Question-6

\(A\) is a square matrix of order \(2\).

Let \(A=\begin{bmatrix}a & b\\ c & d \end{bmatrix}\). We can now compute \(A^{2}\):

\[ \begin{aligned} A^{2} & =\begin{bmatrix}a & b\\ c & d \end{bmatrix}\begin{bmatrix}a & b\\ c & d \end{bmatrix}\\ & =\begin{bmatrix}a^{2}+bc & ab+bd\\ ac+cd & bc+d^{2} \end{bmatrix} \end{aligned} \]

\(A^{2}=0\)

If \(A^{2}=0\), then:

\[ \begin{aligned} a^{2}+bc & =0\\ b(a+d) & =0\\ c(a+d) & =0\\ d^{2}+bc & =0 \end{aligned} \]

One solution is to have \(a=b=c=d=0\). But this is not the only one. The presence of \(a+d\) in two equations suggests that it might be useful to set \(a+d=0\). We can set \(a=1,d=-1\). Since \(bc\) appears in two equations, we can set \(b=1,c=-1\). We have \(A=\begin{bmatrix}1 & 1\\ -1 & -1 \end{bmatrix}.\) Verify that \(A^{2}=0\).

\(A^{2}=I\)

Moving to \(A^{2}=I\), we have to solve the following system:

\[ \begin{aligned} a^{2}+bc & =1\\ b(a+d) & =0\\ c(a+d) & =0\\ d^{2}+bc & =1 \end{aligned} \]

We can set \(a=1,d=-1\) and \(b=d=0\). This results in \(A=\begin{bmatrix}1 & 0\\ 0 & -1 \end{bmatrix}\). Verify that \(A^{2}=I\).

Insights

We can therefore conclude that \(A^{2}=0\) admits solutions other than \(A=0\) and \(A^{2}=I\) admits solutions other than \(A=\pm I\). This question is insightful for the following reason. Consider the corresponding equations in \(\mathbb{R}\):

\[ x^{2}=0\text{ and }x^{2}=1 \]

\(x=0\) is the only solution to the first equation and \(x=\pm1\) are the only two solutions to the second. This shows the difference between simple algebraic equations in one variable and their matrix counterparts.