Let $P_j$ equal matrix $A$ with column $j$ replaced by ones, $1\le j \le n$. Then: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^n \det\paren { P_j }       | r = \sum_{j \mathop = 1}^n \sum_{i \mathop = 1}^n  A_{ij}       | c = Cofactor expansion applied to column $j$ of $P_j$. }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = A       | r = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)       | c = Symbol $A$ is any $2\times 2$ matrix. }} {{eqn | l = J_2       | r = \left(\begin{smallmatrix} 1 & 1 \\ 1 & 1 \end{smallmatrix}\right)       | c = Symbol $J_2$ is the $2\times 2$ matrix of all ones. }} {{eqn | l = \det\paren {A -J_2}       | r = \det\paren { \begin{smallmatrix} a-1 & b-1 \\ c-1 & d-1 \end{smallmatrix} }       | c = Matrix subtraction. }} {{eqn | l =        | r = \det\paren { \begin{smallmatrix} a & b-1 \\ c & d-1 \end{smallmatrix} } - \det\paren { \begin{smallmatrix} 1 &  b-1 \\ 1 & d-1 \end{smallmatrix} }       | c = Determinant as Sum of Determinants }} {{eqn | l =        | r = \det\paren { \begin{smallmatrix} a & b-1 \\ c & d-1 \end{smallmatrix} } - \det\paren { \begin{smallmatrix} 1 &  b \\ 1 & d \end{smallmatrix} }       | c = Determinant rule Replacement Rule for Determinant Columns }} {{eqn | l =        | r = \det\paren { \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} } - \det\paren { \begin{smallmatrix} a & 1 \\ c & 1 \end{smallmatrix} } - \det\paren { \begin{smallmatrix} 1 &  b \\ 1 & d \end{smallmatrix} }       | c = Determinant Linearity in a Column }} {{eqn | l =        | r = \det\paren A - \det\paren {P_2} - \det\paren { P_1 }       | c = Definition of $P_1$, $P_2$. }} {{eqn | l =        | r = \det\paren A - \sum_{j \mathop = 1}^2 \det\paren { P_j }       | c = Equivalent identity verified for $n = 2$. }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = Q_2 | r = \paren { \begin{smallmatrix} 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{smallmatrix} } | c =  Instance $n = 5, m = 2$. }} {{end-eqn}} 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 $m = 1$''': {{begin-eqn}} {{eqn | l = \det\paren { A -J_n }       | r =\det\paren { A -Q_1 }- \det\paren { P_1-Q_1 }       | c = $P_1$ equals $A$ with column $1$ all ones. Determinant as Sum of Determinants }} {{eqn | l = \det\paren { P_1-Q_1 }       | r = \det\paren { P_1 }       | c = Add ones in column 1 to columns $2\cdots n$, Replacement Rule for Determinant Columns }} {{eqn | l = \det\paren { A -J_n }       | r = \det\paren { A -Q_1 }- \det\paren { P_1 }       | c = Combine equations.  }} {{end-eqn}} '''Induction step $m = k$ and $k < n$ implies $m = k+1$''': {{begin-eqn}} {{eqn | l = \det\paren { A - J_n }       | r =\det\paren { A - Q_k }- \sum_{j \mathop = 1}^k \det\paren { P_j }       | c = Induction hypothesis $m = k$. }} {{eqn | l = \det\paren { A -Q_k }        | r = \det\paren { A -Q_{k+1} } - \det\paren { P_{k+1} - Q_{k+1} }       | c = Determinant linearity on column $k+1$. }} {{eqn | l = \det\paren { P_{k+1} - Q_{k+1} }        | r = \det\paren { P_{k+1} }       | c = Add ones in column $k+1$ to columns $k+2\cdots n$.   Replacement Rule for Determinant Columns }} {{eqn | l = \det\paren { A -J_n }       | r = \det\paren { A -Q_{k+1} }-\det\paren { P_{k+1} }- \sum_{j \mathop = 1}^k \det\paren { P_j }       | c = Combine preceding three equations. }} {{eqn | l = \det\paren { A -J_n }       | r = \det\paren { A -Q_{k+1} } - \sum_{j \mathop = 1}^{k+1} \det\paren { P_j }       | c = Induction completed. }} {{end-eqn}} '''Conclusion''': Matrix  $A-Q_n$ equals $A$ because $Q_n$ is the zero matrix. Let $m = n$ in the '''induction identity''', then: :$\displaystyle \det\paren { A - J_n } = \det\paren { A  }- \sum_{j \mathop = 1}^n \det\paren { P_j }$  {{qed}} squ7pcpfrx55ui5k02iztoin9o6q7tv