Let \(F\) be a field and \(D\) a function on \(n \times n\) matrices over \(F\) (with values in \(F\)). Suppose \(D(AB) = D(A)D(B)\) for all \(A, B\). Show that either \(D(A) = 0\) for all \(A\), or \(D(I) = 1\). In the latter case show that \(D(A) \neq 0\) whenever \(A\) is invertible.
Solution
Putting \(A=B=I\) gives \[ D(I)=D(I)^2. \] Since \(F\) is a field, \(D(I)\) is therefore either \(0\) or \(1\).
If \(D(I)=0\), then for every matrix \(A\), \[ D(A)=D(IA)=D(I)D(A)=0. \] Thus either \(D\) is identically zero or \(D(I)=1\).
Assume now that \(D(I)=1\), and let \(A\) be invertible. Then \[ 1=D(I)=D(AA^{-1})=D(A)D(A^{-1}). \] Thus \(D(A)\) has a multiplicative inverse in \(F\), so in particular \(D(A)\ne0\).