From Generators to a Basis
A finite generating list describes a subspace, but it may contain more vectors than are needed. Some generators might be zero, repeated, or combinations of other generators. Removing such redundancy can make the subspace easier to describe: an independent list that still generates the same subspace is a basis, and its number of vectors is the subspace’s dimension.
Throughout, the scalars are real. A linear combination of vectors \(\mathbf{v}_1,\ldots,\mathbf{v}_p\) is a sum \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where each \(c_i\in\mathbb{R}\). The span of the list is the set of all such combinations. The list is a generating list for a subspace \(U\) when its span is exactly \(U\). A basis of \(U\) is an ordered list that spans \(U\) and is linearly independent. A list is linearly independent when the equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) has only the solution \(c_1=\cdots=c_p=0\).
The key test for removing a generator is whether it can be expressed using the others. If \(\mathbf{v}_k\) is such a combination, removing it does not change the span: any combination that used \(\mathbf{v}_k\) can instead substitute its expression in terms of the remaining vectors. The next theorem makes this idea into a finite procedure.
Proof. Begin with a finite spanning list. If it is independent, it is already a basis. If it is dependent, there are coefficients \(c_1,\ldots,c_p\), not all zero, such that \(\sum_{i=1}^{p}c_i\mathbf{v}_i=\mathbf{0}\). Choose an index \(k\) for which \(c_k\ne0\). Solving for \(\mathbf{v}_k\) gives \(\mathbf{v}_k=-\sum_{i\ne k}(c_i/c_k)\mathbf{v}_i\). Thus \(\mathbf{v}_k\) is a combination of the other listed vectors. Any combination involving \(\mathbf{v}_k\) can replace that term with this expression, so deleting \(\mathbf{v}_k\) leaves the span unchanged. Repeat this step whenever the current list is dependent. Each step deletes one vector, and the original list is finite, so the process must stop after finitely many steps. It stops with an independent list, and each deletion preserved the span, so the final list still spans \(U\). It is therefore a basis. By invariance of dimension, the number of vectors in this basis is \(\dim U\). \(\square\)
Row Reduction Finds Which Generators to Keep
For a list of vectors in \(\mathbb{R}^m\), place the vectors as the columns of a matrix \(\mathbf{A}\). Row-reducing this matrix identifies pivot-column indices. The important point is to keep the corresponding columns of the original matrix: row operations change the actual column vectors, even though they preserve the information needed to find which original generators suffice.
Proof. Row reduction applies elementary row operations, so there is an invertible matrix \(\mathbf{E}\) such that \(\operatorname{rref}(\mathbf{A})=\mathbf{E}\mathbf{A}\). Since \(\mathbf{E}\) is invertible, a linear relation among columns of \(\mathbf{A}\) holds exactly when the same relation holds among the corresponding columns of \(\mathbf{E}\mathbf{A}\): multiplying a relation by \(\mathbf{E}\) gives one direction, and multiplying by \(\mathbf{E}^{-1}\) gives the other. In RREF, each pivot column has a \(1\) in its pivot row and zeros in the other rows. Every nonpivot column is a combination of the pivot columns, with coefficients given by its entries in the pivot rows; its entries in zero rows are also zero. Thus the pivot columns of \(\operatorname{rref}(\mathbf{A})\) are independent and span all its columns. The same relations and independence hold for the corresponding columns of \(\mathbf{A}\), by invertibility of \(\mathbf{E}\). Those original columns are therefore independent and span the original columns, so they form a basis of the generated subspace. \(\square\)
This gives a practical route from generators to dimension: row-reduce the matrix with the generators as columns, select the original vectors at pivot-column indices, and count them. A list with no generators presents the zero subspace, whose dimension is zero; the examples below illustrate the more typical case of a nonempty list.
Worked Example: Two Redundant Generators
Find a basis and dimension for the subspace generated by \(\mathbf{v}_1=(1,0,1)\), \(\mathbf{v}_2=(0,1,1)\), \(\mathbf{v}_3=(1,1,2)\), and \(\mathbf{v}_4=(2,1,3)\) in \(\mathbb{R}^3\).
Place the vectors in the columns of \(\mathbf{A}\). Subtract \(R_1\) from \(R_3\), then subtract the new \(R_2\) from \(R_3\):
The pivot columns are 1 and 2, so the corresponding original vectors form a basis: \((\mathbf{v}_1,\mathbf{v}_2)\). The subspace has dimension \(2\). The remaining vectors are redundant: the third column is the sum of the first two, and the fourth is twice the first plus the second.
Check: \(\mathbf{v}_1+\mathbf{v}_2=(1,1,2)=\mathbf{v}_3\), and \(2\mathbf{v}_1+\mathbf{v}_2=(2,1,3)=\mathbf{v}_4\). To check independence of the retained pair, \(a\mathbf{v}_1+b\mathbf{v}_2=(a,b,a+b)=\mathbf{0}\) forces \(a=0\) and \(b=0\). The retained vectors are independent and still generate all four original vectors.
Zero and Repeated Generators
A zero generator contributes nothing to the span, while a repeated or scaled copy of another generator may also be removed. Row reduction handles these cases without needing a separate procedure. Pivot columns need not be the first columns: the initial generators can include zero vectors or redundant vectors before a later independent one.
Worked Example: A Zero Vector and a Nonpivot Generator
Find a basis and dimension for the subspace generated by \(\mathbf{u}_1=(0,0,0)\), \(\mathbf{u}_2=(1,1,0)\), \(\mathbf{u}_3=(2,2,0)\), and \(\mathbf{u}_4=(0,1,1)\) in \(\mathbb{R}^3\).
Reduce the matrix whose columns are the listed vectors. First use \(R_2\to R_2-R_1\), then \(R_3\to R_3-R_2\):
The pivot columns are 2 and 4. Therefore \((\mathbf{u}_2,\mathbf{u}_4)\), using the original vectors, is a basis, and the subspace has dimension \(2\). The first column is zero and has no pivot; the third column is twice the second and has no pivot.
Check: \(\mathbf{u}_1=\mathbf{0}\) and \(\mathbf{u}_3=2\mathbf{u}_2\), so the two retained vectors generate all four. For independence, \(a\mathbf{u}_2+b\mathbf{u}_4=(a,a+b,b)=\mathbf{0}\) gives \(a=0\) from the first coordinate and \(b=0\) from the third. Thus the retained list is a basis.
When the Redundancy Depends on a Parameter
A parameter can change which columns are pivots, so it can change both the basis extracted from a fixed list and the dimension of the generated subspace. The relevant step is to inspect when a potential pivot becomes zero. The row reduction should be read in cases: dividing by a parameter-dependent entry is valid only when that entry is nonzero.
Worked Example: A Parameter Changes the Dimension
For real \(t\), find a basis and dimension for the subspace generated by \(\mathbf{w}_1=(1,0,1)\), \(\mathbf{w}_2=(0,1,1)\), and \(\mathbf{w}_3=(1,1,t)\) in \(\mathbb{R}^3\).
Put the vectors in the columns of a matrix. Subtract \(R_1\) from \(R_3\), then subtract \(R_2\) from the new \(R_3\):
If \(t\ne2\), scale \(R_3\) by \(1/(t-2)\), then use \(R_1\to R_1-R_3\) and \(R_2\to R_2-R_3\). The result is the identity matrix, so every column is a pivot column. A basis is \((\mathbf{w}_1,\mathbf{w}_2,\mathbf{w}_3)\), and the dimension is \(3\).
If \(t=2\), the last row is zero and the first two columns are pivots. A basis is \((\mathbf{w}_1,\mathbf{w}_2)\), and the dimension is \(2\). In this case \(\mathbf{w}_3=\mathbf{w}_1+\mathbf{w}_2\).
Check: For \(t=2\), \((1,0,1)+(0,1,1)=(1,1,2)=\mathbf{w}_3\). For \(t\ne2\), suppose \(a\mathbf{w}_1+b\mathbf{w}_2+c\mathbf{w}_3=\mathbf{0}\). The first two coordinates give \(a=-c\) and \(b=-c\). The third then gives \(-c-c+tc=(t-2)c=0\). Since \(t\ne2\), \(c=0\), and hence \(a=b=0\). So all three vectors are independent in this case.
Common Pitfall: Keeping the Reduced Columns
Row reduction tells us which indices to select, not that the columns of the reduced matrix are the desired original generators. For example, let \(\mathbf{a}_1=(1,0,1)\) and \(\mathbf{a}_2=(0,1,1)\). Their column matrix reduces as follows:
A wrong conclusion is that the columns of the reduced matrix, \((1,0,0)\) and \((0,1,0)\), are the basis to report. They are not in the original subspace: every vector in \(\operatorname{span}\{\mathbf{a}_1,\mathbf{a}_2\}\) has the form \((x,y,x+y)\), and neither \((1,0,0)\) nor \((0,1,0)\) has that form. The correction is to use the pivot indices, 1 and 2, and retain the corresponding original vectors \(\mathbf{a}_1,\mathbf{a}_2\). Those vectors are independent and span the specified subspace.
The same principle applies when some columns are redundant: row operations are used to discover which original columns to keep. The pivot-column theorem justifies the method; the displayed reduced columns are intermediate tools, not replacements for the given generators.
Why Basis Extraction Is Useful
A generating list describes every vector in a subspace, but redundant entries can obscure its structure. An extracted basis gives a shorter description with no redundancy. Its length records the dimension, and its vectors are still drawn from the original list, which can be useful when those generators have a particular meaning. If the aim is only to determine dimension, counting the pivots is enough; if the aim is to give a basis from the supplied generators, report the original vectors at those pivot indices.
For a quick check, verify both parts of the basis claim. Independence rules out hidden redundancy among the retained vectors. Spanning confirms that no generator—and hence no vector of the original subspace—has been lost. When a specific dependence relation is available, substitute the retained vectors into it directly, as in the examples. When there are many generators, the pivot-column result supplies the general justification.
Check Your Understanding
For each problem, identify which vectors to retain and justify that they form a basis for the generated subspace.
- Find a basis and dimension for the subspace generated by \((1,0)\), \((2,0)\), and \((0,1)\) in \(\mathbb{R}^2\).
- Suppose a column matrix has pivot columns 1 and 3. Which original vectors should be selected for a basis?
- Explain why deleting a generator that is a linear combination of the others preserves the span.
- For what real values of \(t\) are \((1,0)\) and \((t,0)\) a dependent generating list? What is the dimension of their span in each case?
- In the row-reduction method, why is it generally incorrect to report the pivot columns of the RREF as the basis vectors?
