Basis of a Vector Space Example

A set of vectors in vector space V is called a basis for V if the vectors in the set are linearly independent and every vector is a linear combination of vectors in this set. The former one equals the vector space being spanned by the set. The dimension of vector space V is the number of vectors in its basis. A vector space is called nite dimensional if it has a nite basis.

Example 1. Let R^3 be the real vector space in three dimensions, and consider the set of vectors S=fv_1; v_2; v_3g where v_1 = (1; 2; 0), v_2 = (0; 1; 0) and v_3 = (1; 0; 1). Let us see whether S is a basis for R^3 or not.

S is linearly independent since the solution for the equation c_1v_1+c_2v_2+c_3v_3 = 0 is unique, and it is c_1 = c_2 = c_3 = 0.

On the other hand, we need to investigate whether the vectors span the whole vector space. Any vector in R^3 is of the form (a; b; c). We need to end coefficients c_1, c_2 and  c_3 in terms a, b and c.

basis of a vector sector example

Once we apply Gauss-Jordan reduction to this equation we get the following solution:

vector sector

The standard basis for R^3 is f(1; 0; 0); (0; 1; 0); (0; 0; 1)g. The dimension of R^3 is three, since there are three vectors in its bases.

Example 2. Now, consider the set of vectors S_2=fv_1; v_2g where v_1 = (1; 0; 2) and v_2 = (2; 1; 0) and consider another set of vectors S_3=fv_3; v_4; v_5g where v_3 = (1; 0; 0); v_4 = (2; 1; 3) and v_5 = (0; 1; 3).

The first set S_1 is not a basis for R^3 since it has two vectors, and a set of two vectors cannot span a vector space with dimension three. The second set S_2 is not a basis for R^3 because it is not linearly independent, since v_42v_3 = 0.

This sample will help you deal with the basis of a vector sector. You will spend much less time on your assignment with the help of the sample. Of course, this basis of a vector sector example was created by an expert who is knowledgeable in the topic. Therefore, you can be sure that it is absolutely correct.

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