Q:

If a function of matrices into the reals distributes over products then it sends any matrix with a row of zeros to zero.

Suppose $D:\mathbb{R}_{2\times 2}\rightarrow \mathbb{R}$ is a function from the space of 2x2 matrices into the reals.  Suppose also that it satisfies $D(AB)=D(A)D(B)$ for all matrices $A,B\in \mathbb{R}_{2\times 2}$.  Call  
$$J = \begin{bmatrix}0&1\\1&0\end{bmatrix}$$ 
and suppose $D(I)\ne D(J)$.  I have already shown that such a function must map $D(I)=1$ and $D(J)=-1$, and $D(0)=0$.  I've also shown that if $A^2=0$ then $D(A)=0$, and that if $A,B$ differ by a row-swap then $D(A)=-D(B)$.  
However, I can't see my way through showing that if any matrix $A$ has a zero-row, then $D(A)=0$.  I have considered squaring $A$, writing it as a product of two other matrices, multiplying it by other matrices, and using the fact that matrices don't commute but real products do.  Nothing has worked and I'm out of ideas.

A:

If $A$ has the first row $0$ iff $A = P \cdot A$, where $P=\begin{bmatrix}0&0\\0&1\end{bmatrix}$. Now it enough to show that $P$ maps to $0$. Consider also $Q= \begin{bmatrix}1&0\\0&0\end{bmatrix}$. We have $Q=JPJ^{-1}$, so $P$, $Q$ map to the same value. Moreover, we have $PQ=0$, the zero matrix, and so $PQ$ maps to $0$.  We conclude $P$ maps to $0$. 

