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A real symmetric matrix has real eigenvalues, orthogonal eigenspaces for distinct eigenvalues, and an orthonormal basis of eigenvectors. From diagonalizability, apply the Gram-Schmidt process within each eigenspace; all its linear combinations stay in that eigenspace. This gives $A=Q\Lambda Q^T$ with an orthogonal matrix $Q$ and real diagonal $\Lambda$.

View a possible complex eigenvector $v$ using the Hermitian inner product. For a real symmetric matrix, $A^{†}=A$, so $v^{†}Av$ is real. Since $v^{†}Av=\lambda v^{†}v$ and $v^{†}v>0$, every eigenvalue is real.

For eigenvectors $v,w$ with distinct eigenvalues $\lambda ,\mu$, symmetry givesso $v^{†}w=0$. In particular, real eigenvectors are orthogonal in the real dot product.

Use the permitted diagonalizability assumption to take bases of the real eigenspaces whose union spans ${\mathbb{R}}^n$. Apply the Gram-Schmidt process within each eigenspace. Its linear combinations remain eigenvectors of that same eigenvalue; vectors in different eigenspaces are already orthogonal. The resulting union is an orthonormal basis of eigenvectors, giving an orthogonal diagonalization of a real symmetric matrix.

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

Use the convention that a Hermitian metric $h$ is complex-linear in its first argument and conjugate-linear in its second. It is a smoothly varying positive-definite Hermitian form on each fibre. A connection on a vector bundle is a complex-linear operator $D:\Gamma (E)\to {\Omega }^1(X,E)$ satisfying $D(as)=da\otimes s+aDs$ for smooth complex functions $a$. Metric compatibility means, for every real vector field $V$,This is the metric-compatible connection condition with the sesquilinear convention fixed.

Apply the smooth Gram-Schmidt process to a local frame to obtain an $h$-orthonormal smooth frame $s_1,\dots ,s_r$. Positivity guarantees that all normalization denominators are nonzero and depend smoothly on the base point. Write $D{s}_j={\sum }_i{A}_{ij}{s}_i$. Differentiating $h(s_i,s_j)={\delta }_{ij}$ and using compatibility givesfor every real $V$. Thus $\boxed{A(V{)}^{†}=-A(V)}$: the connection matrix is skew-Hermitian. This smooth unitary frame for a Hermitian connection is generally not holomorphic; the requested local-frame assertion requires only a smooth frame.

The holomorphic dual vector bundle $E^{\ast }$ is obtained by dualizing fibres and using transition matrices $g_{ij}^{-T}$ when $E$ has transitions $g_{ij}$. Their entries are holomorphic because matrix inversion is holomorphic on $G{L}_r(\mathbb{C})$. Their cocycle property follows from preservation of the fibrewise evaluation pairing, defining the natural holomorphic bundle with fibre $(E_x{)}^{\ast }$.

The conjugate vector bundle $\overline{E}$ has the same underlying real fibres but opposite scalar action: $\lambda \cdot \overline{v}=\overline{\overline{\lambda }v}$. Its smooth transition matrices are $\overline{g_{ij}}$; they need not be holomorphic on $X$. The conjugate bundle is used here as a smooth complex bundle, whereas $E^{\ast }$ is holomorphic.

Define the smooth dual tensor $H$ byIt is complex-linear in each tensor factor, because conjugating the second bundle converts the conjugate-linearity of $h$ into linearity. Thus $H\in \Gamma ((E\otimes \overline{E}{)}^{\ast })$.

The conjugate connection is defined on real vector fields by ${\overline{D}}_V(\overline{s})=\overline{D_{V}s}$ and extended complex-linearly on the conjugate bundle. The tensor product connection isIts dual connection $D_0$ is uniquely characterized byThe Leibniz rule makes this a genuine connection on $(E\otimes \overline{E}{)}^{\ast }$. Evaluate it on the decomposable tensor $u=s\otimes \overline{t}$:Decomposable tensors span every fibre. Hence this tensor-valued one-form vanishes precisely when the compatibility identity holds for every $V,s,t$:This is metric compatibility as parallelism of a Hermitian tensor.

Start with the assumed basis of eigenvectors, grouping them by eigenvalue. In each eigenspace, apply the Gram-Schmidt process, which replaces a linearly independent list by an orthonormal list with the same span. Every new vector stays within that eigenspace, since an eigenspace is a vector subspace.

By part (iv), different eigenspaces are orthogonal. The combined lists therefore form an orthonormal basis of the entire space, still consisting of eigenvectors. Let $U$ have these vectors as columns. Then $U$ is a unitary matrix, soThis gives the required unitary diagonalization of a normal matrix, including repeated eigenvalues.

Throughout this question the Banach spaces and Hilbert spaces are complex. Separability is an additional standing hypothesis, not part of the general definition of a Hilbert space. Here and below unit ball means the closed unit ball. An open unit ball in positive finite dimension is not compact. In finite dimension, equivalence of norms identifies the closed unit ball with a closed bounded subset of Euclidean space, so the Heine-Borel theorem gives compactness in the norm topology.

In an infinite-dimensional Hilbert space, the Gram-Schmidt process constructs an infinite orthonormal sequence $(e_n)$. Put $f_n=e_n/\sqrt{2}$. Then $\Vert f_n\Vert =1/\sqrt{2}\le 1$ and, for $m\ne n$, $\Vert f_n-f_m{\Vert }^2=(1+1)/2=1$. No subsequence is a Cauchy sequence, contradicting sequential compactness. Thus the closed unit ball is norm compact exactly in finite dimension.

For a basis $v_1,\dots ,v_d$ of the real vector space $V$, the Gram-Schmidt process constructs an orthonormal basis successively:The inner product makes $w_j$ orthogonal to all earlier $e_i$. Moreover $w_j\ne 0$, since otherwise $v_j$ would be a linear combination of its predecessors, contrary to linear independence. Induction gives $\mathrm{span}(e_1,\dots ,e_j)=\mathrm{span}(v_1,\dots ,v_j)$. Starting with a basis of a linear subspace $U$ and extending it to a basis of $V$ gives the same construction with the first $r=\dim U$ vectors spanning $U$.

The orthogonal complement is $U^{\perp }=\{v\in V:⟨v,u⟩=0\text{ for every }u\in U\}$. For every $v\in V$, set $u={\sum }_{i=1}^r⟨v,e_i⟩e_i$. Then $u\in U$ and $v-u\in U^{\perp }$. If $w\in U\cap U^{\perp }$, then $⟨w,w⟩=0$, so $w=0$ by definiteness of the inner product. Thus the direct sum isThis also covers $U=0$ and $U=V$.

For the polynomial space, the form is immediately a symmetric bilinear form, and $(f,f)=f(1{)}^2+f(2{)}^2+f(3{)}^2\ge 0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, so this form is an inner product for $n=1,2$. For $n\ge 3$, the nonzero polynomial $(x-1)(x-2)(x-3)$ has $(f,f)=0$. Therefore, among the stipulated positive integers,

This is the Gram-Schmidt process applied degree by degree. Inductively, subtracting projections onto $p_0,\dots ,p_n$ makes $p_{n+1}$ orthogonal to every earlier polynomial. Since every subtracted polynomial has degree at most $n$, the leading coefficient of the monic $q_{n+1}$ remains one, so $p_{n+1}$ is a real monic orthogonal polynomial of degree exactly $n+1$.

Now take $q_{n+1}=x{p}_n$. For $k\le n-2$, symmetry of the inner product givesbecause $x{p}_k$ has degree at most $n-1$. Only the $p_n$ and $p_{n-1}$ projections remain, yielding the three-term recurrence for monic orthogonal polynomialswhereThe second identity uses $x{p}_{n-1}=p_n$ plus lower-degree terms.

For the symmetric Legendre inner product, parity gives ${\alpha }_n=0$. Starting with $p_0=1$, one obtains $p_1=x$, thenThus the first four monic Legendre polynomials are