[proofplan] We construct $\alpha^*$ using the matrix approach: choose orthonormal bases for $V$ and $W$, let $A$ be the matrix of $\alpha$, and define $\alpha^*$ to be the linear map with matrix $A^\dagger = \bar{A}^\top$. The adjoint identity $(\alpha(v), w)_W = (v, \alpha^*(w))_V$ is verified by direct computation in coordinates. Uniqueness follows from the non-degeneracy of the inner product. [/proofplan] [step:Construct $\alpha^*$ via the conjugate transpose matrix in orthonormal bases] Choose orthonormal bases $(e_1, \dots, e_n)$ for $V$ and $(f_1, \dots, f_m)$ for $W$. Let $A \in M_{m \times n}(\mathbb{F})$ be the matrix of $\alpha$ in these bases, so $\alpha(e_j) = \sum_{k=1}^m A_{kj}\,f_k$. Define $\alpha^*: W \to V$ to be the [linear map](/page/Linear%20Map) with matrix $A^\dagger = \bar{A}^\top \in M_{n \times m}(\mathbb{F})$ in the same bases, so $\alpha^*(f_k) = \sum_{j=1}^n \overline{A_{kj}}\,e_j$. [/step] [step:Verify the adjoint identity $(\alpha(v), w)_W = (v, \alpha^*(w))_V$] For $v = \sum_j v_j e_j \in V$ and $w = \sum_k w_k f_k \in W$, the coordinate vectors are $v = (v_1, \dots, v_n)^\top$ and $w = (w_1, \dots, w_m)^\top$. Since the bases are orthonormal, the inner products reduce to standard coordinate inner products: \begin{align*} (\alpha(v), w)_W = \sum_{k=1}^m \overline{(Av)_k}\, w_k = \overline{(Av)}^\top w = \bar{v}^\top \bar{A}^\top w = \bar{v}^\top A^\dagger w. \end{align*} On the other hand: \begin{align*} (v, \alpha^*(w))_V = \bar{v}^\top (A^\dagger w). \end{align*} These are equal, confirming the adjoint identity. [guided] The computation works because orthonormal bases diagonalise the inner product: in orthonormal coordinates, $(x, y) = \bar{x}^\top y$ (the standard Hermitian inner product on $\mathbb{F}^n$). The key algebraic step is the identity $\overline{(Av)}^\top = \bar{v}^\top \bar{A}^\top$. To see this: $(Av)_k = \sum_j A_{kj} v_j$, so $\overline{(Av)_k} = \sum_j \overline{A_{kj}}\,\bar{v}_j$. Then: \begin{align*} \overline{(Av)}^\top w = \sum_k \overline{(Av)_k}\, w_k = \sum_{j,k} \bar{v}_j\, \overline{A_{kj}}\, w_k = \bar{v}^\top (\bar{A}^\top w). \end{align*} Since $A^\dagger = \bar{A}^\top$, we get $(\alpha(v), w)_W = \bar{v}^\top A^\dagger w = (v, \alpha^*(w))_V$. Over $\mathbb{R}$, the conjugate is trivial, and $A^\dagger = A^\top$: the adjoint of a linear map between real inner product spaces is represented by the transpose matrix. Over $\mathbb{C}$, the conjugate transpose appears because the inner product is sesquilinear. [/guided] [/step] [step:Prove uniqueness by the non-degeneracy of the inner product] Suppose $\beta: W \to V$ also satisfies $(\alpha(v), w)_W = (v, \beta(w))_V$ for all $v \in V$ and $w \in W$. Then $(v, \alpha^*(w) - \beta(w))_V = 0$ for all $v \in V$. Taking $v = \alpha^*(w) - \beta(w)$: \begin{align*} \|\alpha^*(w) - \beta(w)\|^2 = 0, \end{align*} so $\alpha^*(w) = \beta(w)$ for every $w \in W$. Therefore $\alpha^* = \beta$. [/step]