Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 10G (Linear Algebra)

Define the determinant by the signed permutation sumIt is multilinear and alternating in the rows, and likewise in the columns. Row multilinearity follows because every product contains exactly one entry from each row; a swap changes the permutation sign, and equal rows give cancellation in pairs. Applying the same argument to the transpose proves the column properties.

Consequently, for either rows or columns, a swap multiplies the determinant by $-1$; multiplication of one row or column by $c$ multiplies it by $c$; addition of a multiple of another row or column leaves it unchanged. For the last rule, multilinearity separates the original determinant and a determinant with two equal rows or columns, which is zero. These arguments also cover singular matrices and zero scaling.

In the determinant expansion of the block upper triangular matrix, a nonzero product must assign each of the bottom $n$ rows to one of the last $n$ columns, because the lower-left block is zero. Those rows therefore occupy all the last $n$ columns, leaving the first $n$ columns for the top rows. The contributing permutations are precisely independent permutations within the two blocks. Their signs multiply, so the sum factors:The off-diagonal block contributes to no surviving term.

For the determinant of a complex block representation, putThe change-of-basis matrix $T$ is invertible, and direct block multiplication givesThus $T^{-1}MT$ is block diagonal. Using multiplicativity of the determinant and the preceding block formula,The identity follows from this constant change of basis for arbitrary complex $A,B$.