How to Pare Down a Set to Form a Basis

If a finite set of vectors spans a space, you can pare it down to a basis by removing vectors that are redundant. The most reliable method is to put the vectors into a matrix as columns, row-reduce, and keep the original vectors whose column positions contain pivots.

A basis must satisfy two conditions: its vectors are linearly independent, and they span the space. Starting with a spanning set, removing a vector that is a linear combination of the others preserves the span; when no more such vectors can be removed, the remaining set is a basis.

The pivot-column method

Suppose your vectors are $v_1,\ldots,v_n$ in $\mathbb{R}^m$. Form the matrix

$$A = \begin{bmatrix} | & | & & | \\ v_1 & v_2 & \cdots & v_n \\ | & | & & | \end{bmatrix}.$$

Row-reduce $A$ to echelon form or reduced row-echelon form. The columns containing leading entries are the pivot columns. Keep the vectors in those positions from the original matrix. Those vectors form a basis for the span of the original set.

Why does this work? Row operations are invertible operations on the equations $Ac=0$, so they preserve which coefficient lists $c$ give linear dependencies among the columns. The pivot-column positions therefore identify a maximal independent subset of the original columns. Also, every non-pivot column is a linear combination of the pivot columns, so keeping the pivot columns preserves the original span.

Worked example

Consider these four vectors in $\mathbb{R}^3$:

$$v_1=\begin{bmatrix}1\\0\\1\end{bmatrix},\quad v_2=\begin{bmatrix}0\\1\\1\end{bmatrix},\quad v_3=\begin{bmatrix}1\\1\\2\end{bmatrix},\quad v_4=\begin{bmatrix}0\\0\\1\end{bmatrix}.$$

Place them in a matrix as columns:

$$A=\begin{bmatrix} 1&0&1&0\\ 0&1&1&0\\ 1&1&2&1 \end{bmatrix}.$$

Row-reduce, for example by replacing the third row first with the third row minus the first, then subtracting the second row from the new third row:

$$\begin{bmatrix} 1&0&1&0\\ 0&1&1&0\\ 1&1&2&1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1&0&1&0\\ 0&1&1&0\\ 0&0&0&1 \end{bmatrix}.$$

The pivot columns are 1, 2, and 4. Therefore keep $v_1$, $v_2$, and $v_4$. These three vectors are a basis for the same space as all four original vectors. The discarded vector is redundant: directly,

$v_3=v_1+v_2.$

Notice that the pivot columns in the reduced matrix are not the vectors to keep. For instance, the reduced matrix’s fourth column is $(0,0,1)^T$, but in other examples a reduced column need not equal the corresponding original vector. Use the pivot positions to select columns from the original matrix.

A quick alternative: remove dependencies directly

If you can see a dependence relation, you can remove a vector without row-reducing. In the example, $v_3=v_1+v_2$, so deleting $v_3$ leaves the span unchanged. Repeating this process until the remaining vectors are independent produces a basis.

For a larger set, row reduction is usually less error-prone. It identifies all pivot positions at once and provides a systematic test for independence. A set is already independent exactly when every column is a pivot column; if it is also known to span the space, it is already a basis.

Common mistakes

• Keeping columns of the reduced matrix. Row operations change the vectors. Keep original columns at the pivot positions.
• Choosing arbitrary vectors to delete. A vector can be removed only if it is a linear combination of the others, or if the remaining vectors otherwise still span the same space.
• Confusing spanning with independence. A spanning set can contain redundant vectors. A basis has no redundancy: neither condition can be dropped.
• Assuming every set can be pared down. This procedure applies directly to a finite spanning set. For an infinite set, the question requires more care; in a finite-dimensional space, a finite subset can be selected as a basis when the set spans the space.

If you have an independent set instead

Sometimes the goal is to make a basis for a larger space, not to pare down a spanning set. If your vectors are independent but do not span the target space, add vectors outside their current span until they do. In $\mathbb{R}^m$, an independent set can be extended this way until it has $m$ vectors. In contrast, if the starting set spans the target space, use the pivot-column method to remove redundancy.