Vector space, direct sums involving linear maps and polynomials

Question

Let be $V$ a vector space over $K$ and lets $a\in K\setminus\{0\}$ and $T:V \to V$ a linear map s. t. $a^2T-3aT^2+T^3=0$. Show that $V = Ker(T) \oplus Im(T)$.

My professor told me an observation:

This result is more general, that is, if I have a polynomial $P$ such with $P(0) = 0$ and $P'(0) \neq 0 $ such that $P(T) = 0 $ then $V = Ker(T) \oplus Im(T)$.

However, I didn't find the relation with the exercise and his observation (of course, that polynomial assets the assumption but???). Can you give me a tip to solve this exercise??

Also I've tried to show that $T$ is a projection, but it didn't work.

Answer 1

Suppose the dimension is finite. we show that if $p$ is a polynomial such that $p(T)=0, p(0)=0, p'(0)\neq 0$ then $V=Ker T\oplus ImT$

Let $y\in Ker T\cap ImT$, $y=T(x)$ implies that $T^2(x)=T(y)=0$, write $p(X)=a_nX^n+...+a_1X$, $p(T)(x)=a_1T(x)=0$, $a_1\neq 0$ since $a_1=p'(0)$, we deduce that $T(x)=y=0$ and for $KerT\oplus ImT=V$ since $dim(KerT)+dim(ImT)=dim(V)$.

Take $p(X)=a^2X-3aX^2+X^3$.

Answer 2

Multiply $a^2T - 3aT^2 + T^3 = 0$ by $v$ to obtain $$Tv = \frac{1}{a^2}T^2\left(3a - T\right) v$$ for all $v \in V$. So $\operatorname{im}T = \operatorname{im}T^2$ and likewise $\ker T = \ker T^2$. Now what happens if $v \in \ker T \cap \operatorname{im}T$?