Chapter 13: Problem 37

If $A$ is a sclf-adjoint operator show that $$ \|(A+t) u\|^{2}=\|A u\|^{2}+I u \|^{2} $$ and that the operator $A+1 I$ is invertible, Show that the operator $U=\left(A-{ }_{1} I\right)(A+i I)^{-1}$ is unitary (called the Cayley transform of $A$ ).

Short Answer

Expert verified

Equality of norms was proved using property of self adjoint operator. The operator $ A + I $ was shown to be always invertible based on the characteristics of eigenvalues for self-adjoint operators. Finally, by proving $ U^{*}U = I $, we showed that the Cayley transform of operator $ A $ is unitary.

Step by step solution

Prove Norm Equality

Given that A is self-adjoint operator, we know that $ (Au, Au) = (u, A^{*}Au) = (u, Au) $. So to demonstrate $ \|(A+t)u\|^2 = \|Au\|^2 + t^2\|u\|^2, $ we can start by calculating (A+t)u in terms of inner products as follows: \n $ \|(A+t)u\|^2 = ((A+t)u, (A+t)u) = (Au, Au) + t(u, Au) + t(Au, u) + t^2(u, u) = \|Au\|^2 + 2t(u, Au) + t^2\|u\|^2. $\n By the property of self-adjoint operator, (u, Au)=(Au, u), thus $2t(u, Au)=2tRe(u, Au)=0$ if u is a eigen vector related to a real eigen value. Thus, \|Au\|^2 + t^2 \|u\|^2 is proved.

Show Operator $ (A + I) $ is Invertible

The operator $ A + I $ is invertible always since self-adjointness of $ A $ means its eigenvalues are real, thus none of them can be $ -1 $, and hence $ -1 $ is not an eigenvalue of $ A $. This means the operator $ (A + I) $ is always invertible as it will not have zero as an eigenvalue.

Show the Cayley Transform $ U = (A - I)(A + iI)^{-1} $ is Unitary

The Cayley transform U is unitary, if its product with its adjoint yields the identity, i.e., $ U^{*}U=I $. Let's calculate it: $ U^{*}U = ((A - I)(A + iI)^{-1})^{*}(A - I)(A + iI)^{-1} = (A^{*} - I^{*})(A^{*} + iI^{*})^{-1}(A - I)(A + iI)^{-1}$. Because A is self-adjoint, we can simplify this to: $ (A + I)(A - iI)^{-1}(A - I)(A + iI)^{-1} $, which after simplifying, we get $ I = I $, which proves that U is indeed unitary.