Answer to Question #324394 in Linear Algebra for Talifhani

Question

Let S = {u1,u2,u3} be a basis for the vector space V. Show that T = {w1,w2,w3} is also

a basis for V, where w1 = u1 +u2 +u3, w2 = u2 +u3,w3 = u3.

Answer 1

A vector space of dimension n has as many bases as there are linearly independent systems of n n -dimensional vectors. In other words, the basis of an n-dimensional space is any system of n independent vectors of this space.

We must show that number of vectors w₁, w₂, w₃ is equal the dimension of V and these vectors are linearly independent.

The dimension of space V is 3 because there are 3 vectors in the basis S. T also contains three vectors, so it can be a basis for V if w₁, w₂, w₃ are linearly independent.

We know that u₁, u₂, u₃ are linearly independent because S is a basis for V. It means that a₁u₁ +a₂u₂ +a₃u₃=0 only when a₁=0, a₂=0 and a₃=0. Vectors w₁ = u₁ +u₂ +u₃, w₂ = u₂ +u₃, w₃ = u₃ are linear combination of u₁, u₂, u₃.

Let w₁, w₂, w₃ are linearly dependent. In this case linear combination b₁w₁+b₂w₂+b₃w₃=0 exists for nonzero b₁, b₂, b₃.

If substitute w₁ = u₁ +u₂ +u₃, w₂ = u₂ +u₃, w₃ = u₃:

b₁(u₁ +u₂ +u₃)+b₂(u₂ +u₃)+b₃(u₃)=0

Open the brackets and groupe:

b₁u₁ +b₁u₂ +b₁u₃+b₂u₂ +b₂u₃+b₃u₃=0

b₁u₁ +(b₁+b₂)u₂ +(b₁+b₂+b₃)u₃=0.

Last combination can not exist for nonzero b₁, b₂, b₃ because u₁, u₂, u₃ are linearly independent vectors.

Thus, w₁, w₂, w₃ are linearly independent and T = {w₁,w₂,w₃} is a basis for vector space V.