[proofplan] The adjoint converts the [inner product](/page/Inner%20Product) of two images under $V$ into an inner product on $H$ involving $V^*V$. Thus the equation $V^*V=I_H$ is exactly the operator form of preservation of inner products. We use the same-run characterization of [Hilbert space](/page/Hilbert%20Space) isometries by inner products to pass between norm preservation and inner product preservation, then use nondegeneracy of the Hilbert inner product to identify the operators. [/proofplan] [step:Convert the isometry hypothesis into preservation of inner products] Assume first that $V$ is a Hilbert space isometry. Since $V\in\mathcal{L}(H,K)$ and $V$ preserves the Hilbert-space norm, [citetheorem:8700] \begin{align*} (Vx,Vy)_K=(x,y)_H \end{align*} for all $x,y\in H$. [/step] [step:Use the adjoint identity to identify $V^*V$ with the identity] By the definition of the Hilbert space adjoint, for every $x,y\in H$, \begin{align*} (Vx,Vy)_K=(x,V^*Vy)_H. \end{align*} Combining this with the inner-product preservation obtained above gives \begin{align*} (x,V^*Vy)_H=(x,y)_H \end{align*} for all $x,y\in H$. Therefore \begin{align*} (x,(V^*V-I_H)y)_H=0 \end{align*} for all $x,y\in H$. Fix $y\in H$ and define $a_y\in H$ by \begin{align*} a_y=(V^*V-I_H)y. \end{align*} Taking $x=a_y$ in the preceding identity gives \begin{align*} \|a_y\|_H^2=(a_y,a_y)_H=0. \end{align*} Hence $a_y=0$. Since $y\in H$ was arbitrary, $(V^*V-I_H)y=0$ for every $y\in H$, and therefore $V^*V=I_H$. [guided] We want to turn the equality of inner products into equality of operators. The adjoint is the mechanism that moves $V$ from the first [inner product space](/page/Inner%20Product%20Space) back to $H$: for every $x,y\in H$, the definition of $V^*:K\to H$ gives \begin{align*} (Vx,Vy)_K=(x,V^*Vy)_H. \end{align*} From the previous step, the isometry hypothesis implies \begin{align*} (Vx,Vy)_K=(x,y)_H \end{align*} for all $x,y\in H$. Substituting the adjoint formula into this equality yields \begin{align*} (x,V^*Vy)_H=(x,y)_H \end{align*} for all $x,y\in H$. By linearity in the first argument and conjugate-linearity in the second argument in the complex case, this is equivalently \begin{align*} (x,(V^*V-I_H)y)_H=0 \end{align*} for all $x,y\in H$. Now fix an arbitrary vector $y\in H$ and define \begin{align*} a_y=(V^*V-I_H)y. \end{align*} The displayed identity says that every vector $x\in H$ is orthogonal to $a_y$ in the sense that $(x,a_y)_H=0$. In particular, we may choose $x=a_y$. This gives \begin{align*} \|a_y\|_H^2=(a_y,a_y)_H=0. \end{align*} The positivity axiom of the Hilbert-space inner product implies $a_y=0$. Since $y$ was arbitrary, $(V^*V-I_H)y=0$ for every $y\in H$, which is precisely the operator identity \begin{align*} V^*V=I_H. \end{align*} [/guided] [/step] [step:Derive norm preservation from the adjoint equation] Conversely, assume \begin{align*} V^*V=I_H. \end{align*} For every $x\in H$, the adjoint identity gives \begin{align*} \|Vx\|_K^2=(Vx,Vx)_K=(x,V^*Vx)_H. \end{align*} Using $V^*V=I_H$, we obtain \begin{align*} \|Vx\|_K^2=(x,x)_H=\|x\|_H^2. \end{align*} Both norms are non-negative [real numbers](/page/Real%20Numbers), so taking the non-negative square root gives \begin{align*} \|Vx\|_K=\|x\|_H \end{align*} for every $x\in H$. Thus $V$ is a Hilbert space isometry. This proves both implications. [/step]