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

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SUMMARY

The discussion centers on determining whether specific sets of vectors are bases for the complex vector space C^3. The first set, {(i,0,-1),(1,1,1),(0,-i,i)}, is confirmed as a basis because it contains three linearly independent vectors, as evidenced by having three leading entries in the corresponding matrix. In contrast, the second set, {(i,1,0),(0,0,1)}, is not a basis for C^3 since it contains only two vectors, which is insufficient to span the three-dimensional space. The conclusion is that a basis for C^3 must consist of exactly three linearly independent vectors.

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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)}

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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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