Question

Prove or disprove: if U₁, U₂, Uz are linearly independent, then uj + U₂ + U₃, U₁ + U₃, U₂ – Uz are also linearly independent.

Answer 1

If U₁, U₂, Uz are linearly independent, then uj + U₂ + U₃, U₁ + U₃, U₂ – Uz are also linearly independent.

Step-by-step explanation:

Claim: If U₁, U₂, Uz are linearly independent, then uj + U₂ + U₃, U₁ + U₃, U₂ – Uz are also linearly independent.

Proof:

We know that a set of vectors is linearly independent if the only solution to the equation:

a₁U₁ + a₂U₂ + a₃Uz = 0

is a₁ = a₂ = a₃ = 0.

Let's assume that uj + U₂ + U₃, U₁ + U₃, U₂ – Uz are linearly dependent. This means that there exist scalars b₁, b₂, b₃, not all zero, such that:

b₁(uj + U₂ + U₃) + b₂(U₁ + U₃) + b₃(U₂ – Uz) = 0.

Expanding the equation, we get:

b₁uj + b₁U₂ + b₁U₃ + b₂U₁ + b₂U₃ + b₃U₂ - b₃Uz = 0.

Now, we can rearrange the terms:

uj(b₁) + U₁(b₂) + U₂(b₁ + b₃) + U₃(b₁ + b₂ + b₃) - Uz(b₃) = 0.

Since U₁, U₂, Uz are linearly independent, the only solution to this equation is:

b₁ = b₂ = b₃ = 0.

This means that uj + U₂ + U₃, U₁ + U₃, U₂ – Uz are indeed linearly independent.