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Question

The following problem illustrate ways in which the algebra of matrices is not analogous to the algebra of real numbers. Find a 2 x 2 matrix A with each element +1 or -1 such that A2=0\mathbf{A}^{2}=\mathbf{0}. The formula of earlier problem may be helpful.

Solution

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Let

A=[abcd]\textbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}

The formula of Problem 29:

A2=(a+d)A−(ad−bc)I\textbf{A}^2 = (a+d) \textbf{A} - (ad-bc) \textbf{I}

Since we want the equality A2=0\textbf{A}^2 = \textbf{0} to hold, it is clearly enough that we have

a+d=0andad−bc=0a+d = 0 \quad \text{and} \quad ad-bc = 0

From the first equation we get that d=−ad = -a. Plugging this into the second equation yields

−a2−bc=0⇒a2=−bc-a^2-bc = 0 \Rightarrow a^2 = -bc

Since a=1a = 1 or a=−1a = -1, a2=1a^2 = 1, so

−bc=1-bc = 1

Again, b=1b = 1 or b=−1b = -1, and c=1c = 1 or c=−1c = -1. If b=cb = c, then

−b2=1-b^2 = 1

which is clearly a contradiction. So, b≠cb \ne c.

We can finally conclude something about A\textbf{A}. Since a≠da \ne d, we pick one of them and immediately know the other. Let a=1a = 1. Since d=1d = 1 or d=−1d = -1 and d≠ad \ne a, we get that d=−1d = -1 (of course, we could have picked a=−1a = -1, d=1d = 1).

Similarly, we can pick b=1b = 1, c=−1c = -1.

So,

A=[11−1−1]\textbf{A} = \begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix}

We can additionally check that

A2=AA=[11−1−1][11−1−1]=[0000]=0\textbf{A}^2 = \textbf{A} \textbf{A} = \begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix} = \textbf{0}

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