What a Row Operation Changes
Row reduction is useful because it replaces a matrix by a simpler one without changing the solutions of its homogeneous system. But this does not mean that every feature of a matrix stays the same. In particular, the row space and null space are preserved by elementary row operations, while the column space can change. Knowing the distinction prevents us from treating a row-equivalent matrix as interchangeable with the original for every purpose.
Work over a scalar field \(\mathbb{F}\), which may be \(\mathbb{R}\) or \(\mathbb{C}\). For a matrix \(\mathbf{A}\in\mathbb{F}^{m\times n}\), its row space is the span in \(\mathbb{F}^n\) of its rows, and its column space is the span in \(\mathbb{F}^m\) of its columns. When we treat a row as an element of \(\mathbb{F}^n\), we identify it with its transpose, so vectors in these spaces are columns. Its null space is \(\operatorname{Nul}\mathbf{A}=\{\mathbf{x}\in\mathbb{F}^n:\mathbf{A}\mathbf{x}=\mathbf{0}\}\).
Each operation changes how the rows are written, and it also changes the equations represented by those rows. The key questions are whether the new rows generate the same row space and whether the new equations have the same solutions. The column space asks a different question: which vectors in \(\mathbb{F}^m\) can be made from the columns? As the examples will show, row operations can change those columns as vectors in \(\mathbb{F}^m\).
The Row Space Is Preserved
Proof. Write the rows of \(\mathbf{A}\) as \(\mathbf{r}_1,\dots,\mathbf{r}_m\), and regard each row as a vector in \(\mathbb{F}^n\). For a row interchange, the list of rows is merely reordered, so the two lists have the same span. For a scaling \(R_i\to cR_i\), where \(c\ne0\), the new row \(c\mathbf{r}_i\) belongs to the old row space, so every new row belongs to the old row space. Conversely, \(\mathbf{r}_i=c^{-1}(c\mathbf{r}_i)\), so every old row belongs to the new row space. For a replacement \(R_i\to R_i+cR_j\), the new \(i\)th row is \(\mathbf{r}_i+c\mathbf{r}_j\), a combination of old rows, and all other rows are unchanged. Thus every new row belongs to the old row space. The operation is reversed by \(R_i\to R_i-cR_j\), so every old row is also a combination of the new rows. In each case, both inclusions hold, and the row spaces are equal. \(\square\)
The proof uses reversibility: showing that each new row is a combination of old rows gives only one inclusion. The inverse operation supplies the other. This is why equality of row spaces follows, rather than just containment.
Worked Example: Three Operations, One Row Space
Consider the rows of \(\mathbf{A}=\begin{bmatrix}1&2\\3&6\\0&1\end{bmatrix}\). Apply a swap, a nonzero scaling, and a row replacement. First interchange \(R_1\) and \(R_3\):
Next scale \(R_2\) by \(\tfrac13\), then replace \(R_3\) by \(R_3-R_2\):
In the original matrix, the rows are \((1,2)\), \((3,6)\), and \((0,1)\), and \((3,6)=3(1,2)\). Thus its row space is \(\operatorname{span}\{(1,2),(0,1)\}\). The final matrix has rows \((0,1)\), \((1,2)\), and \((0,0)\), which span the same space. The original rows and the final rows therefore give the same row space, as the theorem predicts.
Check: The replacement created the zero row because the two rows being subtracted are both \((1,2)\). The row \((3,6)\) in the original list is recovered from the final row \((1,2)\) by multiplying by \(3\); no direction in the row space has been lost.
The Null Space Is Preserved
Proof. The equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\) says that the dot product of each row of \(\mathbf{A}\) with \(\mathbf{x}\) is zero. Interchanging two rows only reorders these equations, so it does not change which \(\mathbf{x}\) satisfy all of them. Multiplying one row by a nonzero scalar multiplies its equation by that scalar; the new equation holds exactly when the old equation holds, since the scalar has a multiplicative inverse. Finally, suppose \(R_i\to R_i+cR_j\). For any fixed \(\mathbf{x}\), the new \(i\)th equation is the old \(i\)th equation plus \(c\) times the \(j\)th equation. If the old equations all hold, the new ones hold. Conversely, if the new equations all hold, the unchanged \(j\)th equation holds, and subtracting \(c\) times that equation from the new \(i\)th equation recovers the old \(i\)th equation. Thus the solution sets agree in every case, proving equality of the null spaces. \(\square\)
This result concerns the homogeneous system: its right-hand side is \(\mathbf{0}\). The same reasoning says that elementary row operations preserve the solution set of a system when they are applied to the entire equation, including its right-hand side. Here the null space is the set of homogeneous solutions, so only the coefficient matrix is needed.
Worked Example: The Same Null Space Before and After Reduction
Let \(\mathbf{A}=\begin{bmatrix}1&2&1\\2&4&2\end{bmatrix}\). Subtract \(2R_1\) from \(R_2\) to obtain \(\mathbf{B}\):
For \(\mathbf{A}\mathbf{x}=\mathbf{0}\), the equations are \(x_1+2x_2+x_3=0\) and \(2x_1+4x_2+2x_3=0\). The second equation is twice the first. For \(\mathbf{B}\mathbf{x}=\mathbf{0}\), the equations are \(x_1+2x_2+x_3=0\) and \(0=0\). In either system, set \(x_2=s\) and \(x_3=t\). Then \(x_1=-2s-t\), so the full solution set is
Check: For \(\mathbf{v}_1=\begin{bmatrix}-2\\1\\0\end{bmatrix}\), \(\mathbf{A}\mathbf{v}_1=\begin{bmatrix}-2+2+0\\-4+4+0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). For \(\mathbf{v}_2=\begin{bmatrix}-1\\0\\1\end{bmatrix}\), \(\mathbf{A}\mathbf{v}_2=\begin{bmatrix}-1+0+1\\-2+0+2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). The corresponding products with \(\mathbf{B}\) are also zero: its first row gives \(-2+2+0=0\) and \(-1+0+1=0\), while its second row is zero. Every displayed combination is therefore a solution of both systems.
The Column Space Can Change
The row space records combinations of the rows, and the null space records which input vectors the rows send to zero. The column space instead consists of combinations of the columns in \(\mathbb{F}^m\). A row operation changes entries down a column, so it may change the vectors that generate this space. Equality of row spaces and null spaces does not imply equality of column spaces.
Proof. Consider \(\mathbf{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\), and obtain \(\mathbf{B}\) by interchanging its rows. The operation gives \(\mathbf{B}=\begin{bmatrix}0&0\\1&0\end{bmatrix}\). The columns of \(\mathbf{A}\) are \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\0\end{bmatrix}\), so \(\operatorname{Col}\mathbf{A}=\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}\). The columns of \(\mathbf{B}\) are \(\begin{bmatrix}0\\1\end{bmatrix}\) and \(\begin{bmatrix}0\\0\end{bmatrix}\), so \(\operatorname{Col}\mathbf{B}=\operatorname{span}\left\{\begin{bmatrix}0\\1\end{bmatrix}\right\}\). These spaces are different: \(\begin{bmatrix}1\\0\end{bmatrix}\) cannot equal \(c\begin{bmatrix}0\\1\end{bmatrix}\) for any scalar \(c\), since the first coordinates would require \(1=0\). Thus this row operation changes the column space. \(\square\)
Worked Example: Comparing All Three Spaces
For the matrices in the proposition, compare their row spaces and null spaces as well as their column spaces. The rows of \(\mathbf{A}\) are \((1,0)\) and \((0,0)\), while those of \(\mathbf{B}\) are \((0,0)\) and \((1,0)\). Hence both row spaces are \(\operatorname{span}\{(1,0)\}\). For \(\mathbf{A}\mathbf{x}=\mathbf{0}\), the only equation is \(x_1=0\); for \(\mathbf{B}\mathbf{x}=\mathbf{0}\), the equations are \(0=0\) and \(x_1=0\). Thus both null spaces are \(\left\{\begin{bmatrix}0\\t\end{bmatrix}:t\in\mathbb{F}\right\}\).
The column spaces, however, are \(\operatorname{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}\) and \(\operatorname{span}\left\{\begin{bmatrix}0\\1\end{bmatrix}\right\}\), respectively. The two displayed nonzero vectors are not scalar multiples, so these one-dimensional subspaces are distinct.
Check: Multiplication confirms the null-space claim. For \(\mathbf{x}=\begin{bmatrix}0\\t\end{bmatrix}\), \(\mathbf{A}\mathbf{x}=\begin{bmatrix}0\\0\end{bmatrix}\) and \(\mathbf{B}\mathbf{x}=\begin{bmatrix}0\\0\end{bmatrix}\) for every \(t\in\mathbb{F}\). Conversely, either matrix equation forces \(x_1=0\), so there are no other null vectors.
The proposition says “not in general,” rather than claiming that every row operation changes the column space. For example, swapping the rows of \(\mathbf{I}_2\) changes its columns from the standard basis vectors to the same two vectors in the opposite order; both column spaces are \(\mathbb{F}^2\). The important point is that preservation of the column space cannot be assumed from row equivalence alone.
Common Pitfall: Carrying Pivot Columns Across a Row Operation
A tempting but incorrect claim is: “The matrices are row equivalent, so their column spaces are equal.” The row-swap calculation above is a direct counterexample. In that example, \(\mathbf{A}\) has first column \(\begin{bmatrix}1\\0\end{bmatrix}\), while \(\mathbf{B}\) has first column \(\begin{bmatrix}0\\1\end{bmatrix}\); the first column of \(\mathbf{A}\) is not in \(\operatorname{Col}\mathbf{B}\). The correction is to keep the spaces separate: row operations preserve row space and null space, but they need not preserve column space.
Another related error is to take the columns of a row-reduced matrix as a basis for the original column space. Those columns belong to the column space of the row-reduced matrix, which may be different. To find a column-space basis, identify the pivot-column positions after row reduction and take those columns from the original matrix; the row reduction locates the positions but does not replace the original column vectors.
| Space | Effect of elementary row operations | Reason |
|---|---|---|
| Row space | Preserved | New rows are combinations of old rows, and the operations are reversible |
| Null space | Preserved | The homogeneous equations have exactly the same solutions |
| Column space | Not necessarily preserved | The column vectors in \(\mathbb{F}^m\) can change |
These distinctions explain how row reduction can safely simplify a homogeneous system and reveal row-space information while still changing the vectors that generate the column space. Before using a row-equivalent matrix, identify which of these spaces the question concerns.
Check Your Understanding
For each question, distinguish the space being discussed and justify the conclusion from the definitions.
- Explain why multiplying one row by \(c\ne0\) preserves the row space. Which reverse operation is needed for the other inclusion?
- For \(\mathbf{A}=\begin{bmatrix}1&0&2\\2&0&4\end{bmatrix}\), apply \(R_2\to R_2-2R_1\). Find the null space before and after the operation.
- Give a row operation that changes the column space of \(\begin{bmatrix}1&0\\0&0\end{bmatrix}\), and state both column spaces.
- Why does reversing an elementary row operation matter when proving equality, rather than only containment, of row spaces?
- Does row equivalence alone guarantee that two matrices have the same column space? Support your answer with a specific calculation.
