Cofactor Sum Identity

THEOREM
Let $J_n$ be the $n\times n$ matrix of all ones.

Let $A$ be an $n\times n$ matrix.

Let $A_{ij}$ denote the cofactor of element $\left(i,j\right)$ in $\det\left( A\right)$, $1\le i,j \le n$. Then:


 * $\displaystyle \det\left(A -J_n\right) = \det\left(A\right) - \sum_{i=1}^n \sum_{j=1}^n A_{ij} $

Proof
Let $P_j$ equal matrix $A$ with column $j$ replaced by ones, $1\le j \le n$, then:

To complete the proof it suffices to prove the equivalent identity:


 * $ \det\left(A -J_n\right) = \det\left(A\right) - \sum_{j=1}^n \det\left( P_j\right)$

Expansion of left side $\det\left( A - J_n\right)$ for the $2\times 2$ case illustrates how determinant theorems will be used:

Let $A$ be $n\times n$.

Let matrix $Q_m$ equal ones matrix $J_n$ with zeros replacing all entries in columns $1$ to $m$.

Example:

Induction on $m$ will be applied to prove the induction identity:


 * $\det\left(A -J_n\right)=\det\left(A -Q_m\right)- \sum_{j=1}^m \det\left(P_j\right)$, $1\le m \le n$

Induction step $m=1$:

Induction step $m=k$ and $k<n$ implies $m=k+1$:

Conclusion: Matrix $A-Q_n$ equals $A$ because $Q_n$ is the zero matrix.

Let $m=n$ in the induction identity, then:


 * $\det\left(A -J_n\right)=\det\left(A \right)- \sum_{j=1}^n \det\left(P_j\right)$