Latest revision as of 10:29, 14 March 2013

When do vectors form a basis?

In linear algebra, vectors $ v_1, v_2... v_n $ form a basis for the subspace V when

• The vectors span V. (in other words, the span of the vectors is V)
• The vectors are linearly independent.

If V is a subspace of $ \mathbb R^m $ it follows that n must be less than or equal to m.

Note that there can be more than one set of vectors that form a basis for the same space. In fact, there are an infinite number of bases (plural of basis) for a subspace provided the subspace is not just $ \vec 0 $. However, all bases for a given subspace have the same number of vectors. This number of vectors is called the dimension of the subspace.

The easiest way to find the basis of a subspace is to first find the span of the subspace and then eliminate any vectors that are not linearly independent. You will then be left with vectors that still span the subspace and are linearly independent. This can be shown with an example:

Suppose that $ \vec v_1,\vec v_2,\vec v_3,\vec v_4 $ span the subspace V. Then any vector in V can be written as

$ \vec x = c_1\vec v_1 + c_2\vec v_2 + c_3\vec v_3 + c_4\vec v_4 $

But if these vectors are not linearly independent and $ \vec v_4 $ can be written as a linear combination of the other three such as:

$ \vec v_4 = d_1\vec v_1 + d_2\vec v_2 + d_3\vec v_3 $

Now any vector in V can be written as: (after gathering like terms)

$ \vec x = (c_1+c_4d_1)\vec v_1 +(c_2+c_4d_2)\vec v_2 + (c_3+c_4d_3)\vec v_3 $

Which is still just a linear combination of the first three vectors. So $ \vec v_1, \vec v_2, \vec v_3 $ now span all of V and, supposing they are now linearly independent (you'd have to check to make sure), they form a basis for V.

For a supplementary explanation of basis, click here

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