I'm trying to understand the proof of a problem, but I'm stuck. In my book they consider that if all lines of a matrix has sum 0 then it's determinant is also 0. I checked some random examples and it's true, but I couldn't proof it. Could you help me?


5 Answers 5


Let your matrix be called $A$. Then set $x$ to be the column vector of all 1's. You have $$Ax=0=0x$$ Hence $x$ is an eigenvector of $A$, while $0$ is an eigenvalue. Since the determinant is the product of the eigenvalues, and one of those is zero, the determinant must be zero.


It doesn't change the determinant of a matrix if we take a column and replace it by the sum of itself and (any scalar multiple of) another column in the matrix.

So take the last column and replace it by the sum of itself with the first column. Next, replace it by the sum of itself with the second column. Continue. Repeating this process till you get to the next-to-last-column, you see that you have replaced each entry the original last column with the sum of all the entries in that row, which is zero. So now the entire last column is zero, and the determinant must be zero.


Hint: The vector $$\begin{bmatrix}1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}$$ is an eigenvector with eigenvalue __?

  • $\begingroup$ I think I would have hinted the opposite side: that multiplication by this vector encodes the information given. $\endgroup$
    – user14972
    Jul 26, 2015 at 4:14
  • 2
    $\begingroup$ True, but using eigenvalues is killing a fly with a jackhammer anyways. The argument about the linear dependence of the columns is much more elegant. $\endgroup$
    – user217285
    Jul 26, 2015 at 4:18
  • $\begingroup$ Up-voted your answer and comment ;) $\endgroup$
    – Carsten S
    Jul 26, 2015 at 11:00

Another way to derive this result:

When the sum of all rows are zero, the row vectors are linearly dependent. Hence the determinant is zero.

In fact, if any two or three or four rows add up to pure zero, the determinant is zero as well.

  • $\begingroup$ This answer was my first thought, and I like it. Specifically, each row is contained in the hyperplane $x_1 + x_2 + \ldots x_n = 0$, hence they cannot form a basis for the space. $\endgroup$
    – pjs36
    Jul 25, 2015 at 16:11

Geometrically speaking, the determinate calculates the area, volume, or hyper-volume of a trapezoid, parallelepiped, of hyper-parallelepiped in $\mathbb{R}^2$, $\mathbb{R}^2$, or $\mathbb{R}^n$ (respectively). A row full of zeros in a $n\times n$ matrix implies that the hyper-parallelepiped is "missing" a dimension that is needed in order to compute the hyper-volume; thereby causing the determinate to become zero.

To better visualize this explanation, visualize how a trapezoid (a two dimensional object) can be represented by a $3\times 3$ matrix. Since a trapezoid does not have a volume, then the determinate must be zero.

Algebraically speaking, apply the technique of co-factor expansion.

  • 1
    $\begingroup$ I don't understand how this answers the question. You seem to be interpreting OP's question to mean that there is a row of all 0s. $\endgroup$
    – Samuel
    Jul 26, 2015 at 21:02

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