5
$\begingroup$

Lately, I´ve been struggling with math homework and came across a question I´m not sure how to answer. I will be glad for any help...

Suppose we have matrix $A$ (size $n\times n$) and its inverse (lets call it $B$). They are both non-negative in the sense that all their elements $A_{ij}$ and $B_{ij}\geq 0$, where $1\leq i,j\leq n$.

The question is, what can we say about these matrices - everything must be justified.

This where I got so far: 1) $A$ is regular (otherwise it wouldn't have and inverse - I don't think I have to justify this statement)

Are there any other features? I think I can justify some of them by using minor matrices, but I'm not sure how :-(

$\endgroup$
  • $\begingroup$ Welcome to math.stackexchange! I just improved tex formatting in your post. $\endgroup$ – Davide Giraudo Oct 6 '12 at 15:41
  • $\begingroup$ At least, when $n=2$ we can say that either $A$ is diagonal or has the form $\pmatrix{0&a\\a'&0}$ where $a>0$, $a'>0$. $\endgroup$ – Davide Giraudo Oct 6 '12 at 15:50
2
$\begingroup$

You know that $AB= I_n$.

Since for all $i \neq j$ you have $\sum_k A_{ik}B_{kj}=0$ it follows that for all $i,j,k$ with $i \neq j$ you have either $A_{ik}=0$ or$B_{kj}=0$.

Now, since $\sum_k A_{ik}B_{ki} \neq 0$, for each $i$ you can find some $k_i$ so that $A_{ik_i} \neq 0$ and $B_{k_i i} \neq 0$. Combining this with the above, you can prove that $B_{k_i j}=0 \forall j\neq i$ and $A_{j k_i}=0$ for all $j \neq i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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