The cofactor expansion formula of a square matrix along a column

Let A be a square matrix over a field F. Formulate and prove the cofactor expansion formula for \mathsf{det}\ A along the jth column.


Let A = [\alpha_{i,j}].

We begin with a definition. If A = \left[ \begin{array}{c|c|c} A_{1,1} & C_1 & A_{1,2} \\ \hline R_1 & \alpha_{i,j} & R_2 \\ \hline A_{2,1} & C_2 & A_{2,2} \end{array} \right], where A_{1,1} has dimension (i-1) \times (j-1), then the (i,j)-minor of A is the matrix A_{i,j} = \left[ \begin{array}{c|c} A_{1,1} & A_{1,2} \\ \hline A_{2,1} & A_{2,2} \end{array} \right].

Recall that the cofactor expansion formula for \mathsf{det}\ A along the ith row is

\mathsf{det}\ A = \sum_{j=1}^n (-1)^{i+j} \alpha_{i,j} \mathsf{det}(A_{i,j}).

The analogous expansion along the jth column is

\mathsf{det}\ A = \sum_{i=1}^n (-1)^{i+j} \alpha_{i,j} \mathsf{det}(A_{i,j}).

which we presently prove to be true. First, note that (A_{i,j})^\mathsf{T} = (A^\mathsf{T})_{j,i}; this follows from our definition of minors and the fact that \left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^\mathsf{T} = \left[ \begin{array}{c|c} A^\mathsf{T} & C^\mathsf{T} \\ \hline B^\mathsf{T} & D^\mathsf{T} \end{array} \right]. Now we have the following.

\mathsf{det}(A)  =  \mathsf{det}(A^\mathsf{T})
 =  \sum_{j=1}^n (-1)^{i+j} \alpha_{j,i} \mathsf{det}((A^\mathsf{T})_{i,j})
 =  \sum_{j=1}^n (-1)^{i+j} \alpha_{j,i} \mathsf{det}((A_{j,i})^\mathsf{T})
 =  \sum_{j=1}^n (-1)^{i+j} \alpha_{j,i} \mathsf{det}(A_{j,i})
 =  \sum_{i=1}^n (-1)^{i+j} \alpha_{i,j} \mathsf{det}(A_{i,j}),

as desired.

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