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 $\paren {i,j}$ in $\det\paren A$, $1\le i,j \le n$.

Then:


 * $\displaystyle \det \paren {A -J_n} = \det \paren A - \sum_{i \mathop = 1}^n \sum_{j \mathop = 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:


 * $\displaystyle \det\paren {A -J_n} = \det\paren A - \sum_{j \mathop = 1}^n \det\paren { P_j }$

Expansion of left side $\det\paren { A - J_n}$ for the $2\times 2$ case illustrates how determinant theorems will be used:

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

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:


 * $\displaystyle \det\paren { A -J_n } = \det\paren { A -Q_m }- \sum_{j \mathop = 1}^m \det\paren { P_j }$, $1\le m \le n$

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:


 * $\displaystyle \det\paren { A - J_n } = \det\paren { A }- \sum_{j \mathop = 1}^n \det\paren { P_j }$