I'm assuming that V is a finite-dimensional vector space with an inner product , so that the adjoint of T is given by . The operator is selfadjoint, and is invertible if and only if T is (reason: take determinants in the equation ). Also, for each ,

So there is a (unique) isometric linear map from the range of R to the range of T defined by . If T and R are invertible then these ranges are the whole of V, and is the unique isometry on V such that .

Now suppose that T is not invertible. Then is still an isometry, but neither its domain nor its range are the whole of V. However, since the ranges of T and R have the same dimension, so do their orthogonal complements. Therefore there exists a (nonzero) isometry from to . The operators and are then distinct isometries on the whole of V, and they both have the property that . So in this case, S is not unique.