(a) Prove that any set of n linearly independent vectors in a vector space V of dimension n constitutes a basis for V. 10 (b)…

(a) n independent vectors form a basis

Let V be a finite-dimensional vector space over a field F with dim V = n. Let S = {v₁, v₂, ..., vₙ} be a linearly independent set in V. We must prove that S spans V; then S will be a basis.

Suppose, on the contrary, that S does not span V. Then there exists a vector w ∈ V such that w ∉ span(S). Consider the set S ∪ {w}. We claim it is linearly independent. Let c₁v₁ + c₂v₂ + ... + cₙvₙ + cw = 0. If c ≠ 0, then w = −(c₁/c)v₁ − (c₂/c)v₂ − ... − (cₙ/c)vₙ, so w ∈ span(S), a contradiction. Hence c = 0. The equation then becomes c₁v₁ + c₂v₂ + ... + cₙvₙ = 0. Since S is linearly independent, c₁ = c₂ = ... = cₙ = 0. Thus S ∪ {w} is linearly independent.

But S ∪ {w} contains n + 1 linearly independent vectors. By the dimension theorem, in an n-dimensional vector space no linearly independent set can contain more than n vectors. This is a contradiction. Therefore S must span V.

Since S is linearly independent and spans V, S is a basis for V. Hence any set of n linearly independent vectors in an n-dimensional vector space V is a basis of V.

Final answer: S is a basis of V.

(b) Finding T(2, 4)

Let e₁ = (1, 0) and e₂ = (0, 1) in ℝ². The given data are T(e₁) = T(1, 0) = (1, 2, 3), T(1, 1) = T(e₁ + e₂) = (−3, 2, 8). By linearity, T(e₁ + e₂) = T(e₁) + T(e₂). Therefore T(e₂) = T(e₁ + e₂) − T(e₁) = (−3, 2, 8) − (1, 2, 3) = (−4, 0, 5).

Now (2, 4) = 2e₁ + 4e₂. Hence, again using linearity, T(2, 4) = 2T(e₁) + 4T(e₂) = 2(1, 2, 3) + 4(−4, 0, 5) = (2, 4, 6) + (−16, 0, 20) = (−14, 4, 26).

Final answer: T(2, 4) = (−14, 4, 26).

(c) Limit of (eˣ + x)^(1/x)

Let L = lim (x → ∞) (eˣ + x)^(1/x). For x > 0, eˣ + x > 0, so we may take logarithms. Put y = (eˣ + x)^(1/x). Then ln y = (1/x) ln(eˣ + x). Now eˣ + x = eˣ(1 + x e⁻ˣ), so ln(eˣ + x) = ln(eˣ) + ln(1 + x e⁻ˣ) = x + ln(1 + x e⁻ˣ). Therefore ln y = 1 + [ln(1 + x e⁻ˣ)]/x.

As x → ∞, we have x e⁻ˣ → 0. Using the standard limit ln(1 + t) ~ t as t → 0, ln(1 + x e⁻ˣ) ~ x e⁻ˣ. Thus [ln(1 + x e⁻ˣ)]/x ~ e⁻ˣ → 0. Hence ln y → 1. By continuity of the exponential function, y → e¹ = e.

Final answer: lim (x → ∞) (eˣ + x)^(1/x) = e.

(d) Convergence of the integral

We examine I = ∫₀² dx/(2x − x²). The denominator is 2x − x² = x(2 − x). Thus the integrand has singularities at both endpoints x = 0 and x = 2. The integral is improper at both ends.

First consider the behaviour near x = 0. For 0 < x ≤ 1, 2 − x ≤ 2, so x(2 − x) ≤ 2x. Therefore 1/[x(2 − x)] ≥ 1/(2x). But ∫₀¹ dx/(2x) diverges to +∞ because ∫ dx/x diverges near 0. Hence by the comparison test, the part of the integral near x = 0 diverges.

Similarly, near x = 2, put t = 2 − x. Then for small t > 0, 2x − x² = x(2 − x) = (2 − t)t ≤ 2t, so 1/[x(2 − x)] ≥ 1/(2t). Again ∫₀^δ dt/(2t) diverges to +∞. Hence the part near x = 2 also diverges.

For completeness, using partial fractions, 1/[x(2 − x)] = (1/2)(1/x + 1/(2 − x)). So an antiderivative is F(x) = (1/2) ln(x/(2 − x)). Then ∫_ε^(2−δ) dx/(2x − x²) = F(2 − δ) − F(ε) = (1/2) ln(((2 − δ)(2 − ε))/(δε)). As ε → 0⁺ and δ → 0⁺, the expression ln(1/(δε)) tends to +∞. Therefore the improper integral does not converge.

Final answer: The integral diverges to +∞; it is not convergent.

(e) Locus of the centre of the sphere

Let the variable plane meet the coordinate axes at A = (α, 0, 0), B = (0, β, 0), C = (0, 0, γ), where α, β, γ are nonzero. The equation of the plane in intercept form is X/α + Y/β + Z/γ = 1. Since the plane passes through the fixed point (a, b, c), a/α + b/β + c/γ = 1. (1)

Let the centre of the sphere through O, A, B, C be P = (x, y, z). Since the sphere passes through O = (0, 0, 0) and A = (α, 0, 0), the centre is equidistant from O and A: x² + y² + z² = (x − α)² + y² + z². This gives x² = x² − 2αx + α², so 2αx = α², hence x = α/2, that is, α = 2x.

Similarly, equidistance from O and B gives y = β/2, that is, β = 2y, and equidistance from O and C gives z = γ/2, that is, γ = 2z.

Substitute α = 2x, β = 2y, γ = 2z into (1): a/(2x) + b/(2y) + c/(2z) = 1. Multiplying by 2, a/x + b/y + c/z = 2.

Thus the locus of the centre (x, y, z) of the sphere is a/x + b/y + c/z = 2, with xyz ≠ 0. Equivalently, multiplying by xyz, a yz + b zx + c xy = 2xyz.

Final answer: The locus is a/x + b/y + c/z = 2, or equivalently a yz + b zx + c xy = 2xyz.