rank of a matrix if one component of each row is a linear combination of the others I am sure that this problem is really simple but now I can not see the solution. Let $A$ be a matrix $m\times m$ with full rank. Then, we change all the components $A(j,j)$ by $-\sum_{k,k\neq j} A(j,k)$, where $j$ is the row number. I have made simulations with random matrix and after the change the rank of the new matrix is $m-1$. Why?. I see that in every row we have a term that is a linear combination of the term from the other rows. But I do not see how to demonstrate the reduction of the rank.
 A: Denote column $i$ by $A_i$. The matrix has full rank, so it's columns are linearly independent. Namely $ \sum_i A_i \neq 0$, where $0$ is the zero vector.
Now for every diagonal entry $A(j,j)$, replace it with negative the sum of all other entries in that row, call this new matrix $B$.  Then clearly $ \sum_i B_i = 0$ (can you see why?).  This means that the columns are no longer linearly independent.  In other words, $A$ has reduced in rank when it becomes $B$:
$$ rank(B) < m . $$
To compute the rank exactly, we apply elementary column operations to kill the last column. Recall that the rank of a matrix is unchanged by elementary operations on rows or columns:
$$ \begin{cases} \mbox{Add column } 1 \mbox{ to column } m \\
 \mbox{Add column } 2 \mbox{ to column } m \\
 \dots \\
\mbox{Add column } m-1 \mbox{ to column } m \end{cases}$$
Check yourself that $B$ now has columns $B_1 , B_2 , B_{m-1}, 0 $.  But the first $m-1$ columns are linearly independent, since the columns of $A$ were linearly independent. To see this, assume there were coefficients $\beta_1, ..., \beta_{m-1}$, not all $0$, such that
$$ \sum_{i=1}^{m-1} \beta_i B_i = 0 .$$
This would mean that the first row of $B$ sums to $0$:
$$ \beta_1 B(1,1)  + \beta_2 B(1,2) + \dots + \beta_{m-1} B(1, m-1) = 0  $$
$$ \implies - \beta_1 \left( \sum_{k=2}^{m} A(1,k) \right)  + \beta_2 A(1,2) + \dots + \beta_{m-1} A(1, m-1) = 0  $$
$$ \implies  - \beta_1 A(1,m) + \sum_{k=1}^{m-1} \beta_k A(1,k) = 0 $$
By full rankness of $A$, this last equality is true if and only if the coefficients were all $0$, a contradiction.
Therefore the first $m-1$ columns of $B$ are linearly independent. Namely,
$$ rank(B)=m-1 .$$
