You will need

- paper;
- - handle.

Instruction

1

Using only basic definitions to check linear independence

*of a system of*vector-columns, respectively, and give a conclusion on the existence of the basis, is very difficult. Therefore, in this case, you may be able to use some special signs.2

It is known that the vectors are linearly independent if derived from the determinant is not zero.Therefore, it is possible enough to explain the fact that the system

**of vectors**forms a basis. So, in order to prove that the vectors form a basis, should be the coordinates of their determinant and make sure it is not zero.Further, in order to reduce and simplify records, the representation of the vector-column matrix column will replace the transpose by the matrix-line.3

*Example 1. Whether form a basis in R^3 vector-columns (1, 3, 5)^T, (2, 6, 4)^T, (3, 9, 0)^T. the Solution. Make up a determinant |A|, whose rows are the elements of the specified column (see Fig.1).Expanding this determinant by the rule of triangles, you get: |A| = 0+90+36-90-36-0=0. Therefore, these vectors cannot form a basis.*

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*Example. 2. The system*

**of vectors**consists of (10, 3, 6)^T, (1, 3, 4)^T, (3, 9, 2)^T. if they Can form a basis?Solution. By analogy with the first example, make a key (see Fig.2): |A| =60+54+36-54-360-6=270, ie is not zero. Therefore, this system is a vector-column suitable for use as the basis in R^3.

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Now obviously becomes clear, that for finding of the basis

*system of*the vector-column is sufficient to take the determinant of any suitable dimension different from zero. Elements its columns form a base system. Moreover, always it is desirable to have a simple basis. Since the determinant of the identity matrix is always nonzero (for any dimension), as the basis you can always choose a system (1, 0, 0,...,0)^T, (0, 1, 0,...,0)^T, (0, 0, 1,...,0)^T,..., (0, 0, 0,...,1)^T.