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 $\mathbf{A}^{2}=\mathbf{0}$. The formula of earlier problem may be helpful.

Solution

VerifiedLet

$\textbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

The formula of Problem 29:

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

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

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

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

$-a^2-bc = 0 \Rightarrow a^2 = -bc$

Since $a = 1$ or $a = -1$, $a^2 = 1$, so

$-bc = 1$

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

$-b^2 = 1$

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

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

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

So,

$\textbf{A} = \begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix}$

We can additionally check that

$\textbf{A}^2 = \textbf{A} \textbf{A} = \begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0& 0 \\ 0 & 0 \end{bmatrix} = \textbf{0}$

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