Is Set a Basis for C^3 as a C-Vector Space?

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    Basis Space
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Homework Help Overview

The discussion revolves around determining whether specific sets of vectors form a basis for the complex vector space C^3. The original poster presents two sets of vectors for evaluation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use matrix representation to assess linear independence and basis status for the given sets of vectors. Questions arise regarding the adequacy of the number of vectors in part (b) and the implications for forming a basis.

Discussion Status

Participants are engaging in a dialogue about the requirements for a basis in C^3, with some confirming the original poster's approach for part (a) while questioning the validity of part (b) due to the insufficient number of vectors. There is an acknowledgment of the need for three linearly independent vectors to form a basis.

Contextual Notes

There is a focus on the definitions of linear independence and basis, as well as the implications of having fewer vectors than the dimension of the space being considered.

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Homework Statement



determine whether or not the given set is a basis for C^3 ( as a C-vector space)

(a) {(i,0,-1),(1,1,1),(0,-i,i)}
(b) {(i,1,0),(0,0,1)}

Homework Equations





The Attempt at a Solution



All I did was to put the 3 vectors in part (a) into a matrix as 3 columns. Then I determined that the matrix has 3 leading entries, hence it is a basis. But when I tried using the same method for part (b), it doesn't work. Why is that so?
 
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The set in b has only two vectors, which isn't enough for a basis for C^3. There are some vectors in C^3 that aren't any linear combination (i.e., a sum of (complex) scalar multiples of (i, 1, 0) and (0, 0, 1).
 
But what i did for part(a) is right?
 
Assuming your work is correct, yes. A basis for C^3 has to have three vectors in it. If you have three vectors that are linearly independent, that's a basis.
 

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