I have the following exercise:
Let $V : H_A → H_A ⊗ H_E$ denote an isometry and $|ψ_E⟩ ∈ H_E$ a normalized vector. Show that there exists a unitary $U : H_A ⊗ H_E → H_A ⊗ H_E$ such that $$U(1_{H_A} ⊗ |ψ_E⟩) = V.$$ My question is how do i choose to decompose my general isometry and what is the strategy of choosing the specific Unitary?
$V$ is a linear map from $H_A \rightarrow H_A \otimes H_E$, so we can pick arbitrary orthonormal bases $\{|i_A\rangle\}_{i=1}^{\text{dim}(H_A)}$ and $\{|i_E\rangle\}_{i=1}^{\text{dim}(H_E)}$ for $H_A$ and $H_E$ respectively and write \begin{equation} V = \sum_{ijk} V_{ij,k} |i_A j_E\rangle \langle k_A| \tag{1} \end{equation} where $V_{ij,k} = \langle i_A j_E | V |k_A \rangle$. We can write $U$ in a similar way, but to make things easy we use another, different basis for $H_E$. Let us construct an orthonormal basis which contains $|\psi_E\rangle:= |\psi_E^1\rangle$: $$ \{|\psi_E^1\rangle, |\psi_E^2\rangle, \dots |\psi_E^{ \text{dim}(H_E) }\rangle\}. \tag{2} $$ We can always do this via Gram-Schmidt procedure, and we have that $\langle \psi_E | \psi_E^{i}\rangle = 0$ for $i>1$. Now, we express any $U$ uniquely in terms of this basis: \begin{align} U = \sum_{ijk\ell} U_{ij,k\ell}|i_A j_E\rangle \langle k_A \psi_E^{\ell}|, \tag{3} \end{align} with $U_{ij,k\ell} = \langle i_A j_E |U| k_A \psi_E^{\ell}\rangle $. Using the basis of Eq. (2) we compute: \begin{align} U(\mathbb{I}_A \otimes |\psi_E\rangle) &= \sum_{ijk\ell} \langle\psi_E^{\ell}| \tag{4}\psi_E\rangle U_{ij,k\ell}|i_A j_E\rangle \langle k_A| \\ &=\sum_{ijk} U_{ij,k1}|i_A j_E\rangle \langle k_A|. \tag{5} \end{align} This means that you can start with an operator $U$ that satisfies $U_{ij,k1} = V_{ij,k}$, and then fill out the remaining elements of $U$ ($\ell=2,3,\dots$) in any manner such that $U$ as a whole is unitary.
