Eigenspace decomposition of $T$ such that $T^3=T$

Question

Suppose $T : V \rightarrow V$ is a linear operator such that $T^3=T$. Show that $T$ is diagonalizable, and $V = \text{im }T \oplus\ker T $, $\text{im }T = E_1\oplus E_2 $.

Note: $E_\lambda$ is the eigenspace corresponding to eigenvalue $\lambda$.

I am stuck on how to decompose $V$ into image and kernel. It is easy to show that $V = \text{im }T +\ker T $, now I have to show that the intersection is $0$.

I tried by contradiction, supposing that there exists a $v \neq 0$ such that $Tv=0$. It seems that I need to use the fact $T^3=T$, but I can't seem to get any further.

Any help will be appreciated. Thanks in advance!

Answer 1

Let $v\in Ker(T)\cap Im(T), T(u)=v$ and $T(v)=0$, this implies that $T^2(u)=T(v)=0$ and $T^3(u)=T^2(T(u))=v=0$, since $dimKer(T)+dimIm(T)=n$, we deduce that $Ker(T)\oplus Im(T)=V$.

The restriction $U$ of $T$ to $Im(T)$ verifies also $U^3=U$, this implies that that the eigenvalues of $U$ are $0,1$ or $-1$, since $Ker(T)\cap Im(T)=0$, we deduce that the eigenvalues are $1$ or $-1$ and $U^2-I=0$, (to see this, if $x=U(y), U^2(x)=U^3(y)=U(y)=x$) this implies that $U$ is a projection and $Im(T)=V_1\oplus V_2$, $V_1=\{x:U(x)=x\}, V_2=\{x:U(x)=-x\}$.

Answer 2

Since $T^3=T$, if $v\in V$ you have $Tv=T^3v$ and so $T(v-T^2v)=0$. In particular, $v-T^2v\in\ker T$. Hence $$ v=(v-T^2v)+T^2v \tag{*} $$ and therefore $V=\ker T+\operatorname{im}T$. If $v\in\ker T+\operatorname{im}T$, then $v=Tw$ for some $w\in V$; since $Tv=0$, also $T^2w=0$ and therefore $T^3w=0$. Thus $v=Tw=T^3w=0$.

Now you just have to prove that $\operatorname{im}T=E_{1}\oplus E_{-1}$.

Let $v\in\operatorname{im}T$. Then $v-T^2v=0$ because of the decomposition (*).

Now use the fact that $$ v=\frac{1}{2}(v+Tv)+\frac{1}{2}(v-Tv) $$ Can you finish?