Sum of Elements of Invertible Matrix

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Theorem

Let $\mathbf J_n$ be the $n \times n$ square ones matrix.

Let $\mathbf B$ be an $n\times n$ invertible matrix with entries $b_{i j}$, $1 \le i, j \le n$.

Then:

$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{i j} = 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}$

Proof

Lemma

Let$\mathbf J_n$ be the $n \times n$ square ones matrix.

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

Let $\mathbf A_{ij}$ denote the cofactor of element $a_{ij}$ in $\map \det {\mathbf A}$, $1 \le i, j \le n$.

Then:

\(\ds \map \det {\mathbf A - \mathbf J_n}\)

\(=\)

\(\ds \map \det {\mathbf A} - \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{i j}\)

Cofactor Sum Identity

$\Box$

Let $\mathbf A = \mathbf B^{-1}$.

Then:

\(\ds \mathbf A \mathbf B\)

\(=\)

\(\ds \mathbf I\)

Definition of Invertible Matrix

\(\ds \mathbf B\)

\(=\)

\(\ds \dfrac {\adj {\mathbf A} } {\map \det {\mathbf A} }\)

Matrix Product with Adjugate Matrix with $\mathbf B = \mathbf A^{-1}$

Then:

\(\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n b_{ij}\)

\(=\)

\(\ds \dfrac 1 {\map \det {\mathbf A} } \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{ij}\)

\(\ds \)

\(=\)

\(\ds \map \det {\mathbf B} \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n \mathbf A_{ij}\)

Determinant of Matrix Product and $A B = I$

\(\ds \)

\(=\)

\(\ds \map \det {\mathbf B} \paren {\map \det {\mathbf A} - \map \det {\mathbf A - \mathbf J_n} }\)

Apply the Lemma

\(\ds \)

\(=\)

\(\ds \map \det {\mathbf B} \map \det {\mathbf A} - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}\)

Definition $A = B^{-1}$

\(\ds \)

\(=\)

\(\ds 1 - \map \det {\mathbf B} \map \det {\mathbf B^{-1} - \mathbf J_n}\)

Determinant of Matrix Product and $\mathbf B \mathbf A = \mathbf I$

$\blacksquare$