Let \(V\) be a complex inner product space and \(T\) a self-adjoint linear operator on \(V\). Show that
\(||\alpha + iT\alpha|| = ||\alpha - iT\alpha||\) for every \(\alpha\) in \(V\).
\(\alpha + iT\alpha = \beta + iT\beta\) if and only if \(\alpha = \beta\).
\(I + iT\) is non-singular.
\(I - iT\) is non-singular.
Now suppose \(V\) is finite-dimensional, and prove that \[ U = (I - iT)(I + iT)^{-1} \] is a unitary operator; \(U\) is called the Cayley transform of \(T\). In a certain sense, \(U = f(T)\), where \(f(x) = (1 - ix)/(1 + ix)\).
Solution
Goal. For a self-adjoint operator \(T\), prove the five facts about \(I \pm iT\) and the Cayley transform \(U = (I - iT)(I + iT)^{-1}\).
1. (a) \(\|\alpha + iT\alpha\| = \|\alpha - iT\alpha\|\).
1.1. \(\|\alpha + iT\alpha\|^2 = \langle \alpha + iT\alpha, \alpha + iT\alpha\rangle\).
definition of the norm.
1.2. Expanding: \(\|\alpha\|^2 + \langle \alpha, iT\alpha\rangle + \langle iT\alpha, \alpha\rangle + \|T\alpha\|^2\).
bilinearity of the inner product.
1.3. \(\langle \alpha, iT\alpha\rangle = -i\langle \alpha, T\alpha\rangle\) and \(\langle iT\alpha, \alpha\rangle = i\langle T\alpha, \alpha\rangle = i\langle \alpha, T\alpha\rangle\) (since \(T\) is self-adjoint).
conjugate-linearity in the first slot and self-adjointness.
1.4. Hence \(\|\alpha + iT\alpha\|^2 = \|\alpha\|^2 + \|T\alpha\|^2\).
the cross terms cancel.
1.5. Similarly \(\|\alpha - iT\alpha\|^2 = \|\alpha\|^2 + \|T\alpha\|^2\).
same computation with \(-i\) in place of \(i\).
1.6. Hence the two norms are equal.
2.4 and 2.5.
2. (b) \(\alpha + iT\alpha = \beta + iT\beta\) iff \(\alpha = \beta\).
2.1. \(\alpha + iT\alpha = \beta + iT\beta\) implies \((\alpha - \beta) + iT(\alpha - \beta) = 0\).
subtract.
2.2. Then \(\|(\alpha - \beta) + iT(\alpha - \beta)\| = 0\), so by (a) \(\|(\alpha-\beta) - iT(\alpha-\beta)\| = 0\).
apply (a) to \(\alpha - \beta\).
2.3. Adding the two equations \((\alpha-\beta) + iT(\alpha-\beta) = 0\) and \((\alpha-\beta) - iT(\alpha-\beta) = 0\) gives \(2(\alpha - \beta) = 0\), so \(\alpha = \beta\).
the second equation follows from 2.2 (a vector of norm zero is zero).
3. (c) \(I + iT\) is nonsingular.
3.1. If \((I + iT)\alpha = 0\), then \(\alpha + iT\alpha = 0\), so \(\alpha = 0\) by (b).
- with \(\beta = 0\).
3.2. Hence \(I + iT\) is injective, so nonsingular.
a linear operator with trivial kernel is nonsingular.
4. (d) \(I - iT\) is nonsingular.
4.1. If \((I - iT)\alpha = 0\), then \(\alpha - iT\alpha = 0\), so \(\alpha = 0\) by (b) (with \(T\) replaced by \(-T\), which is also self-adjoint).
same argument as (c).
5. (e) \(U = (I - iT)(I + iT)^{-1}\) is unitary (for \(V\) finite-dimensional).
5.1. \(U^* = (I + iT)^{-1}(I - iT)^* = (I + iT)^{-1}(I + iT)\).
\((I - iT)^* = I + iT\) (since \(T^* = T\)), and \((I + iT)^{-1}\) commutes with \(I - iT\) (both are polynomials in \(T\)).
5.2. \(U^* U = (I + iT)^{-1}(I + iT)(I - iT)(I + iT)^{-1} = I\).
\((I + iT)\) and \((I - iT)\) commute (polynomials in \(T\)), so the middle terms cancel.
5.3. Hence \(U\) is unitary.
\(U^* U = I\) (and \(U U^* = I\) similarly).
6. Q.E.D.
<1>1–<1>5 prove (a)–(e).