Suppose that each of the row sums of an $n\times n$ matrix $A$ is equal to zero. Show that $A$ must be singular.

Suppose that each of the row sums of an $n\times n$ matrix $A$ is equal to zero. Show that $A$ must be singular.

I can show this for $2\times2$ by using the fact that $det(A)=0$, whenever $A$ is singular. But do not know how to make it valid of for all square matrices.

Thanks!

Let $A$ be such a $n\times n$ matrix and let $\textbf{1}_n$ be the $n\times1$ vector with each element $1$. Then $$A\textbf{1}_n=0$$ so $\textbf{1}_n$ belongs to null space of $A$. So $rank(A)\leq n-1$. So $A$ is singular.

Adding the remaining $n-1$ columns to the first column does not change the value of the determinant of $A$. However since all the row sums are equal to $0$, the first column has all entries equal to $0$ in the new matrix. Expanding the determinant in the first column, we see that the value of the determinant is $0$ so $\det(A)=0$. Hence $A$ is singular.

Row operations preserve the property that rows sum to zero.

If the matrix is non-singular, then it is row equivalent to the identity matrix.

However, the rows in the identity matrix do not sum to zero. Thus, the matrix must be singular.