Proof of Uniqueness of the Adjoint, and Existence in Finite Dimensions

Throughout we use the four conditions of Complex Inner Product Space, the facts in Elementary Properties of a Complex Inner Product, and the properties of finite sums in Properties of Finite Sums of Vectors and Properties of Finite Sums. Write $0_{V}$ for the zero vector. In claim 2, $[n]$ denotes the initial segment determined by $n$, and by the definition of a tuple the $n$-tuple $e$ is a map from $[n]$ to $V$ with components $e_{k}$, so the finite sums below are formed from maps on $[n]$ as required. Neither claim uses completeness of $V$ or boundedness of $T$.

Claim 1. Suppose $A$ and $B$ are both adjoints of $T$, so that $\langle A(u),v\rangle=\langle u,T(v)\rangle=\langle B(u),v\rangle$ for all $u,v\in V$. Fix $u\in V$ and put $x=A(u)$ and $y=B(u)$, so that $\langle x,v\rangle=\langle y,v\rangle$ for every $v\in V$. 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,

Taking $v=x+(-y)$ gives $\langle x+(-y),x+(-y)\rangle=0$, so $x+(-y)=0_{V}$ by claim 4 of Elementary Properties of a Complex Inner Product, that is $x=y$. As $u$ was arbitrary, $A=B$.

Claim 2. Define a map $A:V\to V$ by

$A$ is a linear operator. Let $u,u'\in V$ and $\lambda\in\mathbb{C}$. By additivity in the second argument (condition 2 of Complex Inner Product Space) we have $\langle T(e_{k}),u+u'\rangle=\langle T(e_{k}),u\rangle+\langle T(e_{k}),u'\rangle$ for every $k\in[n]$, so by the vector space identity $(\alpha+\beta)x=\alpha x+\beta x$ and claim 2 of Properties of Finite Sums of Vectors,

Similarly, by homogeneity in the second argument (condition 3) and the identity $(\alpha\beta)x=\alpha(\beta x)$ together with claim 3 of Properties of Finite Sums of Vectors,

Thus $A$ is a linear map from $V$ to $V$, hence a linear operator on $V$.

$A$ is an adjoint of $T$. Let $u,v\in V$. By the second identity of claim 6 of Properties of Finite Sums of Vectors, applied with the scalars $c_{k}=\langle T(e_{k}),u\rangle$ and the vectors $e_{k}$, and then conjugate symmetry (condition 1 of Complex Inner Product Space),

On the other hand, since $e$ is an orthonormal basis, claim 1 of Orthonormal Expansion and Parseval's Identity in Finite Dimensions gives $v=\sum_{k=1}^{n}\langle e_{k},v\rangle e_{k}$; applying $T$ and using claim 4 of Properties of Finite Sums of Vectors with the homogeneity of $T$,

so by the first identity of claim 6 of Properties of Finite Sums of Vectors,

The two $n$-tuples in $\mathbb{C}$ being summed coincide by the commutativity of multiplication in the field of complex numbers, so their finite sums coincide and $\langle A(u),v\rangle=\langle u,T(v)\rangle$. Thus $A$ satisfies the condition of the definition of an adjoint.

By claim 1 the adjoint is unique, so $T^{*}=A$ and the displayed formula holds. Finally, that condition involves only $\langle\cdot,\cdot\rangle$ and $T$; hence the operator obtained from any other orthonormal basis of $V$ by the same construction is also an adjoint of $T$ and therefore equals $T^{*}$.

Boundedness. Since $V$ has the orthonormal basis $e$, Every Linear Operator on a Space with a Finite Orthonormal Basis is Bounded applies to every linear operator on $V$; in particular $T$ and $T^{*}$ are bounded linear operators.