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Suppose v₁,v₂, vₘ is linearly independent in V and w ∈ V . Prove that dim span(v+w,v₂+w,..vₘ+w)≥ m-1

User Bobzhang
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Final answer:

To prove dim span(v+w,v₂+w,..vₘ+w)≥ m-1, we can show that the vectors v₁+w, v₂+w, ..., vₘ+w are linearly independent.

Step-by-step explanation:

To prove dim span(v+w,v₂+w,..vₘ+w)≥ m-1, let's start by considering a linear combination of the vectors v₁+w, v₂+w, ..., vₘ+w:

(c₁(v₁+w) + c₂(v₂+w) + ... + cₘ(vₘ+w)), where c₁, c₂, ..., cₘ are scalars.

Expanding this expression, we get: c₁v₁ + c₂v₂ + ... + cₘvₘ + (c₁ + c₂ + ... + cₘ)w.

Since v₁, v₂, ..., vₘ are linearly independent, c₁v₁ + c₂v₂ + ... + cₘvₘ = 0 only if all the scalars c₁, c₂, ..., cₘ are zero. This implies that (c₁ + c₂ + ... + cₘ)w ≠ 0.

Therefore, the vectors v₁+w, v₂+w, ..., vₘ+w are linearly independent, and the dimension of their span is at least m-1.

User Ajerneck
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