Vector Space Basis Completion Theorem

If a vector space basis is incomplete, it's always possible to expand and complete the vector space basis by adding other linearly independent vectors.

The Definition

In a finite-dimensional vector space V of dimension n over the field K, given v₁,...,v_k linearly independent vectors in V, with k<n, there exist n-k vectors v_k+1,...,v_n in V, which together with the previous ones form a basis of V.
$$ B = \{ v_1,...,v_k,v_{k+1},...,v_n \} $$

An Example

In the vector space V=R³ over the field R, consider two vectors:

$$ v_1=(2,1,0) $$ $$ v_2=(1,1,0) $$

These two vectors are linearly independent, but the basis B is incomplete.

$$ B = \{ v_1 , v_2 , ? \} $$

To find the missing vector, I represent the two vectors as a matrix

$$ \begin{pmatrix} 2 & 1 \\ 1 & 1 \\ 0 & 0 \end{pmatrix} $$

Then, I augment this matrix with the canonical basis of the vector space V=R³.

$$ \begin{pmatrix} 2 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$

Note. Any other basis of the vector space V=R³ would work. However, it's more practical to use the canonical basis due to its diagonal of ones.

Now, I apply the Gauss-Jordan elimination method to transform it into a staircase matrix.

$$ \begin{pmatrix} 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$

The pivots of the staircase matrix are located in the first, second, and fifth columns.

Therefore, I select the first, second, and fifth columns from the original matrix.

This way, I've identified the third linearly independent vector of the basis, namely v₃=(0,0,1).

The complete basis is as follows:

$$ B = \{ v_1 = (2,1,0) , v_2 = (1,1,0) , v_3=(0,0,1) \} $$

Now, the number of elements in the basis matches the dimension of the vector space V (n=3).