Let
\[ A = \left[ \begin{array}{c c} a & b \\ c & d \end{array} \right] \]
be a \(2 \times 2\) matrix with complex entries. Suppose that \(A\) is row-reduced and also that \(a + b + c + d = 0\) . Prove that there are exactly three such matrices.
Solution
The three matrices are \[ \boxed{ \begin{pmatrix}0&0\\0&0\end{pmatrix},\qquad \begin{pmatrix}1&-1\\0&0\end{pmatrix},\qquad \begin{pmatrix}0&0\\1&-1\end{pmatrix}.} \]
Recall that in this section a matrix is row-reduced when the first nonzero entry of each nonzero row is \(1\), and every column containing such a leading \(1\) has all its other entries equal to \(0\) [@HK71]. No condition orders the nonzero rows; that stronger requirement belongs to row-reduced echelon form.
1. If exactly one row is nonzero, there are exactly two possibilities.
Suppose first that the first row is nonzero and the second row is zero. Then \(c=d=0\). If \(a\ne0\), row-reducedness forces the leading entry \(a\) to equal \(1\), and the sum condition gives \(1+b=0\), so \[ A=\begin{pmatrix}1&-1\\0&0\end{pmatrix}. \] If instead \(a=0\), then the nonzero row begins with \(b\), so \(b=1\); the sum of the entries is then \(1\), impossible.
Now suppose the first row is zero and the second row is nonzero. The same argument gives either \(c=1\) and then \(d=-1\), or else \(c=0,d=1\), the latter again contradicting the zero-sum condition. Thus the only possibility in this case is \[ A=\begin{pmatrix}0&0\\1&-1\end{pmatrix}. \]
2. If both rows are nonzero, no matrix satisfies the sum condition.
Each nonzero row has a leading \(1\). The two leading \(1\)’s cannot occur in the same column, because a pivot column must have all other entries zero.
If the first row leads in column \(1\), then \(a=1\) and the pivot-column condition forces \(c=0\). Since the second row is nonzero, it must then lead in column \(2\), so \(d=1\); the pivot condition in column \(2\) forces \(b=0\). Hence \(A=I_2\), whose entries sum to \(2\).
If instead the second row leads in column \(1\), then \(c=1\) and \(a=0\). The first row must therefore lead in column \(2\), so \(b=1\), and the pivot condition in column \(2\) forces \(d=0\). Hence \[ A=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \] whose entries also sum to \(2\). Therefore no matrix with two nonzero rows works.
3. The zero matrix supplies the third example, and all three displayed matrices are row-reduced.
The zero matrix is row-reduced vacuously and has entry sum zero. In each of the two rank-one matrices displayed above, the only nonzero row has leading entry \(1\), and its leading column has zero in the other row; their entry sums are \(1-1=0\). Thus all three satisfy the hypotheses. Steps 1 and 2 show that there are no others.