Proof of Self-Adjointness, Unitarity and Orthogonal Projections Through the Adjoint in Finite Dimensions

Every linear operator $X$ on $V$ has an adjoint $X^{*}$ by claim 2 of Uniqueness of the Adjoint, and Existence in Finite Dimensions, unique by claim 1 of the same theorem and characterised by

Write $0_{V}$ for the zero vector and $x-y$ for $x+(-y)$.

A preliminary. If $x,y\in V$ satisfy $\langle x,v\rangle=\langle y,v\rangle$ for every $v\in V$, then $x=y$. Indeed, by additivity in the first argument (claim 1 of Elementary Properties of a Complex Inner Product) and conjugate homogeneity in the first argument (claim 2 of the same lemma) applied to $-y=(-1)y$, where $(-1)y=-y$ by Elementary Identities in a Vector Space and $\overline{-1}=-1$ because $-1$ is a real number and claim 1 of Properties of Complex Conjugation and Modulus applies, we get $\langle x-y,v\rangle=\langle x,v\rangle-\langle y,v\rangle=0$ for every $v\in V$. Taking $v=x-y$ and applying claim 4 of Elementary Properties of a Complex Inner Product gives $x-y=0_{V}$, that is $x=y$.

Claim 1. Suppose $T^{*}=T$. Then for all $u,v\in V$,

so $T$ is self-adjoint. Conversely, suppose $T$ is self-adjoint, that is $\langle T(u),v\rangle=\langle u,T(v)\rangle$ for all $u,v\in V$. Then $T$ itself satisfies the property characterising the adjoint of $T$, so the uniqueness part of claim 1 of Uniqueness of the Adjoint, and Existence in Finite Dimensions gives $T=T^{*}$.

Claim 2. Suppose first that $T^{*}T=\mathrm{id}_{V}$ and $TT^{*}=\mathrm{id}_{V}$; by the definition of the product and the identity operator this says $T^{*}(T(x))=x$ and $T(T^{*}(x))=x$ for every $x\in V$. Applying the characterising property of $T^{*}$ with the vector $T(u)$ in place of $u$,

and every $v\in V$ satisfies $v=T\bigl(T^{*}(v)\bigr)$, so $T$ is surjective. Hence $T$ is unitary.

Conversely suppose $T$ is unitary. For all $u,v\in V$, the characterising property and the preservation of the inner product give

Since $v\in V$ was arbitrary, the preliminary gives $T^{*}(T(u))=u$; as $u$ was arbitrary, $T^{*}T=\mathrm{id}_{V}$. Now let $v\in V$. By surjectivity there is $w\in V$ with $v=T(w)$, and then $T^{*}(v)=T^{*}(T(w))=w$, so $T\bigl(T^{*}(v)\bigr)=T(w)=v$. As $v$ was arbitrary, $TT^{*}=\mathrm{id}_{V}$.

Claim 3. By claim 1 applied to $P$, the condition $P^{*}=P$ holds if and only if $P$ is self-adjoint. Hence the conjunction of $PP=P$ and $P^{*}=P$ holds if and only if $P$ is self-adjoint and $PP=P$, which by that definition is exactly the condition that $P$ is an orthogonal projection.