Prove that if each row of a matrix sums to zero, then it has no inverse.

Question

Could anyone help me with this proof without using determinant? I tried two ways.

Let $A$ be a matrix. If $A$ has the property that each row sums to zero, then there does not exist any matrix $X$ such that $AX=I$, where $I$ denotes the identity matrix.

I then get stuck. The other way was to prove by contradiction, and I failed too.

Answer 1

Hint: You can sum the elements of a row by multiplying this row with a vector of $1$'s. Can you find now a matrix $X$ (with appropriate columns) such that $AX=Ο$?

Answer 2

If the sum of the rows is zero, then the matrix has the eigenvalue $0$. As a result its $\ker$ is of dimension $\ge1$, i.e. there is a nonzero solution to $AX=0$, hence it's noninvertible

Answer 3

If each row of $A$ sums to zero, then each row of the column vector that is the sum of the column vectors constituting $A$ is zero. So the columns of $A$ are not linearly independent, and therefore the matrix is singular (i.e. it has no inverse).

Answer 4

Hint: For a matrix $A$ having such a property has vector $(1,1,...,1)$ in its kernel thereby giving $\dim(\ker A)>0$.

Answer 5

Assuming $A$ is an $n \times n$ matrix, let $v_0 = (0,0,\dots,0)$ and $v_1 = (1,1,\dots,1)$ be $n$-element column vectors. Since each row of $A$ sums to zero, it follows that $$A v_1 = (0,0,\dots,0) = A v_0,$$ showing that $A$ cannot have a (left) inverse.