Hoffman Problem. Section 5, Problem 13. Let  $\mathbb{R}$ be the field  of real  numbers,  and  let  D  be  a function  on  2x2  matrices over  $\mathbb{R}$,  with  values  in  $\mathbb{R}$,  such  that  $D(AB) = D(A)D(B)$  for  all  $A, B$.  Suppose also  that
$$D\begin{pmatrix} 0     & 1\\ 1 & 0 \end{pmatrix} \neq D\begin{pmatrix} 
  1 & 0\\ 0 & 1 \end{pmatrix}$$  
If we already know that:
$1.D(0)=0$
$2.D(A)=0$ if $A^2=0.$
$3.D(A)=-D(B)$ if $B$ is obtained  by interchanging  the  rows  (or  columns) 
of  $A$.
Prove that:
(a) $D(A)=0$  if  one  row  (or  one  column)  of  $A$  is  $0$.
(b) $D(A)  =  0$  whenever $A$  is singular.
 A: Consider 
$$D\left(\begin{bmatrix}
x&y\\
0&0
\end{bmatrix} \right) = D\left(\begin{bmatrix}
1&0\\
0&0
\end{bmatrix} \begin{bmatrix} x&y\\x&y\end{bmatrix}\right) = D\left(\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}\right)D\left(\begin{bmatrix} x&y\\x&y\end{bmatrix}\right)$$
and use the fact that property (3) implies 
$$D\left(\begin{bmatrix} x&y\\x&y\end{bmatrix}\right) = 0.$$
It should be easy to fill in the remaininig details from there.
A: To begin this problem observe that if $A$ is nilpotent or if $A$ is zero then part (a) is completed for free by properties $(1)$ and $(2)$; as such, assume that $A$ has a row of zeroes (it is sufficient to solve this only for row space; since $A$ is a finite square matrix, the case where we are interested in columns of $A$ is taken care of by looking at $A^t$, the transpose of $A$). So, let us assume that $A$ has a row of zeroes and that $A$ is neither nilpotent nor zero. Then set $B \in \mathcal{M}_2(\mathbb{R}) := \left\lbrace \begin{bmatrix} a & b \\ c & d \end{bmatrix} \; \big| \; a,b,c,d \in \mathbb{R} \right\rbrace$ so that $B$ is obtained by a row swap from $A$. Then by property $(3)$ we have that $$D(A) = -D(B).$$
Now, consider the case when $B$ is nilpotent (this arises from when $A$ takes the form
$$\begin{bmatrix}
0 & 0 \\
0 & a
\end{bmatrix}$$
for nonzero $a \in \mathbb{R}$). Now, since $B$ is nilpotent, direct evaluation gives us that $B^2 = 0$. Thus
$$ 0 = D(B^2) = D(B)D(B) = -D(A)D(B) = D(A)D(A). $$
Since $D(A)$ is a real number and the real numbers have the property that $ab = 0 \iff a = 0$ or $b = 0$, we have that $D(A) = 0$ and the case is complete.
Now consider the case where $B$ is not nilpotent; that is, $A$ takes the form
$$\begin{bmatrix}
a & b \\
0 & 0
\end{bmatrix}$$
for nonzero $a,b \in \mathbb{R}$. Then direct calculation gives us that
$$ AB = \begin{bmatrix}
a & b \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ a & b \end{bmatrix} = \begin{bmatrix} ba & b^2 \\ 0 & 0 \end{bmatrix} = bA.$$
This gives us that
$$D(AB)=D(A)D(B)=-D(A)D(A)=-D(A^2)$$
and since
$$D(AB)=D(bA)=\alpha D(A)$$
for some appropriate nonzero $\alpha$, we have
$$D(A^2)=-\alpha D(A).$$
From here we see that
$$ -\frac{D(A^2)}{\alpha} = D(A),$$
implying that $D(A) = 0$, completing the case and giving us part (a).
Now we show part (b). To do this, let $A \in \mathcal{M}_2(\mathbb{R})$ and assume that $A$ is singular; that is, there is no product 
$$ E = \prod\limits_{i = 1}^n E_i $$
of elementary matrices $E_i$ so that $AE = I_2$, where $I_2$ is the $2\times 2$ identity matrix. As such, taking the reduced row echelon form of $A$ gives us that $A$ has at most one leading one; call $R$ the row-reduced form of $A$. Now, we have that
$$ R = AF$$
where $F = \prod_{j=1}^m E_j$ is the (finite) product of elementary matrices $E_j$. Now, since $R$ has at most one leading one, $R$ has a row of zeroes, giving us by part (a) that
$$ 0 = D(R) = D(A)D(F) = D(A)\prod\limits_{i=1}^mD(E_i).$$
Each elementary matrix has that $D(E_i) \ne 0$; as such, $D(F) = a \ne 0$. This gives us that
$$ 0 = D(R) = D(A)D(F) = D(A)a,$$
from whence we see that $D(A) = 0$. This completes the proof.
A: Hints.
(a) We consider a matrix with a zero row. The case for a matrix with a zero column is similar. Let $A=\pmatrix{a&b\\ 0&0}$ and $B=\pmatrix{0&0\\ a&b}$, so that $D(B)=-D(A)$. There are two possibilities:


*

*$A=0$. Why is $D(A)=0$?

*$A\ne0$. Then there exist some $\theta$ such that $(\cos\theta,\sin\theta)=\frac1{\sqrt{a^2+b^2}}(a,b)$. Let $R=\pmatrix{\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta}$. Verify that $R^TR=I$ and $ARB=0$.


(b) If $A$ is singular, its row reduced echelon form must have a zero row.
In other words, $A=E_kE_{k-1}\ldots E_1F$ for some elementary
matrices and some matrix $F$ with a zero row.
