Analysis Theorems

Proposition 8.1. Every endomorphism \(T:V\to V\) on a finite dimensional inner product space \(V\) has a unique adjoint \(T^*\) so that \((Tu,v)=(u,T^*v)\).

Proof. Choose an orthonormal basis, yielding an isometry with the standard inner product space on \(\CC ^n\). Now if \(T\) is represented by the matrix \(A\), \(T^*\) is represented by \(\bar {A}^t\) as we have \((Au,v) = (Au)^t\cdot \bar {v} = u^t\overline {\bar {A}^tv} = (u,\bar {A}^tv)\). □

Proposition 8.2. If \(V\) is a finite dimensional inner product space, \(T\) a linear operator, \(\ker (T) = T^*V^\perp \), \((W^\perp )^\perp = W, W \subset V\).

Proof. For the first part, \(v \in \ker (T)\) iff \(0 = (Tv,u) = (v,T^*u)\) for all \(u\), and the second part follows from the fact that we can have an orthonormal basis for \(W\), and extend it to an orthonormal basis for \(V\), the remaining vectors making up a basis for \(W^\perp \). □

Definition 8.3. An endomorphism is normal if it commutes with its adjoint, and is self-adjoint if it is its own adjoint.

Theorem 8.4. A normal endomorphism \(T\) on a finite dimensional vector space \(V\) satisfies \(\Vert Tv\Vert = \Vert T^*v\Vert \), \(\ker T = \ker T^*\), \(TV = T^*V\), \(T-\lambda 1_V\) is normal, \(Tv = \lambda v \leftrightarrow T^*v = \bar {\lambda }v\), and if \(u,v\) are eigenvectors of \(T\) with different eigenvalues, then \((u,v) =0\)

Proof. For the first, we have \((Tv,Tv) = (v,T^*Tv)=(v,(T^*)^*T^*v) = (T^*v,T^*v)\). The second immediately follows, and the third follows from the second and Proposition 8.2. For the fourth we have \((T-\lambda 1_V)(T-\lambda 1_V)^* = T^*T-\bar {\lambda }T-\lambda T^*+|\lambda |^21_V = (T-\lambda 1_V)^*(T-\lambda 1_V)\). The fifth follows from the fourth and the second, and the last follows from the fact that \(\lambda _1(u,v) = (Tu,v) = (u,T^*v) = \lambda _2(u,v)\). □

Proposition 8.5. If \(T\) is an endomorphism on a finite dimensional vector space \(V\) and \(\Vert Tv\Vert = \Vert T^*v\Vert \) then \(T\) is normal.

Theorem 8.6 (Spectral Theorem for \(\CC \)). An endomorphism \(T\) of an finite dimensional complex inner product space \(V\) is normal iff there is an orthonormal basis of eigenvectors, ie. there is a diagonal matrix representing it.

Proof. If there is an orthonormal basis of eigenvectors \(v_1,\dots v_n\), \(T\) with respect to this basis is diagonal, so commutes with its adjoint. Conversely if \(T\) is normal, its minimal polynomial has a root so it has a nontrivial eigenvector \(v\). Then by Theorem 8.4, if \(u\) is in the orthogonal complement, \((Tu,v) = (u,T^*v)= \bar {\lambda }(u,v) = 0\), so \(T\) acts within the orthogonal complement and we can repeat until we get an orthonormal basis of eigenvectors. □

Lemma 8.7. The eigenvalues of a self-adjoint endomorphism \(T\) on a vector space \(V\) are real.

Theorem 8.8 (Spectral Theorem for \(\RR \)). An endomorphism \(T\) of a finite dimensional real inner product space \(V\) is self-adjoint iff there is an orthonormal basis of eigenvectors, ie. there is a diagonal matrix representing it.

Proof. If there is an orthonormal basis of eigenvectors, \(T\) is represented by a diagonal matrix, which is self-adjoint. Conversely if \(T\) is self-adjoint, we can tensor with \(\CC \) and find an eigenvector, but its eigenvalue is real, and \(T\) is real, so its conjugate is an eigenvector as well, but then either their sum is a real eigenvector or it is \(0\) in which case we can multiply our original eigenvector by \(i\) to get a real eigenvector. In either case, there must be a real eigenvector, so by Theorem 8.4 \(T\) acts within the orthogonal complement again, so we can repeat until we get an orthonormal basis of eigenvectors. □