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If $A$ is matrix from $M_{n-1,n}$ such that sum of all elements in row is zero.

Prove det$(AA^T)=nk^2$, for some $k\in\mathbb{Z}$.

I tried to solve it, but I’m really confused.

I know that $A[1, 1, \dots, 1]^t= 0 * [1, 1,\dots, 1]^t$

So zero is eigenvalue of $A$, so it's singular and $\det A=0$.

So $\det(AA^{T})=\det(A)\det(A^T)=\det(A)\det(A)=\det(A^2)=0$

Is this correct? Is there any other way to solve this? Thank you in advance.

user10354138
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techno
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    It's not a square matrix so how can it have eigenvalues? – Jyrki Lahtonen May 31 '19 at 16:11
  • Yeah, you are right. det(AB)=det(A)det(B) also doesn't work. I don't know how to start solving this. – techno May 31 '19 at 16:35
  • See https://math.stackexchange.com/questions/1819118/we-have-matrix-a-in-m-n-1-times-n-mathbb-z-so-that-the-sum-of-entries-in-e?rq=1 (funnily, the OP there also tried eigenvalues). – darij grinberg May 31 '19 at 16:40
  • Thank you @darijgrinberg. I saw it in related field few minutes after i posted this. Sorry for spaming. In the class we don't study Cauchy Binet theorem. I don't know is there any way to do it without using this teorem? ps:Thank you for response. – techno May 31 '19 at 19:45

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