Question

Qno1. Please solve this.
Let vi = {(a1, a2, a3, an)}; a1, a2, a3, a4€R,
a1+a2+a3+a4=0}
Then with usual Vector Space.
Prove or dsprove​

Answer 1

Answer:

which is again solved only by λ = µ = 0.

Step-by-step explanation:

Problem 1. In each part, use the Lagrange interpolation formula to construct the polynomial of

smallest degree whose graph contains the following points:

(b) (−4, 24),(1, 9),(3, 3)

(d) (−3, −30),(−2, 7),(0, 15),(1, 10)

Solution. (b) The polynomial is

24 (x − 1)(x − 3)

(−4 − 1)(−4 − 3) + 9

(x + 4)(x − 3)

(1 + 4)(1 − 3) + 3

(x + 4)(x − 1)

(3 + 4)(3 − 1) = −3x + 12.

(d) The polynomial is

−30

(x + 2)(x)(x − 1)

(−3 + 2)(−3)(−3 − 1) + 7

(x + 3)(x)(x − 1)

(−2 + 3)(−2)(−2 − 1) + 15

(x + 3)(x + 2)(x − 1)

(0 + 3)(0 + 2)(0 − 1)

+ 10

(x + 3)(x + 2)(x)

(1 + 3)(1 + 2)(1) = 2x

3 − x

2 − 6x + 15.

Problem 2. Let u and v be distinct vectors of a vector space V . Show that if { u, v } is a basis

for V and a, b ∈ F are non-zero scalars, then both { u + v, au } and { au, bv } are also bases for V .

Solution. In each case, we have the correct number of vectors for a basis, so it suffices to check

linear independence. In the first case, if

λ(u + v) + µ(au) = (λ + aµ)u + λv = 0,

then because { u, v } is a basis, hence linearly independent,

λ + aµ = 0 and λ = 0.

This is solved only by λ = µ = 0.

In the second case, if

λ(au) + µ(bv) = (aλ)u + (bµ)v = 0,

then by the same reasoning,

aλ = 0 and bµ = 0,

which is again solved only by λ = µ = 0.