A good property of trace 0 matrices

Question

I found out an interesting property of a trace $0$ matrix.
Suppose $A$ be a $n\times n$ matrix whose trace is $0$. Then $A^2$ will be a scalar matrix (that is $A^2$ is of the form $\lambda I_n$, where $\lambda$ is any real constant and $I_n$ is the identity matrix ) iff $n=2$.But however, this result does not hold true for $n \ge 3$.
For $n=2$ it is simple to prove if we assume the form of the matrix and just use the restriction $\operatorname{tr}(A)=0$.
But how to approach this using eigenvalues? Also, how can we infer that this result does not hold for higher $n$ using eigenvalues?

Answer 1

Suppose $A$ is a $2\times 2$ matrix with trace $0$. Then the characteristic polynomial of $A$ is given by $p_A(x)=x^2-\operatorname{tr}(A)x+\det(A)=x^2+\det A$. Now by Cayley-Hamilton Theorem, it follows that $A^2=\lambda\cdot I_2$ for some $\lambda\in\Bbb R$.

Other direction is not true. For example, consider the matrix $$ A= \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} .$$ We have $A^2=0=0\cdot I_3$ and $\operatorname{tr}(A)=0$, but $A$ is a $3\times 3$ matrix.