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If the row sums of a symmetric matrix of size 4 by 4 are all $0$, then why are all the cofactors of the matrix equal?

Thanks in advance for any helpful answers.

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While googling I found this article at www.maths.qmul.ac.uk/~pjc/odds/zero.pdf –  pritam May 29 '13 at 19:54

2 Answers 2

There is no reason to consider a $4\times 4$ matrix in particular. When $n=1$, there is only one cofactor so nothing to prove. The result is easy to figure out for $2\times 2$ matrices, and that's maybe a good starting point.

Then one realizes that more generally, as observed by user1551, it suffices that $A$ be an $n\times n$ matrix over an arbitrary field such that the row sums and the column sums be null (the latter being equivalent to the former for symmetric matrices).

In short: we use the adjugate formula $A\cdot\mbox{adj}(A)=(\det A)I_n$ to prove that the rows and the columns of $C=\mbox{adj}(A)^T$, the matrix of cofactors, are all in the span of $(1,\ldots,1)$, which is the nullspace of $A$ and $A^T$ in the only nontrivial case.

Proof: the row condition means that the sum of the columns of $A$ is null. Whence the columns the columns are linearly dependent and $\mbox{rank} A\leq n-1$. In particular, $\det A=0$. Let $C$ be the matrix of the cofactors of $A$. Then the transpose $C^T$ is the adjugate of $A$ and, as is well-know $AC^T=(\det A)I_n$. Therefore $$ AC^T=0\quad\iff \quad \sum_{k=1}^na_{ik}c_{jk}=0\quad \forall 1\leq i,j\leq n. $$

If $\mbox{rank} A\leq n-2$, then all the cofactors are zero and the result follows. So we assume that $\mbox{rank}\, A=n-1$. That is, the nullspace of $A$ has dimension $1$. But we already know that $v=(1,\ldots,1)$ is in $\ker A$ by assumption. So $\ker A$ is the one-dimensional span of $v$. Note that $\ker A^T$ also has dimension $1$ and is spanned by $v$ as-well by assumption.

Now looking at the above, we see that it means that every row $R_j$ of $C$ is in the nullspace of $A$, whence $R_j=r_jv=(r_j,\ldots,r_j)$.

Since the adjugate matrix of $A^T$ is the transpose of the adjugate matrix of $A$, we get $A^TC=0$ (or simply by taking the tranpose of $C^TA=0$). From that, we deduce that the columns $C_j$ of $C$ are of the form $C_j=c_jv=(c_j,\ldots,c_j)$.

Starting from $c=c_{1j}$ for every $j$ (first row), we see that $c_{ij}=c$ ($j$th column) for every $i$. That is, all the coefficients of the cofactor matrix are equal as desired. QED.

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An elegant proof. +1 –  user1551 May 29 '13 at 23:16

We can prove something more general:

If $A$ is an $n\times n$ matrix over any field such that all its row sums and column sums are zero, then its cofactors are identical to each other. That is, all entries of $\operatorname{adj}(A)$ are equal.

Denote by $m_{ij}$ the $(i,j)$-th minor of $A$ and the corresponding submatrix $M_{ij}$ (hence $M_{ij}$ is obtained by deleting the $i$-th row and $j$-th column of $A$ and $m_{ij}=\det M_{ij}$). If $i<n$, add the first $n-2$ rows of $M_{ij}$ to the last row. Since all column sums of $A$ are equal to $0$, the last row becomes $(-a_{11}, -a_{12}, \ldots, -a_{i,j-1}, -a_{i,j+1}, \ldots, a_{in})$. By interchanging rows, move this row up to row $i$, while maintaining the order other rows (there are $n-i$ interchanges in total). Then pull out the $-1$ from this new row $i$. So, we see that $$m_{ij} = (-1)^{n-i+1}m_{nj}.\tag{1}$$ By similar reasoning, but using column operations, we have $m_{ij} = (-1)^{n-j+1}m_{in}$. Put $i=n$ and we get $$m_{nj} = (-1)^{n-j+1}m_{nn}.\tag{2}$$ Combine $(1)$ and $(2)$, we have $m_{ij}=(-1)^{i+j}m_{nn}$, i.e. $(-1)^{i+j}m_{ij}=m_{nn}$. Hence the result.

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Thank you. I've never been very talented with row/column operations. +1 for doing it like this. –  1015 May 29 '13 at 23:31

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