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

Then:


 * $\ds \map \det {A -J_n} = \map \det 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 by the Expansion Theorem for Determinants applied to column $j$ of $P_j$:
 * $\ds \sum_{j \mathop = 1}^n \map \det {P_j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = 1}^n A_{ij}$

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


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

Expansion of $\map \det {A - J_n}$ 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 $m$.

For example, for $n = 5$ and $m = 2$:
 * $Q_2 = \begin {pmatrix}

0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ \end {pmatrix}$

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


 * $\ds \map \det {A - J_n} = \map \det {A - Q_m} - \sum_{j \mathop = 1}^m \map \det {P_j}$

for $1 \le m \le n$.


 * Induction step $m = 1$:


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

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

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


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