Proof of Algebraic Properties of the Adjoint in Finite Dimensions

Every linear operator on $V$ has an adjoint, and by the uniqueness part of claim 1 of Uniqueness of the Adjoint, and Existence in Finite Dimensions a linear operator $A$ on $V$ equals the adjoint of a linear operator $X$ on $V$ as soon as

Each of the operators $S^{*}+T^{*}$, $\overline{\lambda}\,S^{*}$, $T^{*}S^{*}$ and $\mathrm{id}_{V}$ is a linear operator on $V$ by Sums, Scalar Multiples, Composites and the Identity are Linear Operators, so it suffices in each case to verify the displayed identity. Let $u,v\in V$.

Claim 1. Using the definition of the sum of operators, additivity in the first argument (claim 1 of Elementary Properties of a Complex Inner Product), the defining property of $S^{*}$ and $T^{*}$, and additivity in the second argument (condition 2 of Complex Inner Product Space),

Hence $S^{*}+T^{*}=(S+T)^{*}$.

Claim 2. Using conjugate homogeneity in the first argument (claim 2 of Elementary Properties of a Complex Inner Product), the identity $\overline{\overline{\lambda}}=\lambda$ from claim 1 of Properties of Complex Conjugation and Modulus, the defining property of $S^{*}$, and homogeneity in the second argument (condition 3 of Complex Inner Product Space),

Hence $\overline{\lambda}S^{*}=(\lambda S)^{*}$.

Claim 3. Using the definition of the product of operators and the defining property of $T^{*}$ and then of $S^{*}$,

Hence $T^{*}S^{*}=(ST)^{*}$.

Claim 4. By conjugate symmetry (condition 1 of Complex Inner Product Space) applied twice, the defining property of $S^{*}$, and the identity $\overline{\overline{z}}=z$ from claim 1 of Properties of Complex Conjugation and Modulus,

Thus $S$ satisfies the identity characterising the adjoint of $S^{*}$, so $S=(S^{*})^{*}$.

Claim 5. Since $\mathrm{id}_{V}(u)=u$ and $\mathrm{id}_{V}(v)=v$, we have $\langle \mathrm{id}_{V}(u),v\rangle=\langle u,v\rangle=\langle u,\mathrm{id}_{V}(v)\rangle$, so $\mathrm{id}_{V}=(\mathrm{id}_{V})^{*}$.