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# Why the determinant of a matrix with the sum of each row's elements equal 0 is 0?

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?

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

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.

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# Why the determinant of a matrix with the sum of each row's elements equal 0 is 0?

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?

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

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. - Nitin Jul 26 at 4:18

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