if $T$ is self-adjoint and $T^2=T\Rightarrow T=Id$?

Question

V is a finite dimension inner product space. $T\in End(V)$ is a self adjoint operator for which $T^{2}=T$ and I'm asked to prove a subspace $U\subset V$ exists for which T is the orthogonal projection onto U.

My line of thought is that since T is self adjoint it must be invertible, and since $T^{2}=T\Rightarrow T^{-1}TT=T^{-1}T\Rightarrow T=Id_V$. But then no such subspace U exists. What is the mistake in my reasoning?

Answer 1

It's not true that self-adjoint implies invertible. Self-adjoint only means that $T = T^*$ (that is, $(Tx, y) = (x, Ty)$ for all $x, y \in V$.

All of the following matrices are self-adjoint with respect to the standard inner product and satisfy the equation $A^2 = A$: $$ \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix} $$ where $r = \operatorname{Rank}(A) \le n$. In fact, if $T^2 = T$ then there is a basis in which $T$ has the above form. If $T$ is self-adjoint, then there is an orthonormal basis in which $T$ has the above form.

The key property of self-adjoint operators is that if $U$ is $T$-invariant, then so is $U^\perp$. To see this, take $y \in U^\perp$. Then by definition, $(x,y) = 0$ for all $x \in U$. Now apply $T$. This gives $(x,Ty) = (Tx, y) = 0$ because $Tx \in U$. Since this works for all $x$, $Ty \in U^\perp$. So $U^\perp$ is $T$-invariant.

Now, if $T$ is the projection onto $U$ then $U = \operatorname{im}(T)$. So you need to show that $T : V \to \operatorname{im}(T)$ is an orthogonal projection.

Answer 2

Define $U:=TV$.

From $T^2=T$ you get, that $T$ is a projection from $V$ into $U$, since for $u\in U$ there exists $v\in V$ such that $u=Tv$ and $$Tu=T^2v=Tv=u.$$

To prove that $T$ is an orthogonal projection, you consider $u^\bot\in U^\bot$ and an arbitrary $v\in V$ $$ \langle Tu^\bot,v\rangle = \langle u^\bot,\underbrace{ Tv}_{\in U}\rangle=0. $$ Since $v$ was arbitrary it holds for all $v$ and you get $Tu^\bot=0$. Therefore $TU^\bot=\{0\}$ and your projection is orthogonal.

As you see from the definition of $U$ you have a special case if $T\equiv 0$ then $U=\{0\}$ and if $T$ is invertible then $U=V$. If you need a nontrivial subspace $\{0\}\subsetneq U\subsetneq V$ then you need the restriction that $T$ is neither $0$ nor invertible.

Answer 3

$T$ is self-adjoint means: There exists unitary $U$ and diagonal matrix $D$ such that $T=U^*DU$ (By spectral theorem).

Hence if $d_i$ are the diagonal entries of $D$ (= eigenvalues of $T$), the $(d_i) ^2=d_i$ for all $i$. This says $d_i =0\,$ or $\,1$ (NOT necessarily $1$).