Here is task: $$A: R^{n} \rightarrow R^{n}$$ $A^{3}$ - projection.
(a) - What eigenvalues this linear operator have?
(b) - Is it true that A will have a diagonal matrix in some basis $R^{n}$?
============
I have solved (a) here it is: $$ A^{3} - projection \rightarrow A^{6} = A^3. $$ We can find minimal polynomial $f(A) = 0$. $$ f(x)=x^6 - x^3 = x^3(x^3 - 1) $$ Here we can find spectrum $A$: $$ x^3(x^3 - 1) = x^3(x^2 + x + 1)(x - 1) $$ And spectrum will be: $$ λ_1=0 $$ $$ λ_2=1 $$ $$ λ_3=\frac{-1 + \sqrt{3}i}{2} $$ $$ λ_4=\frac{-1 - \sqrt{3}i}{2} $$ Can't find next step for solve (b).