I need to prove that $Ker(u - Id_E)\oplus Ker(u - aId_E)\oplus Ker(u - a^2Id_E)=E$

Question

Let be E a $\mathbb{C}$ vectorial space and $u\in\mathcal{L}(E,E)$ a linear operator such that $u^3 = Id_E$. Prove that

$$E = Ker(u - Id_E)\oplus Ker(u - aId_E)\oplus Ker(u - a^2Id_E)$$

where $a = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}, \qquad i = \sqrt{-1}$.

R: I prove that sum is direct by the following

Let $\lambda\in\sigma(u)$ then existe $0 \neq x \in E$ such that $u(x) = \lambda x$ using that $u^3 = Id_E$ we have $x = \lambda^3 x\Longrightarrow (\lambda^3 - 1)x = 0$ like $x \neq 0$ then $$0 = (\lambda^3 - 1) = (\lambda - 1)(\lambda^2 +\lambda + 1)$$

Therefore $$\lambda = 1,\ \lambda = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2},\ \lambda = -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$$ are eigenvalues. As they are different then the sum of their associated spaces are direct.

With this result is clear that $$Ker(u - Id_E)\oplus Ker(u - aId_E)\oplus Ker(u - a^2Id_E)\subset E$$.

We need to prove the other inclusion but don't have idea.

Answer 1

Hint: if $\lambda_i$ are distinct, you can prove that $$I = \alpha_1 (u-\lambda_1I)(u-\lambda_2I) + \alpha_2 (u-\lambda_1I)(u-\lambda_3I) + \alpha_3 (u-\lambda_3I)(u-\lambda_2I)$$ for some complex $\alpha_1,\alpha_2,\alpha_3$.

Then you conclude using $(u-\lambda_1I)(u-\lambda_2I)(u-\lambda_3I)=0$.

Answer 2

Define linear transformations $$\phi:E\rightarrow Ker(u-Id_E)\oplus Ker(u-aId_E)\oplus Ker(u-a^2Id_E)$$ by

$\phi(e)=\left(\frac{1}{3}u^2(u-aId_E)(u-a^2Id_E)(e), \frac{1}{3}u^2(u-Id_E)(u-a^2Id_E)(e), \frac{1}{3}u^2(u-aId_E)(u-aId_E)(e)\right)$

and $$\psi: Ker(u-Id_E)\oplus Ker(u-aId_E)\oplus Ker(u-a^2Id_E)\rightarrow E$$ by $\psi(x,y,z)=x+y+z$. Then $\psi\circ\phi=Id_E$ and $\phi\circ\psi=Id_{Ker(u-Id_E)\oplus Ker(u-aId_E)\oplus Ker(u-a^2Id_E)}$.

So, these linear transformations are isomorphisms.

Since $Ker(u-Id_E)\oplus Ker(u-aId_E)\oplus Ker(u-a^2Id_E)\subset E$, we have the equality of these $\Bbb{C}$ vector spaces.