How to prove that the vectors form a basis

Basis in n-dimensional space is called the system of n vectors, when all the other vectors in the space can be represented as a combination of vectors included in the basis. In three-dimensional space, any basis consists of three vectors. But not any three form a basis, therefore, there is the problem of system test vectors on the opportunity to build their base.

How to prove that the vectors form a basis
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You will need

- the ability to calculate the determinant of a matrix

Instruction

Suppose that in a linear n-dimensional space there exists a system of vectors e1, E2, E3, ... , EN. Their coordinates are: e1 = (e11; e21; e31; ... ; en1), E2 = (E12; E22; е32; ... ; EP2), ... , EN = (e1n; e2n; e3n; ... ; enn). To find out whether they form a basis in this space, make a matrix with columns e1, E2, E3, ... , EN. Find its determinant and compare it with zero. If the determinant of the matrix of these vectors is not zero, then these vectors form a basis in this n-dimensional linear space.

For example, suppose you are given three vectors in three dimensions a1, a2 and a3. Their coordinates: A1 = (3; 1; 4), A2 = (-4; 2; 3) and A3 = (2; -1; -2). Need to find out whether these form a vector basis in three-dimensional space. Make a matrix from the vectors, as shown in the figure.

Calculate the determinant of the resulting matrix. The figure shows a simple method of calculating the determinant of the matrix 3 by 3. Elements are connected by a line should be multiplied. The works outlined in red are included in the total amount with the sign "+", and the United blue line - with the sign "-". det A = 3*2*(-2) + 1*2*3 + 4*(-4)*(-1) - 2*2*4 - 1*(-4)*(-2) - 3*3*(-1) = -12 + 6 + 16 - 16 - 8 + 9 = -5 -5≠0, therefore, A1, A2 and A3 form a basis.

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