From Euclidean to Hilbert Spaces - Edoardo Provenzi

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      The residual vector of the projection of v on S is r = v − PSv = (2i, 0, 0) and thus d(v, PSv)2 = ‖r‖2 = 4. The most general vector in S is image and image. This confirms that PSv is the vector in S with the minimum distance from v in relation to the Euclidean norm.

      5) r is orthogonal to S, which is generated by u and w, hence (u, w, r) is a set of orthogonal vectors in

3, that is, an orthogonal basis of
3. To obtain an orthonormal basis, we then simply divide each vector by its norm:

image

      6) Decomposition: image.

      The vector with the heaviest weight in the reconstruction of a is thus r̂: this vector gives the best rough approximation of a. By calculating the vector sum of this rough representation and the other two vectors, we can reconstruct the “fine details” of a, first with ŵ and then with û.

       Exercise 1.2

      Let M(n,

) be the space of n × n complex matrices. The application ϕ : M(n,
) × M(n,
) →
is defined by:

image

      where image denotes the adjoint matrix of B and tr is the matrix trace. Prove that ϕ is an inner product.

       Solution to Exercise 1.2

      The distributive property of matrix multiplication for addition and the linearity of the trace establishes the linearity of ϕ in relation to the first variable.

      Now, let us prove that ϕ is Hermitian. Let A = (ai,j)1≼i,j≼n and B = (bi,j)1≼i,j≼n be two matrices in M(n,

). Let image be the coefficients of the matrix B† and let image be the coefficients of A†.

      This gives us:

image

      Thus, ϕ is a sesquilinear Hermitian form. Furthermore, ϕ is positive:

image image

      Thus, ϕ is an inner product.

       Exercise 1.3

      Let E = ℝ[X] be the vector space of single variable polynomials with real coefficients. For P, Q ∈ E, take:

image

      1) Remember that image means that Ǝ a, C > 0 such that |t − t0| < a |f(t)| ≼ C |g(t)|. Prove that for all P, Q ∈ E, this is equal to:

image

      and:

image

      Use this result to deduce that Φ is definite over E × E.

      2) Prove that Φ is an inner product over E, which we shall note 〈 , 〉.

      3) For n ∈ ℕ, let Tn be the n-th Chebyshev polynomial, that is, the only polynomial such that ∀θ ∈ ℝ, Tn(cos θ) = cos(nθ). Applying the substitution t = cos θ, show that (Tn)n∈ℕ is an orthogonal family in E. Hint: use the trigonometric formula [1.13]:

      4) Prove that for all n ∈ ℕ, (T0, . . . , Tn) is an orthogonal basis of ℝn[X], the vector space of polynomials in ℝ[X] of degree less than or equal to n. Deduce that (Tn)n∈ℕ is an orthogonal basis in the algebraic sense: every element in E is a finite linear combination of elements in the basis of E.

       Solution to Exercise 1.3

      1) We write image. Since P and Q are polynomials, the function image is continuous in a neighborhood V1(1) and thus, according to the Weierstrass theorem, it is bounded in this neighborhood, that is, Ǝ C1 > 0 such that image. Similarly, the function image is continuous in a neighborhood V2(−1), thus Ǝ C2 > 0 such that image. This gives us:

image

      and:

image

      This implies that

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