Let \(\RR\) be the field of real numbers, and let \(D\) be a function on \(2 \times 2\) matrices over \(\RR\), with values in \(\RR\), such that \(D(AB) = D(A)D(B)\) for all \(A, B\). Suppose also that \[ D\left(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\right) \neq D\left(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right). \] Prove the following.
\(D(0) = 0\);
\(D(A) = 0\) if \(A^2 = 0\);
\(D(B) = -D(A)\) if \(B\) is obtained by interchanging the rows (or columns) of \(A\);
\(D(A) = 0\) if one row (or one column) of \(A\) is 0;
\(D(A) = 0\) whenever \(A\) is singular.
Solution
Let \[ S=\begin{bmatrix}0&1\\1&0\end{bmatrix}. \] The hypothesis \(D(S)\ne D(I)\) rules out the identically zero function. By the preceding multiplicativity argument, \(D(I)=1\). Since \(S^2=I\), \[ D(S)^2=D(I)=1. \] Over \(\mathbb R\) this gives \(D(S)=\pm1\), and the hypothesis excludes \(+1\). Hence \[ D(S)=-1. \]
For (a), multiplicativity gives \(D(0)=D(0)^2\), so \(D(0)\in\{0,1\}\). If \(D(0)=1\), then for every \(A\), \[ 1=D(0)=D(0A)=D(0)D(A)=D(A), \] contradicting \(D(S)=-1\). Thus \(D(0)=0\).
For (b), if \(A^2=0\), then \[ D(A)^2=D(A^2)=D(0)=0, \] so \(D(A)=0\).
For (c), interchanging the rows of \(A\) replaces \(A\) by \(SA\), while interchanging its columns replaces it by \(AS\). In either case multiplicativity and \(D(S)=-1\) give \[ D(B)=-D(A). \]
For (d), suppose first that the first row of \(A\) is zero, so \[ A=\begin{bmatrix}0&0\\a&b\end{bmatrix} =\begin{bmatrix}0&0\\1&0\end{bmatrix} \begin{bmatrix}a&b\\0&0\end{bmatrix}. \] The first factor squares to zero, so by (b) its \(D\)-value is zero; hence \(D(A)=0\). If the second row is zero, interchange the rows and use (c). The column statement follows identically, or by the analogous factorization.
For (e), let \(A\) be singular. If a row is zero, (d) applies. Otherwise its two rows are linearly dependent, so the second row is \(\lambda\) times the first for some \(\lambda\in\mathbb R\). Writing the first row as \((a,b)\), \[ A= \begin{bmatrix}1&0\\\lambda&0\end{bmatrix} \begin{bmatrix}a&b\\0&0\end{bmatrix}. \] The second factor has a zero row, so (d) gives it \(D\)-value zero. Multiplicativity therefore gives \(D(A)=0\).