Past exam of the linear algebra course of the University of Oxford / 2017 / 3

Use the convention that the inner product is conjugate-linear in its first argument and linear in its second, as in part (b). The adjoint operator is characterized byExistence and uniqueness follow from the finite-dimensional representation of linear functionals by the inner product. More concretely, choose an orthonormal basis $e_i$ and let $a_{ij}=⟨e_i,T{e}_j⟩$. Thenwhich defines $T^{\ast }$ uniquely and verifies the required identity by expansion in that basis. Thus its matrix is the conjugate transpose, proved here rather than assumed. A self-adjoint operator satisfies $T^{\ast }=T$, equivalently $⟨Tv,w⟩=⟨v,Tw⟩$ for every pair of vectors.

Proceed by induction on $\dim V$. The result is immediate for dimension zero. In positive dimension, the characteristic polynomial has a root over the complex numbers, so there is a nonzero eigenvector $v$ with eigenvalue $\lambda$. Normalize it to have norm one. Self-adjointness givesso $\lambda$ is real.

The orthogonal complement $v^{\perp }$ is an invariant subspace: if $w\perp v$, thenThe restriction of $T$ to $v^{\perp }$ is again self-adjoint, since the same inner product identity holds on that subspace. By induction it has an orthonormal basis of eigenvectors. Adjoining $v$ givesThe existence of the first eigenvector is the essential use of the complex numbers in this proof of the finite-dimensional spectral theorem.

Let $e_1,\dots ,e_d$ be the given orthonormal basis, with $T{e}_j={\lambda }_j{e}_j$. From the defining identity of the adjoint operator,These coordinates prove $T^{\ast }{e}_j={\overline{\lambda }}_j{e}_j$ without assuming a matrix formula. Thereforefor every basis vector. Thus $T$ is a normal operator, andThe condition of possessing an orthonormal basis of eigenvectors is already part of the hypothesis; real eigenvalues alone would not imply self-adjointness for an arbitrary linear transformation.

The integral is conjugate-linear in $f$, linear in $g$, and conjugate-symmetric, by the corresponding properties of the integral and of complex conjugation. Also,If a polynomial $f$ is not zero, there is a real point $x_0$ with $f(x_0)\ne 0$: a nonzero polynomial cannot vanish at every real point. By continuity, $|f|$ has a positive lower bound on a sufficiently small closed interval about $x_0$. The continuous function $e^{-x^2}$ also has a positive lower bound there. The integral over this interval is strictly positive. HenceTogether with the finiteness allowed in the question, these properties prove that the formula defines an inner product on $V=\mathbb{C}[x]$.

Differentiate the expression from the source before applying any finite-dimensional result. The exponential factors cancel:This is a polynomial whenever $f$ is a polynomial, and its polynomial degree is at most that of $f$. Differentiation and multiplication are linear, so $T$ is a linear transformation $V\to V$, and every $V_n$ is an invariant subspace.

For self-adjointness, put $w(x)=e^{-x^2}$. Then $w^{\prime }=-2xw$ and $wTf=(w{f}^{\prime }{)}^{\prime }-2wf$. Integration by parts therefore givesThe discarded endpoint terms are $w\overline{f^{\prime }}g$ and $w\overline{f}{g}^{\prime }$. They vanish at both infinities because exponential decay dominates every polynomial. The same decay justifies the intermediate integrals. ConsequentlyHere the question's self-adjointness is in the sense of an adjoint on an inner product space: $T$ itself represents the adjoint on the polynomial space. Each finite-dimensional restriction $T{|}_{V_n}$ is a self-adjoint operator. No choice of operator domain in a completed Hilbert space is needed for this argument. The resulting Gaussian polynomial operator will produce Hermite polynomials in part (iv).

If $f\in V_n\setminus V_{n-1}$, write $f(x)=a_n{x}^n+\cdots$ with $a_n\ne 0$. In the expression $Tf=f^{\prime \prime }-2x{f}^{\prime }-2f$, the first term has polynomial degree at most $n-2$, while the other two contributeThus, if $Tf=\lambda f$, comparison of the leading coefficients givesFor $n=0$ take $V_{-1}=\{0\}$; the same result follows from $T1=-2$. This identifies the eigenvalue conditional on the existence of the eigenvector; existence is established next.

Construct compatible choices in the nested spaces, rather than choosing unrelated orthonormal bases for different $V_n$. Set $V_{-1}=\{0\}$ and defineThe orthogonal complement is taken for the inner product from part (i). Since $\dim V_n=n+1$ and $\dim V_{n-1}=n$, the space $L_n$ has dimension one and $V_n=V_{n-1}\oplus L_n$.

It is an invariant subspace. Indeed, if $v\in L_n$ and $u\in V_{n-1}$, part (ii) givesbecause $Tu\in V_{n-1}$. Also $Tv\in V_n$. Hence $Tv\in L_n$. Any nonzero vector in this one-dimensional space is consequently an eigenvector. Choose a unit vector $g_n\in L_n$. It has polynomial degree exactly $n$: membership in $V_{n-1}$ would make it orthogonal to itself and thus zero. Part (iii) now supplies its eigenvalue.

For $m<n$, $g_m\in V_{n-1}$ and $g_n\perp V_{n-1}$, so the choices are pairwise orthogonal. We have provedAn explicit compatible construction is the Gram-Schmidt process on $1,x,x^2,\dots$:The leading term of $r_n$ is $x^n$, so it is never zero. This fixes the positive real leading coefficient and yields the normalized Hermite polynomials. For example, the Gaussian integral givesEvery polynomial is a finite linear combination of these eigenvectors, since $g_0,\dots ,g_n$ form an orthonormal basis of $V_n$ for each $n$.