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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?)

Vincent Tjeng
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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
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    @Taladris: more directly, it means that the rows are linearly dependent: the bottom one is minus the sum of the others. –  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

1 Answers1

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Yes, the determinant must be zero.

Proof 1: Since the sum of each column is zero, the sum of all the rows is the zero vector. Hence the rows are not linearly independent, and the determinant is zero.

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).

Caleb Stanford
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