Let $A$ be an $10\times 10$ invertible matrix with real entries such that the sum of each row is $1$. Then

$A$. The sum of entries of each row of the inverse of $A$ is $1$

$B$. The sum of entries of each column of the inverse of $A$ is $1$

$C$. The trace of the inverse of $A$ is non-zero

$D$. None of the above

I don't see how to proceed or how to use the sum of the elements of of rows. Please help.

I know option $A$ is correct and option $C$ is not.
Why is option $B$ not correct?


Hint: Let $x = (1,\dots,1)^T$ (the column vector with ten $1$s). The fact that the row sum of each matrix of $A$ is $1$ is the same as saying $Ax = x$.

What can we say about $A^{-1}x$?

The statement B amounts to saying that $(A^{-1})^Tx = x$ (where $A^T$ is the transpose of $A$). Note that, in general, $A$ and $A^T$ will not have the same eigenvectors.

In particular, consider the example $$ A = \pmatrix{1/2 & 1/2\\0&1} \implies\\ A^{-1} = \pmatrix{2&-1\\0&1} $$ We indeed have $A^{-1}x = x$, but not $(A^{-1})^Tx = x$.

  • $\begingroup$ $A^{-1} x$=$x$. $1$ is an eigenvalue of $A$ . What next ? Sorry I still cannot get it. $\endgroup$ – user118494 Aug 24 '15 at 16:04
  • 1
    $\begingroup$ Oh I got it . Option $A$ is correct . $\endgroup$ – user118494 Aug 24 '15 at 16:07
  • $\begingroup$ See my latest edit. Note that the "@" notification system doesn't work in answers. $\endgroup$ – Omnomnomnom Aug 24 '15 at 16:30

$C$ is false, by the way. A $2\times 2$ counterexample is $$A = \left(\begin{array}{cc} 0 & 1\\ 1 & 0\end{array}\right)$$Notice that $A^{-1} = A$ has trace zero. I'll let you come up with a similar $10 \times 10$ example.


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