Respuesta :
Answer:
All vectors in C³ are linear combinations of (1,0,-1), (0,1,1) and (1,1,1), as long as we think C³ as a C-vector space. If we think C³ as a real vecor space, only those vectors with real components only will be a linear combination of them.
Step-by-step explanation:
C³ has dimension 3 as C-vector space. We have been given 3 vectors. If they were linaerly independent, then they should span a vector subspace of dimension 3, hence they should generate all C³, and as a result, every vector in C³ is a linear combination of (1,0,-1), (0,1,1) and (1,1,1). In order to find if the vectors are linearly independent, we can put them as the rows of a matrix and triangulate it
[tex]\left[\begin{array}{ccc}1&0&-1\\0&1&1\\1&1&1\end{array}\right] \rightarrow \left[\begin{array}{ccc}1&0&-1\\0&1&1\\0&1&2\end{array}\right] \rightarrow \left[\begin{array}{ccc}1&0&-1\\0&1&1\\0&0&1\end{array}\right][/tex]
In the first step of the triangulation, we took the third row and substract from it the first one. Then we substract from it the second row, obtaining a upper triangular matrix without zeros in the diagonal. This shows that the three vectors are linearly independent, and as a result, every vector of C³ is obtained as a linear combination of them. (Note that if we think of C³ as a R vector space, it should have dimension 6 instead of 3. In this case, the subspace generated by the 3 vectors is R³, because it has dimension 3 and the vectors have, all of them, real components only).
