If a matrix A square is 0, does it follow that A = 0?

Question

Let A be a square matrix. If $A^2 = 0$, then it follows that $A = 0$.

Is there a counterexample for this? If there isn't, what kinds of explanation can I make to justify this statement?

Answer 1

$$A = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)$$ is a classic example.

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More theoretically: if $A^2 = 0$, then $\det A = 0$. Hence $A$ in Jordan Canonical Form (JCF) is either $$J_1 = \left( \begin{matrix} \lambda & 0 \\ 0 & -\lambda \end{matrix} \right) \ \text{ or } \ J_2 = \left( \begin{matrix} \lambda & 1 \\ 0 & -\lambda \end{matrix} \right)$$

As $J_1^2 = 0$ implies $\lambda = 0$ it follows $J_1 = 0$ and hence the JCF must be of the form $J_2$. Now, $$J_2^2 = \left( \begin{matrix} \lambda^2 & 0 \\ 0 & \lambda^2 \end{matrix} \right)$$ and hence $\lambda = 0$.

Therefore any $2 \times 2$ matrix $A$ such that $A \neq 0$ and $A^2 = 0$ must have Jordan Canonical Form $J_2 = \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right)$. This can be confirmed directly for the other examples given in other responses.

Answer 2

This is not true. Consider $\eta = \begin{pmatrix} 1 & 1 \\ -1 & -1 \\ \end{pmatrix}$.

Answer 3

Another example: \begin{align} \left[ \begin{matrix} 2 & 1 \\ -4 & -2 \end{matrix}\right] \end{align} or more general \begin{align} \left[ \begin{matrix} ab & a^2 \\ -b^2 & -ab \end{matrix}\right] \end{align}