can consider the question in view of linear algebra. Using the language of matrices,it is that:

Given two matrices $M_1\in R^{m\times n_1}$ and $M_2\in R^{m\times n_2}$, where $m>n_1>n_2$ and their columns are linearly independent respectively. If there exist two positive vectors $\alpha\in R^{n_1}$, $\alpha_i>0$, and $\beta\in R^{n_2}$, $\beta_i>0$, s.t. $$M_1 \alpha=M_2 \beta,$$ Then can we find one more matrix $N\in R^{n_1\times n_2}$ makes $M_1N=M_2$ hold?

$\endgroup$
1
  • $\begingroup$ I believe we can. Any $N$ mapping $\beta$ to $\alpha$ should do it. Maybe as an example you can look at $f(x,y) = x+y$ and $g(x) = 2x$. That should tell you something. $\endgroup$
    – Vishesh
    Commented Oct 10, 2015 at 8:19

1 Answer 1

1
$\begingroup$

No. Counterexample: $$M_1 = \begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & 0\\ 0 & 0\\ 1 & 0\\ -1 & 1 \end{pmatrix}, \alpha = \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, \beta = \begin{pmatrix}1 \\ 1\end{pmatrix}$$ Clearly there's no $N$ so that $M_1N=M_2$, as $M_1N$ has the last line identically zero for any $N$.

$\endgroup$
3
  • $\begingroup$ I assume all $\alpha_i$ and $\beta_j$ are nonzero. $\endgroup$
    – peter
    Commented Oct 10, 2015 at 20:17
  • $\begingroup$ @peter: See edit $\endgroup$
    – celtschk
    Commented Oct 10, 2015 at 20:25
  • $\begingroup$ You are right. THX! $\endgroup$
    – peter
    Commented Oct 10, 2015 at 20:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .