Suppose $V$ is a finite dimensional Hilbert Space with a subspace $W$ Suppose $T:W\to V$ be a linear operator which preserves inner product i.e $\forall w_1,w_2\in W$ we have $\langle T(w_1),T(w_2)\rangle=\langle w_1,w_2\rangle$ I need to show the existence of a linear map $U:V\to V$ which is unitary and extends $T$.
suppose $\dim V=n,\dim W=k$
I chose a basis for $W$ say $\{w_1,\dots,w_k\}$ by extension theorem we can extend this basis to a basis of $V$ say $\{w_1,\dots,w_k,v_1,\dots,v_{n-k}\}$
Then I can define $U(w_i)=T(w_i)\forall i=1,\dots,k,U(v_i)=0\forall i=1,\dots,n-k$.
Is my extension correct?will be glad to have a hints or solution.
I know this is an old question, but I figured I'd post my answer for future people like me. I initially thought to tensor spaces together, but we want to add the dimensions of spaces, not multiply them.
As T preserves inner products, it must be unitary, so its eigenvectors $\{|i⟩\}$ span W. The key is to look at at the operator as the sum of its tensored eigenvectors, $$T=\sum_{i=1}^{k}|i⟩⟨i|$$ If we want our extended operator U to behave equivalently on space W, then its subset of eigenvectors spanning W must be the eigenvectors of T. Therefore, $$U=\sum_{j=1}^{n}|j⟩⟨j|\ where |j⟩ = |i⟩ for\ j=i=1..k$$ The question remains what acceptable values are for eigenvectors $k+1$ through $n$. Any choice that keeps $\{|j⟩\}$ an orthonormal set is fine, such as unit vectors along dimensions $k+1,...,n$. This will result in U acting as identity everywhere on V except subspace W.