Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 17F (Quadratic Mathematics)

Use an inner product linear in its first argument. The adjoint operator ${\alpha }^{\ast }$ is the unique linear operator such thatIn a finite-dimensional inner product space, existence follows by choosing an orthonormal basis: the matrix of the adjoint is the conjugate transpose of the matrix of $\alpha$. Uniqueness follows from nondegeneracy of the inner product.

If $\alpha (W)\subseteq W$, then for $w\in W$ and $v\in W^{\perp }$,Thus ${\alpha }^{\ast }v\in W^{\perp }$. Conversely, if ${\alpha }^{\ast }(W^{\perp })\subseteq W^{\perp }$, the same equality gives $⟨\alpha w,v⟩=0$ for every $v\in W^{\perp }$, so $\alpha w\in (W^{\perp }{)}^{\perp }=W$. This proves the adjoint criterion for an invariant orthogonal complement.

The equality $(W^{\perp }{)}^{\perp }=W$ uses the finite-dimensional setting appropriate here. In a general Hilbert space it holds for closed subspaces, and an adjoint is assured for bounded operators. Without closedness the converse only implies invariance of the closure. For example in ${\ell }^2$, take $W$ to be the finite-support sequences and $\alpha v=⟨v,e_1⟩(2^{-j}{)}_{j\ge 1}$. Then $W^{\perp }=\{0\}$ is automatically adjoint-invariant, but $\alpha e_1\notin W$. Thus the finite-dimensional or closed-subspace hypothesis cannot be omitted from an infinite-dimensional reading.

A normal operator satisfies ${\alpha }^{\ast }\alpha =\alpha {\alpha }^{\ast }$. Moving an operator across the inner product givesOne of these norms is zero exactly when the other is zero. Therefore

For any complex $\lambda$, the shifted linear operator $T=\alpha -\lambda I$ is normal: expansion of $T^{\ast }T-T{T}^{\ast }$ leaves just ${\alpha }^{\ast }\alpha -\alpha {\alpha }^{\ast }=0$. Apply part (i) to this shifted normal operator:Hence a nonzero vector is an eigenvector of $\alpha$ with eigenvalue $\lambda$ if and only if it is an eigenvector of ${\alpha }^{\ast }$ with eigenvalue $\bar{\lambda }$. The eigenspaces coincide, with conjugate eigenvalues.

Part (ii) shows that ${\alpha }^{\ast }v=\bar{\lambda }v$ for every $v\in E_{\lambda }$, so $E_{\lambda }$ is invariant under ${\alpha }^{\ast }$. Apply the adjoint criterion for an invariant orthogonal complement with ${\alpha }^{\ast }$ in place of $\alpha$, using $({\alpha }^{\ast }{)}^{\ast }=\alpha$. It follows thatThe same argument interchanging $\alpha$ and ${\alpha }^{\ast }$ shows that $E_{\lambda }^{\perp }$ is invariant under both operators, a fact needed for the induction in part (iv).

Proceed by induction on $\dim V$. The zero-dimensional case has the empty orthonormal basis. In positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a unit eigenvector $e$ of $\alpha$, say with eigenvalue $\lambda$. Part (ii) makes it an eigenvector of ${\alpha }^{\ast }$ as well. Thus the line spanned by $e$ and its orthogonal complement are invariant under both operators by the introductory adjoint argument.

On $\{e{\}}^{\perp }$, the restriction of ${\alpha }^{\ast }$ is the adjoint of the restriction of $\alpha$, and the restrictions still commute. The restricted linear operator is therefore normal. By induction it has an orthonormal basis of eigenvectors. Adjoining $e$ proves an orthonormal eigenbasis exists for $V$, establishing the finite-dimensional spectral theorem by construction.

For the final deduction, use this orthonormal basis $e_j$ and write $\alpha e_j={\lambda }_j{e}_j$. DefineThe real nonnegative diagonal entries make $\beta$ a Hermitian operator, and the modulus-one diagonal entries make $\gamma$ a unitary operator. Both are diagonal in the same orthonormal basis, so they commute and $\beta \gamma e_j={\lambda }_j{e}_j$ even when ${\lambda }_j=0$. This proves the required factorization without assuming invertibility.

Conversely, if $\alpha =\beta \gamma$ with ${\beta }^{\ast }=\beta$, ${\gamma }^{\ast }\gamma =\gamma {\gamma }^{\ast }=I$, and $\beta \gamma =\gamma \beta$, then ${\alpha }^{\ast }={\gamma }^{\ast }\beta$ andThus $\alpha$ is normal. Normality is equivalent to a commuting Hermitian–unitary factorization, with the Hermitian factor even selectable nonnegative. This is the polar decomposition with a unitary extension on the kernel.