If $A$ is a square real matrix and $A^3 = 2I$, how can I prove that matrix $A - 2^{\frac{1}{3}}I$ is not invertible?
I know it can be solved using the characteristic polynomial of matrix $A$, but I was looking for a way to solve it by some clever techniques, like matrix multiplication. I am trying to avoid using determinants and eigenvalues if possible.
Note that
$$ (A^2+ 2^{1/3} A + 2^{2/3} I) ( A- 2^{1/3} I) = A^3 - 2 I = 0$$
If $A-2^{1/3} I$ was invertibile, you could deduce that
$$A^2+ 2^{1/3}A + 2^{2/3}I = 0$$
And now we deduce... Nothing!!
Indeed, consider the matrix $A = 2^{1/3} T_{2 \pi/3}$, where $T_{\theta}$ is the matrix corresponding to the rotation of an angle $\theta$ in the plane. This satisfies $A^3= 2I$ , but $A-2^{1/3}I = 2^{1/3}( T_{2 \pi/3} - I) $ is invertible! Indeed, the only vector satisfying the system
$$ (T_{2\pi/3} - I) x = 0 $$ $$ T_{2\pi/3} x = x $$
Is the zero vector. This happens because the rotation has no other fixed points.