I'd like to prove following:

$\det(M+U)=m·s $ where $m$ is a cofactor of $M$ and s is the sum of all entries in $U$ where $M$ is square matrix and sum of each row in $M$ is zero and $U$ is a square matrix with rank $1$

It would've been better if the $\det $ operation is multiplicative, but it looks not.

I think this proof start from the analogy of the case where we usually add the matrix $J$ where its all entries are $1$. $J$ also has rank 1. I'd like to use this fact so first prove $\det(M+J)=m\cdot n^2$ and then generalize it to show $\det(M+U)=m·s$.

  1. Is it true that $\det(M+J)=m\cdot n^2?$
  2. If not, any other approach to prove it?

First, a remark on something you implicitly stated: the hypothesis that $M$ has all its row sum zero implies that all entries of $\mbox{adj}(M)$ are equal (i.e. all its cofactors are equal).

Then as $U$ is rank $1$, there exists two column vectors $u,v$ such that $U=v\cdot {}^t\!u$.

Finally we use the Matrix determinant lemma, which states that $$\det(M+U)=\det(M+v\cdot{}^t\!u)=\det(M)+{}^t\!v \cdot\mbox{adj}(M)\cdot u.$$

As $M$ is not invertible (all its row are in some hyperplane), and as all entries of $\mbox{adj}(M)$ are $m$, it is clear that $\det(M+U)=0+m\cdot {}^t\!v\cdot u=m\cdot s$.


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