Past exam of the algebra and analysis I for mathematics and philosophy course of the University of Oxford / 2013 / Algebra / 2

The adjoint operator $T^{\ast }$ is characterized byThis identity works with either convention for which argument of the complex inner product is linear, provided the convention is used consistently.

For uniqueness, subtract the two given identities to get $⟨v,(S-T^{\ast })w⟩=0$ for every $v,w$. Fix $w$ and choose $v=(S-T^{\ast })w$. Then positive definiteness givesso $(S-T^{\ast })w=0$. As $w$ was arbitrary, $\boxed{S=T^{\ast }}$. The decisive fact is that a vector orthogonal to every vector, including itself, must be zero.

First suppose $E=E^{\ast }$. For $u\in U=\mathrm{im}E$, choose $v$ with $Ev=u$. For $w\in W=\ker E$,Thus $U\subseteq W^{\perp }$. The direct sum of vector spaces gives $\dim U=\dim V-\dim W$, while the orthogonal complement has $\dim W^{\perp }=\dim V-\dim W$. Equal finite dimensions imply $U=W^{\perp }$.

Conversely, suppose $U=W^{\perp }$. Decompose arbitrary vectors as $v=u+w$ and $v^{\prime }=u^{\prime }+w^{\prime }$, with $u,u^{\prime }\in U$ and $w,w^{\prime }\in W$. Since $E$ is the indicated projection, $Ev=u$ and $E{v}^{\prime }=u^{\prime }$. Orthogonality makes the cross terms vanish, soTherefore $E$ satisfies the defining identity of its own adjoint operator. By uniqueness from part (a), $E=E^{\ast }$. HenceThis is the characterization of an orthogonal projection among projections along a complementary subspace.