Does there exist a $2 \times 2$ non-diagonal matrix $A$ such that $A^3 = I$ ?
Suppose it exists then $\lambda^3 = 1$ where $\lambda$ is the eigenvalue of $A$
Now a $2 \times 2$ matrix can have $2$ eigenvalues.
Those look like cube roots of unity
$1,\omega , \omega ^ 2$.
Now, can I generate a $2 \times 2$ matrix whose eigen values are 1,1 and it's non diagonal?
I think yes
$A = \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix}$
But I see that cubing this does not give me $I$, even though it satisfies the determinant and trace conditions.
Is this approach correct?
Any interesting problem can be formulated from this?
Best Answer
Note that $(x-\omega)(x-\omega^2)=x^2+x+1$ so look for a matrix with characteristic function $x^2+x+1$, that is with trace $-1$ and determinant $1$, for instance $$\pmatrix{-1&1\\-1&0}.$$