Row Reduction Changes the Columns
Column-space questions ask which vectors can be built from the columns of a matrix. If a matrix has columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\), its column space is the set of all their linear combinations. Row reduction is useful for solving equations involving those columns, but it changes the entries in each column. The columns of the reduced matrix are therefore not generally the original vectors, and their span need not be the original column space.
Throughout, vectors and matrices have real entries. The span of vectors \(\mathbf{a}_1,\dots,\mathbf{a}_n\) is the set of all vectors \(c_1\mathbf{a}_1+\cdots+c_n\mathbf{a}_n\), where the coefficients \(c_1,\dots,c_n\) range over \(\mathbb{R}\). For a matrix \(\mathbf{A}\), its column space \(\operatorname{Col}\mathbf{A}\) is the span of its columns. A row operation changes the rows of a matrix; because each column has an entry in every row, that operation changes the columns as well.
Worked Example: One Row Operation Changes the Column Space
Consider the matrix whose columns are the first two standard basis vectors in \(\mathbb{R}^3\):
Apply \(R_3\to R_3+R_1\). The resulting matrix is
The columns of \(\mathbf{B}\) are \((1,0,1)^{\top}\) and \((0,1,0)^{\top}\). A general vector in their span is
For instance, \((1,0,1)^{\top}\) is in \(\operatorname{Col}\mathbf{B}\), since it is the first column of \(\mathbf{B}\). It is not in \(\operatorname{Col}\mathbf{A}\), whose vectors all have third coordinate \(0\). Conversely, \((1,0,0)^{\top}\) is in \(\operatorname{Col}\mathbf{A}\) but not in \(\operatorname{Col}\mathbf{B}\): matching the first coordinate in a combination of the columns of \(\mathbf{B}\) forces \(c_1=1\), so its third coordinate would also be \(1\), not \(0\).
Check: \(\operatorname{Col}\mathbf{A}\) consists of vectors with third coordinate \(0\), while combinations of the columns of \(\mathbf{B}\) have equal first and third coordinates. The two descriptions differ, so the column spaces are not equal.
What a Row Operation Does to a Span
The change in the example is systematic. A single row operation applies the same transformation to every column. For example, swapping two rows swaps those coordinates in every column; multiplying a row by a nonzero scalar scales that coordinate in every column; and adding a multiple of one row to another changes the corresponding coordinate in every column. Thus, the operation transforms the vectors themselves, not just the way they are displayed.
Proof. Write the columns of \(\mathbf{A}\) as \(\mathbf{a}_1,\dots,\mathbf{a}_n\). The columns of \(\mathbf{E}\mathbf{A}\) are \(\mathbf{E}\mathbf{a}_1,\dots,\mathbf{E}\mathbf{a}_n\), because left multiplication applies the row operation to each column. Therefore every vector in \(\operatorname{Col}(\mathbf{E}\mathbf{A})\) has the form \(\sum_{j=1}^{n}c_j\mathbf{E}\mathbf{a}_j\), where each \(c_j\in\mathbb{R}\). By distributivity of matrix multiplication over addition, this is \(\mathbf{E}(\sum_{j=1}^{n}c_j\mathbf{a}_j)\), which belongs to \(\{\mathbf{E}\mathbf{v}:\mathbf{v}\in\operatorname{Col}\mathbf{A}\}\). In the other direction, if \(\mathbf{v}\in\operatorname{Col}\mathbf{A}\), then \(\mathbf{v}=\sum_{j=1}^{n}c_j\mathbf{a}_j\) for some coefficients \(c_j\). Hence \(\mathbf{E}\mathbf{v}=\sum_{j=1}^{n}c_j\mathbf{E}\mathbf{a}_j\), which belongs to \(\operatorname{Col}(\mathbf{E}\mathbf{A})\). Both sets contain exactly the same vectors. \(\square\)
This result describes the relationship between the two spans; it does not say that they are equal. They are equal only if applying the transformation happens to leave that particular subspace unchanged. In the example, \(R_3\to R_3+R_1\) changes vectors in the original \(xy\)-plane into vectors whose first and third coordinates agree. The transformed plane is different from the original one. A row-equivalent matrix can therefore be useful for calculation without having the same column space as the matrix from which it came.
Membership Is an Equation with a Target
To ask whether a target vector \(\mathbf{b}\in\mathbb{R}^m\) belongs to \(\operatorname{Col}\mathbf{A}\) is to ask whether some combination of the columns equals \(\mathbf{b}\). If \(\mathbf{A}\) has columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\), then its matrix-vector product gives \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). This gives an exact translation of the membership question.
Proof. If \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\), then by the definition of column space there are coefficients \(x_1,\dots,x_n\) such that \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). With \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\), the definition of matrix multiplication gives \(\mathbf{A}\mathbf{x}=\mathbf{b}\), so the system is consistent. Conversely, if the system is consistent, there is a vector \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\) satisfying \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Expanding the product expresses \(\mathbf{b}\) as \(x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\), a linear combination of the columns. Thus \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\). \(\square\)
The target \(\mathbf{b}\) is part of this equation. When using row reduction to test membership, set up the augmented matrix \([\mathbf{A}\mid\mathbf{b}]\) and apply every row operation to the whole matrix, including the target column. Each row operation changes the equations, so the entries on the right-hand side must change along with the coefficients on the left. Reducing \(\mathbf{A}\) while leaving \(\mathbf{b}\) untouched instead compares transformed columns with the original target; that is a different question.
Worked Example: Test Membership by Reducing the Augmented Matrix
Use the matrix \(\mathbf{A}\) from the first example to test whether \(\mathbf{b}=(1,0,0)^{\top}\) lies in its column space. The original augmented matrix is
Apply \(R_3\to R_3+R_1\) to the entire augmented matrix, including the target column:
The last matrix gives \(x_1=1\) and \(x_2=0\), so the candidate coefficients are \(\mathbf{x}=(1,0)^{\top}\). Check them in the original system, not only in the transformed one:
Thus \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\). The check confirms membership using the original columns and the original target.
Common Mistake: Reducing Only the Coefficient Matrix
Suppose we row-reduce \(\mathbf{A}\) from the examples to \(\mathbf{B}\), then append the original target \((1,0,0)^{\top}\). This produces a misleading calculation:
The final row says \(0=-1\), so this calculation appears to show that the original target is not in the column space. But the first example already showed that \((1,0,0)^{\top}\in\operatorname{Col}\mathbf{A}\). The mistake is that the row operation was applied to \(\mathbf{A}\) but not to the target. The resulting augmented matrix does not represent the original membership equation.
The same error can produce a false positive. The vector \((1,0,1)^{\top}\) is a column of \(\mathbf{B}\), so it is in \(\operatorname{Col}\mathbf{B}\); it is not in \(\operatorname{Col}\mathbf{A}\), because every vector in the latter has third coordinate \(0\). Treating \(\mathbf{B}\)'s columns as though they were the original columns would wrongly answer “yes” for this target. Row reduction can change the set of vectors under consideration, so membership must always be checked against the original matrix and target together.
A Reliable Membership Procedure
Use the augmented matrix to carry the original question through elimination. The point is not that the reduced coefficient columns themselves represent the original column space. Rather, the target column must undergo the same row operations as the coefficient columns so the calculation continues to test the original equation.
For \(\mathbf{A}\in\mathbb{R}^{m\times n}\), specify the target \(\mathbf{b}\in\mathbb{R}^m\). Membership asks whether \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has a solution.
Form \([\mathbf{A}\mid\mathbf{b}]\), keeping the target as the last column.
Do not reduce the coefficient matrix separately and then append the unchanged target.
A row with all zero coefficients and a nonzero right-hand side shows inconsistency. If the system is consistent, substitute a solution into the original equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\).
This distinction is useful whenever elimination is used to answer a question about span. Row reduction is an operation on a matrix, not a guarantee that its original columns and its new columns generate the same subset of \(\mathbb{R}^m\). To test whether a particular target is in the original column space, encode that target in the augmented matrix and carry it through every operation. The coefficient vector obtained this way describes a combination of the original columns, which is the membership question we meant to answer.
Check Your Understanding
For each question, distinguish the column space of the original matrix from that of a row-equivalent matrix.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\), describe \(\operatorname{Col}\mathbf{A}\). After \(R_3\to R_3+R_1\), give a vector in the new column space that is not in the original one.
- Explain why the columns of a row-reduced matrix cannot generally be treated as the original vectors when testing span membership.
- Set up the augmented matrix to test whether \((2,-1,0)^{\top}\) belongs to the column space of \(\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\). What coefficients give the target?
- Describe the error in applying row operations to \(\mathbf{A}\) while leaving the target column unchanged.
- State the relationship between \(\operatorname{Col}(\mathbf{E}\mathbf{A})\) and \(\operatorname{Col}\mathbf{A}\) when \(\mathbf{E}\) represents one elementary row operation.
