Is having the sum of each column zero a sufficient condition for a matrix to be singular?

If I have a square matrix,

$$\begin{pmatrix} a_{1,1} & \ldots & a_{1,n} \\ \vdots & \ddots & \vdots \\ a_{n,1} & \ldots & a_{n,n} \\ \end{pmatrix}$$

Is a sufficient condition for the matrix to have determinant zero that

$$\sum_{i=1}^n a_{i, k}=0 \space\forall\space k$$

If so, can we produce a clear proof why?

(If not, can we make it a sufficient condition by strengthening the condition to include that only the main diagonal contains negative terms?)

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My inspiration for this question was my observation that the transition matrices for Markov processes that I was working with always had an eigenvalue 1. I'd be happy to see counter-examples, though - I'm not 100% sure that my conjecture is correct. – Vincent Tjeng Apr 28 '14 at 15:51
Yes, because it means that the columns are linearly dependant. – Taladris Apr 28 '14 at 15:52
@Taladris: more directly, it means that the rows are linearly dependent: the bottom one is minus the sum of the others. – user64687 Apr 28 '14 at 15:52
@AsalBeagDubh: Yes, you're right. I read "the sum of columns is zero" instead of "the sum of each column". – Taladris Apr 28 '14 at 22:51

Proof 2: Notice that the transpose of the matrix ($A^T$) sends the vector $v = (1,1,1,\ldots,1)$ to the zero vector. Therefore, $A^Tv = 0v$, and $0$ is an eigenvalue of $A^T$. Since the determinant is the product of the eigenvalues, the determinant of $A^T$ (and hence the determinant of $A$) is zero.
Proof 3: Induction on $n$. The base case is trivial. For the inductive step, first add the top row of the matrix $A$ to the second row, making it so that the sum of the $2$nd through $n$th values in any column are zero. Then evaluate the determinant by row-reduction across the top row, and notice that each cofactor is zero since the $n-1 \times n-1$ submatrix has determinant zero (by the induction hypothesis).
Are you sure it's $Av = 0$ and not $vA = 0$? Just take the transpose anyway. – Najib Idrissi Apr 28 '14 at 15:56
Ah good point. ${}{}$ – 6005 Apr 28 '14 at 15:56