(a) Let H be a subspace of R⁴ spanned by the vectors v₁ = (1, –2, 5, –3), v₂ = (2, 3, 1, –4), v₃ = (3, 8, –3, –5). Then find a basis and dimension of H, and extend the basis of H to a basis of R⁴. (10 marks)

(b) Let T : R³ → R³ be a linear operator

Question

(a) Let H be a subspace of R⁴ spanned by the vectors v₁ = (1, –2, 5, –3), v₂ = (2, 3, 1, –4), v₃ = (3, 8, –3, –5). Then find a basis and dimension of H, and extend the basis of H to a basis of R⁴. (10 marks)

(b) Let T : R³ → R³ be a linear operator and B = {v₁, v₂, v₃} be a basis of R³ over R. Suppose that Tv₁ = (1, 1, 0), Tv₂ = (1, 0, –1), Tv₃ = (2, 1, –1). Find a basis for the range space and null space of T. (10 marks)

(c) Discuss the continuity of the function

f(x) = { 1/(1–e^(–1/x)), x ≠ 0
       { 0, x = 0

for all values of x. (10 marks)

(d) Expand ln(x) in powers of (x–1) by Taylor's theorem and hence find the value of ln(1·1) correct up to four decimal places. (10 marks)

(e) Find the equation of the right circular cylinder which passes through the circle x² + y² + z² = 9, x – y + z = 3. (10 marks)

UPSC Answer Check · Accepted Answer

(a) Let c₁v₁ + c₂v₂ + c₃v₃ = 0. Using the coordinates of v₁, v₂, v₃:

c₁ + 2c₂ + 3c₃ = 0
–2c₁ + 3c₂ + 8c₃ = 0
5c₁ + c₂ – 3c₃ = 0
–3c₁ – 4c₂ – 5c₃ = 0

From the first equation, c₁ = –2c₂ – 3c₃. Substitute in the second:

–2(–2c₂ – 3c₃) + 3c₂ + 8c₃ = 0
⇒ 4c₂ + 6c₃ + 3c₂ + 8c₃ = 0
⇒ 7c₂ + 14c₃ = 0
⇒ c₂ = –2c₃.

Then c₁ = –2(–2c₃) – 3c₃ = c₃. The third and fourth equations are then automatically satisfied. Hence

c₃v₁ – 2c₃v₂ + c₃v₃ = 0
⇒ v₁ – 2v₂ + v₃ = 0
⇒ v₃ = 2v₂ – v₁.

So v₃ is dependent on v₁ and v₂, while v₁ and v₂ are not scalar multiples, hence independent. Therefore

Basis of H = {v₁, v₂} = {(1, –2, 5, –3), (2, 3, 1, –4)},
dim H = 2.

To extend this basis to R⁴, take e₃ = (0, 0, 1, 0), e₄ = (0, 0, 0, 1). Consider

αv₁ + βv₂ + γe₃ + δe₄ = 0.

Comparing the first two coordinates:

α + 2β = 0
–2α + 3β = 0.

The determinant is 1·3 – 2(–2) = 7 ≠ 0, so α = β = 0. Then the third and fourth coordinates give γ = 0 and δ = 0. Thus the four vectors are independent, and since R⁴ has dimension 4,

Basis of R⁴ = {v₁, v₂, e₃, e₄}.

(b) Since B = {v₁, v₂, v₃} is a basis of R³, every vector is of the form a v₁ + b v₂ + c v₃. By linearity,

T(a v₁ + b v₂ + c v₃) = aTv₁ + bTv₂ + cTv₃
= a(1, 1, 0) + b(1, 0, –1) + c(2, 1, –1)
= (a + b + 2c, a + c, –b – c).

The range of T is spanned by Tv₁, Tv₂, Tv₃. But

Tv₃ = (2, 1, –1) = (1, 1, 0) + (1, 0, –1) = Tv₁ + Tv₂.

Hence Tv₃ is dependent on Tv₁ and Tv₂, which are independent. Therefore

Basis of range T = {(1, 1, 0), (1, 0, –1)}, rank T = 2.

For the null space, set T(a v₁ + b v₂ + c v₃) = 0:

a + b + 2c = 0
a + c = 0
–b – c = 0.

From a + c = 0, a = –c. From –b – c = 0, b = –c. The first equation is then automatically satisfied. Thus

kernel T = {–c v₁ – c v₂ + c v₃ : c ∈ R} = span{v₃ – v₁ – v₂}.

Therefore

Basis of null space T = {v₃ – v₁ – v₂}, nullity T = 1.

Rank-nullity theorem checks: 2 + 1 = 3 = dim R³.

(c) For x ≠ 0, the function is a quotient of continuous functions,

f(x) = 1/(1 – e^(–1/x)).

The denominator vanishes only if e^(–1/x) = 1, i.e. –1/x = 0, which is impossible for finite x ≠ 0. Hence f is continuous for every x ≠ 0.

Now examine x = 0.

Right-hand limit:

x → 0⁺ ⇒ –1/x → –∞ ⇒ e^(–1/x) → 0.

Therefore

lim_x→0⁺ f(x) = 1/(1 – 0) = 1.

Left-hand limit:

x → 0⁻ ⇒ –1/x → +∞ ⇒ e^(–1/x) → +∞.

Therefore

lim_x→0⁻ f(x) = 1/(1 – ∞) = 0.

The left and right limits are different, so lim_x→0 f(x) does not exist. Since f(0) = 0, the function is discontinuous at x = 0. It is continuous on (–∞, 0) and (0, ∞), and is left-continuous at 0 but not right-continuous.

Hence f is continuous for all x ≠ 0 and has a jump discontinuity at x = 0.

(d) Let f(x) = ln x. By Taylor’s theorem about x = 1,

f(x) = f(1) + f′(1)(x – 1) + f″(1)(x – 1)²/2! + f‴(1)(x – 1)³/3! + …

Now f(1) = 0. Also,

f′(x) = 1/x ⇒ f′(1) = 1,
f″(x) = –1/x² ⇒ f″(1) = –1,
f‴(x) = 2/x³ ⇒ f‴(1) = 2.

In general,

fⁿ(x) = (–1)^(n–1)(n – 1)!/xⁿ,
so fⁿ(1) = (–1)^(n–1)(n – 1)!.

Hence

ln x = (x – 1) – (x – 1)²/2 + (x – 1)³/3 – (x – 1)⁴/4 + …
= Σ (n=1 to ∞) (–1)^(n+1)(x – 1)ⁿ/n.

This expansion is valid for |x – 1| < 1, and at x = 2 it converges conditionally.

Put x = 1.1, so x – 1 = 0.1. Then

ln(1.1) = 0.1 – 0.1²/2 + 0.1³/3 – 0.1⁴/4 + 0.1⁵/5 – 0.1⁶/6 + 0.1⁷/7 – …

= 0.1 – 0.005 + 0.0003333333 – 0.000025 + 0.000002 – 0.0000001667 + 0.0000000143 – …

Summing,

ln(1.1) ≈ 0.0953101798.

Therefore, correct up to four decimal places,

ln(1·1) = 0.0953.

(e) The given circle is the intersection of the sphere

x² + y² + z² = 9

and the plane

x – y + z = 3.

The plane has normal vector n = (1, –1, 1). The distance of the plane from the origin is

|3|/√(1² + (–1)² + 1²) = 3/√3 = √3.

The sphere has radius 3, so the radius of the circle of intersection is

r = √(3² – (√3)²) = √(9 – 3) = √6.

The centre of this circle is the foot of the perpendicular from the origin to the plane:

c = (3/3)(1, –1, 1) = (1, –1, 1).

For a right circular cylinder to contain this circle, the plane of the circle must be perpendicular to the cylinder axis. Hence the axis is parallel to n = (1, –1, 1), and it passes through c = (1, –1, 1). The radius of the cylinder is √6.

Let X = (x, y, z). The perpendicular distance from X to the axis line through c in direction n is given by

d² = |(X – c) × n|² / |n|².

Now

X – c = (x – 1, y + 1, z – 1),

and

(X – c) × n = (y + z, z – x, –x – y).

Also |n|² = 3. Since d = √6, we get

(y + z)² + (z – x)² + (x + y)² = 6 × 3 = 18.

Simplifying,

2(x² + y² + z² + xy + yz – xz) = 18,

so

x² + y² + z² + xy + yz – xz = 9.

(a) Let H be a subspace of R⁴ spanned by the vectors v₁ = (1, –2, 5, –3), v₂ = (2, 3, 1, –4), v₃ = (3, 8, –3, –5). Then find a…

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