Let \(V\) be a finite-dimensional inner product space and \(E\) the orthogonal projection of \(V\) onto some subspace.
Prove that, for any positive number \(c\), the operator \(cI + E\) is positive.
Express in terms of \(E\) a self-adjoint linear operator \(T\) such that \(T^2 = I + E\).
Solution
1. For every \(c>0\), the operator \(cI+E\) is positive.
Because \(E\) is an orthogonal projection, \[ E^*=E. \] Hence \(cI+E\) is self-adjoint. Also, for every nonzero \(x\), \[ \begin{aligned} ((cI+E)x\mid x) &=c\|x\|^2+(Ex\mid x)\\ &=c\|x\|^2+\|Ex\|^2\\ &>0. \end{aligned} \] Thus \(cI+E\) is positive.
2. Construct a self-adjoint square root of \(I+E\).
Seek a polynomial in \(E\) of the form \[ T=I+bE \] with real \(b\). Since \(E^2=E\), \[ T^2 =I+(2b+b^2)E. \] We want \(T^2=I+E\), so \(b\) must satisfy \[ b^2+2b=1. \] Taking \[ b=\sqrt2-1 \] gives the desired coefficient. Therefore \[ \boxed{T=I+(\sqrt2-1)E}. \] Since \(E^*=E\) and the coefficient is real, \(T\) is self-adjoint. Finally, \[ T^2 =I+\bigl(2(\sqrt2-1)+(\sqrt2-1)^2\bigr)E =I+E. \]