Solving for x in Vector Space (1,2)^T; (-1,1)^T

[Dustinsfl]

Find the basis of the vector space (1,2)^T; (-1,1)^T

When I solve the matrix, I obtain x₁=0 and x₂=0

x=(0,0)^T.

Can a basis be two 0 column vectors? Thanks for the help.

 

[VeeEight]

No, a basis cannot be two 0 column vectors.
What matrix are you talking about? Are you trying to solve (1,2)x₁ + (-1,1)x₂ = 0? If so, what does this tell you?
And what does the vector space look like? Is it the span of those two vectors? Is this span minimal?

 

[Dustinsfl]

All the question says is to determine the basis. I am then given two column vectors (1,2) and (-1,1).

I solved the augmented matrix and obtained that both x1=0 and x2=0.

 

[VeeEight]

And what does x₁ = 0 = x₂ tell you? You are trying to determine if the two vectors are linearly independent, so you are solving the system (1,2)x₂ + (-1,1)x₂ = 0. Thus, a solution of (x₁, x₂) = (0, 0) tells you something about the vectors (1,2) and (-1,1).

 

[Dustinsfl]

I know they are independent from the det not equaling 0. I am trying to find the basis of the vector space.

 

[VeeEight]

What does the vector space look like? You have only identified two vectors. Is it the set of linear combinations of these two vectors?

 

[Dustinsfl]

I have giving you all the information. What are you asking for? It is in R^2 but that is also mentioned in the title.

 

[vela]

The phrase "the basis of the vector space (1,2)^T; (-1,1)^T" doesn't make sense. R² is the vector space; (1,2)^T; (-1,1)^T are two vectors, not a vector space. It's not really clear what you're trying to do.