self-adjoint operator, such that $T^2=T^3,$ is orthogonal projection.

Question

Let $T$ be a self-adjoint operator on a Hilbert space $H$, such that $T^2=T^3.$ Prove that $T$ is an orthogonal projection. Is that true if we had $T^4=T^3$?

Thank you for the help.

Answer 1

Because $T$ is selfadjoint, $$\ker T=(\text{ran}\, T^*)^\perp=(\text{ran}\,T)^\perp.$$This implies that $T$ is injective on its range: if $0=T(Tv)$, then $Tv\in\ker T\cap\text{ran}\,T=\{0\}$. So if $$0=T^3-T^2=T(T^2-T),$$ we conclude that $T^2=T$.

A similar reasoning works for $0=T^4-T^3=T(T^3-T^2)$, so $T^3=T^2$.

Answer 2

HINT:

Use the following lemma : $A$ self adjoint, $A^2=0$ implies $A=0$. Generalize to $A^{2^n}=0$ implies $A=0$.

Now consider $A= T(T-I)$.

Answer 3

${T^2}^2=T^4=T^3T=T^2T=T^3=T^2$. This implies that $T^2$ is a projection.

Let $x\in ker(T^2), y\in H$, $\langle x,T^2(y)\rangle=\langle T^2(x),y\rangle=0$ implies that $ker(T^2)$ is orthogonal to $Im(T^2)$ and $T^2$ is an orthogonal projection.

$ker(T)=ker(T^2)$. $ker(T)\subset ker(T^2)$, let $x\in ker(T^2), \langle T(x),T(x)\rangle=\langle x,T^2(x)\rangle=0$. This implies that $T(x)=0$ and $Ker(T^2)=ker(T)$.

Let's show that $T^2=T$.

Let $x\in H$, we can write $x=x_1+x_2, x_1\in ker(T^2), x_2\in Im(T^2)$, there exists $y$ such that $x_2=T^2(y)$, $T(x)=T(x_1+x_2)=T(x_2)=T^3(y)=T^2(y)$; $T^2(x)=T^3(y)=T^2(y)=T(x)$. This implies that $T^2=T$ and henceforth, $T$ is also an orthogonal projection.

Answer 4

Note that $p(T) = 0$ where $p(z) = z^3 - z^2 = z^2(z - 1)$. Thus the only possible eigenvalues are $0$ and $1$.

Let $W$ be the eigenspace corresponding to eigenvalue $1$. I claim that $T = P_W$. To show this, we show that $T$ maps onto $W$ and that $Tv$ and $Tv \perp v - Tv$ for all $v \in H$.

Since $T$ is self-adjoint, we can form an orthonormal basis of eigenvectors: $$B = (w_1, w_2, \ldots, w_n, u_1, u_2, \ldots, u_m)$$ where $w_1, \ldots w_n$ span $W$, and $u_1, \ldots, u_n$ correspond to eigenvalue $0$. If we take some $v \in H$, we have $$v = \alpha_1 w_1 + \ldots + \alpha_n w_n + \beta_1 u_1 + \ldots \beta_m u_m,$$ and \begin{align*} Tv &= \alpha_1 T w_1 + \ldots + \alpha_n T w_n + \beta_1 T u_1 + \ldots \beta_m T u_m \\ &= \alpha_1 w_1 + \ldots + \alpha_n w_n \in W. \end{align*} We also have $$v - Tv = \beta_1 u_1 + \ldots \beta_m u_m \in W^\perp,$$ as $B$ is orthonormal. Hence $T$ is an orthogonal projection. The same argument works for $T^4 = T^3$.

Answer 5

If $T$ is self-adjoint and $T^2=T$, then $T$ is an orthogonal projection operator. Therefore it suffices to show that $T^2=T$ or equivalently $T(T-I)=0$. \newline \newline Note that, for any $x\in E$ \begin{align*} \|T^nx\|^2&=\langle T^nx,\,T^nx\rangle\\ &=\langle T^{2n}x,\,x\rangle\\ &\leq\|T^{2n}x\|\|x\| \end{align*} which implies that \begin{equation} \label{SS} \mathcal{N}(T^n)\subset\mathcal{N}(T^{2n}). \end{equation} We can write $T^4=T^3$ as ${T^3(T-I)=0}$ which is equivalent to $T^4(T-I)=0$ and we have \begin{equation*} T^4(T-I)=0\implies T^2(T-I)=0\implies T(T-I)=0 \end{equation*} and we have shown that $T^4=T^3$ implies that $T$ is an orthogonal projection operator. When $T^3=T^2$ can be written as $T^2(T-I)=0$ which implies that $T(T-I)=0$ and we have shown that $T$ is an orthogonal projection operator in in this case as well. Note that the above easily can be made more general.