Rows as Equations on Coordinates
A row of a matrix records the coefficients in a linear equation on the coordinates of a column vector. For example, the row \((1,2,0,1)\) gives the expression \(x_1+2x_2+x_4\). A matrix may contain several such rows, and some of their equations may be consequences of others. The row space collects all linear combinations of the coefficient rows; it therefore describes which coefficient equations can be formed from the given rows.
Throughout this tutorial, matrices and vectors have real entries. Vectors in \(\mathbb{R}^n\) are columns. To treat a matrix row as a vector in \(\mathbb{R}^n\), transpose it: if the \(i\)th row of \(\mathbf{A}\in\mathbb{R}^{m\times n}\) is \((a_{i1},\dots,a_{in})\), write its transposed row as \(\mathbf{r}_i=(a_{i1},\dots,a_{in})^{\top}\). The dot product of \(\mathbf{r}_i\) with \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\) is \(\mathbf{r}_i^{\top}\mathbf{x}=\sum_{j=1}^{n}a_{ij}x_j\).
Row space is about the rows, not the columns. Its vectors have \(n\) coordinates, matching the number of columns of \(\mathbf{A}\). When we write \(\mathbf{A}\mathbf{x}=\mathbf{0}\), each row gives one homogeneous equation, and that equation says that the corresponding transposed row has dot product zero with \(\mathbf{x}\).
Why Row Operations Preserve Row Space
Elementary row operations are swapping two rows, multiplying a row by a nonzero scalar, or adding a multiple of one row to another. These operations are used to produce an echelon form: a matrix in which any zero rows are at the bottom, and the first nonzero entry of each successive nonzero row lies farther to the right. The row space does not depend on which row-equivalent matrix is used to display it.
Proof. It is enough to consider one row operation. A swap only rearranges the rows, so the span is unchanged. Multiplying a row by a nonzero scalar keeps the new row in the old row space; the reverse operation multiplies it by the reciprocal scalar, so the old row is also in the new row space. If a multiple of one row is added to another, the new row is a linear combination of old rows, and every unchanged row is still in the new list. Thus every new row belongs to the old row space. The inverse operation subtracts the same multiple, so every old row belongs to the new row space as well. In each case the two row spaces are equal. Applying this argument at each step proves the result for any finite sequence of row operations. \(\square\)
Worked Example: Finding a Row-Space Basis
Find a basis for the row space of \(\mathbf{A}=\begin{bmatrix}1&2&0&1\\2&4&1&3\\1&2&1&2\end{bmatrix}\).
Use row operations to produce an echelon form. First replace \(R_2\) by \(R_2-2R_1\) and \(R_3\) by \(R_3-R_1\). Then replace \(R_3\) by \(R_3-R_2\):
The two nonzero rows of the echelon form, transposed, are \(\mathbf{u}_1=(1,2,0,1)^{\top}\) and \(\mathbf{u}_2=(0,0,1,1)^{\top}\). They span the original row space by the row-operation theorem. They are independent: if \(c_1\mathbf{u}_1+c_2\mathbf{u}_2=\mathbf{0}\), the first coordinate gives \(c_1=0\), and the third coordinate then gives \(c_2=0\). Thus they form a basis.
Check: The original rows are \(\mathbf{u}_1^{\top}\), \(2\mathbf{u}_1^{\top}+\mathbf{u}_2^{\top}=(2,4,1,3)\), and \(\mathbf{u}_1^{\top}+\mathbf{u}_2^{\top}=(1,2,1,2)\). Each is in the span of the two proposed basis vectors, as required.
This example illustrates a general way to compute a row-space basis. The rows of an echelon form span the same space as the original rows. The nonzero rows of that echelon form are also independent, because their first nonzero entries occur in distinct, successively later columns.
Proof. Since \(\mathbf{E}\) is obtained from \(\mathbf{A}\) by elementary row operations, the row-operation theorem gives \(\operatorname{Row}\mathbf{E}=\operatorname{Row}\mathbf{A}\). The nonzero rows of \(\mathbf{E}\) span \(\operatorname{Row}\mathbf{E}\), because zero rows contribute nothing. It remains to prove they are independent. Suppose a linear combination of the nonzero rows is zero, with at least one nonzero coefficient. Select the first row, in echelon order, whose coefficient is nonzero, and consider its first nonzero column. Every later row has a zero in that column, since its first nonzero entry is farther right. Earlier rows have coefficient zero by the choice of row. Consequently, the entry of the linear combination in that column is the selected nonzero coefficient multiplied by the selected row’s nonzero leading entry. This entry cannot be zero, contradicting that the combination was zero. The rows are independent and span \(\operatorname{Row}\mathbf{A}\), so their transposes form a basis. \(\square\)
Different Matrices, the Same Row Space
Row-equivalent matrices need not have the same rows. Their row spaces are equal because row operations replace a list of rows by another list spanning the same subspace. This equality is useful: an echelon form may make a row-space basis easier to read than the original matrix does.
Worked Example: Comparing Two Row-Equivalent Matrices
Let \(\mathbf{u}_1=(1,2,0,1)^{\top}\) and \(\mathbf{u}_2=(0,0,1,1)^{\top}\). Consider the matrix whose rows are \(2\mathbf{u}_1^{\top}+\mathbf{u}_2^{\top}\), \(\mathbf{u}_1^{\top}+2\mathbf{u}_2^{\top}\), and \(3\mathbf{u}_1^{\top}\): \(\mathbf{B}=\begin{bmatrix}2&4&1&3\\1&2&2&3\\3&6&0&3\end{bmatrix}\). Find its row space.
The stated rows are, entry by entry, \((2,4,1,3)\), \((1,2,2,3)\), and \((3,6,0,3)\). Reduce \(\mathbf{B}\): swap \(R_1\) and \(R_2\), then replace \(R_2\) by \(R_2-2R_1\) and \(R_3\) by \(R_3-3R_1\):
Now multiply \(R_2\) by \(-\tfrac13\), replace \(R_3\) by \(R_3+6R_2\), and replace \(R_1\) by \(R_1-2R_2\):
The nonzero rows in the final echelon form are \(\mathbf{u}_1^{\top}\) and \(\mathbf{u}_2^{\top}\). They form a basis for \(\operatorname{Row}\mathbf{B}\). Indeed, the original rows were built from \(\mathbf{u}_1^{\top}\) and \(\mathbf{u}_2^{\top}\), and the reduction confirms that the row space contains both. Hence \(\operatorname{Row}\mathbf{B}=\operatorname{span}\{\mathbf{u}_1,\mathbf{u}_2\}\).
Check: The second row of \(\mathbf{B}\) is \(\mathbf{u}_1^{\top}+2\mathbf{u}_2^{\top}=(1,2,2,3)\), and its third row is \(3\mathbf{u}_1^{\top}=(3,6,0,3)\). Both agree with the stated combinations. The reduced matrix has exactly the two nonzero rows used in the basis.
What the Row Space Says About Equations
For \(\mathbf{A}\in\mathbb{R}^{m\times n}\), the equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\) means that every transposed row of \(\mathbf{A}\) has dot product zero with \(\mathbf{x}\). Any linear combination of those rows also has dot product zero with \(\mathbf{x}\), by distributivity of the dot product. Conversely, each row belongs to the row space. Thus the full row space captures exactly the coefficient vectors whose equations are consequences of the original homogeneous equations.
Proof. Suppose first that \(\mathbf{A}\mathbf{x}=\mathbf{0}\). Each row of \(\mathbf{A}\), transposed, has dot product zero with \(\mathbf{x}\). By linearity, every vector in \(\operatorname{Row}\mathbf{A}\) has dot product zero with \(\mathbf{x}\). Equality of the row spaces implies that every row of \(\mathbf{B}\), transposed, also has dot product zero with \(\mathbf{x}\), so \(\mathbf{B}\mathbf{x}=\mathbf{0}\). For the reverse implication, start with \(\mathbf{B}\mathbf{x}=\mathbf{0}\) and use \(\operatorname{Row}\mathbf{B}=\operatorname{Row}\mathbf{A}\) in the same way. Therefore the two systems have the same solutions. \(\square\)
Worked Example: Reading the Coordinate Equations
For the matrix \(\mathbf{A}\) in the first example, write its homogeneous equations and find a basis for the row space. Then express all solutions using two parameters.
Writing \(\mathbf{x}=(x_1,x_2,x_3,x_4)^{\top}\), the original rows give
The second row is the sum of the first and third rows. Subtract the first equation from the third to obtain \(x_3+x_4=0\). Together, the first equation and this difference are equivalent to the first and third original equations; the second original equation is their sum. The resulting coefficient rows are \(\mathbf{u}_1^{\top}=(1,2,0,1)\) and \(\mathbf{u}_2^{\top}=(0,0,1,1)\), the row-space basis already found.
Set \(x_2=s\) and \(x_4=t\), where \(s,t\in\mathbb{R}\). The two equations give \(x_1=-2s-t\) and \(x_3=-t\). Therefore every solution is \(\mathbf{x}=(-2s-t,s,-t,t)^{\top}\). Substitution checks both equations: \(x_1+2x_2+x_4=(-2s-t)+2s+t=0\), and \(x_3+x_4=-t+t=0\). The original second equation is also satisfied because it is the sum of these two equations.
Check: The original rows satisfy \(\mathbf{r}_2=2\mathbf{r}_1+\mathbf{r}_3\), and the third row satisfies \(\mathbf{r}_3=\mathbf{r}_1+\mathbf{u}_2^{\top}\). The displayed coordinate equations and their two-equation reduction are therefore consistent with the row combinations.
The row space describes homogeneous equations through their coefficients. This statement does not mean that the coefficient row space alone determines an arbitrary nonhomogeneous system \(\mathbf{A}\mathbf{x}=\mathbf{b}\): the right-hand side matters too. Row operations applied to a system must be applied to the right-hand side as well.
Common Pitfall: Treating Every Original Row as a Basis Vector
A set of rows may span the row space without being a basis. A basis must also be linearly independent. For the first matrix, a tempting but incorrect answer is that all three original rows form a basis for \(\operatorname{Row}\mathbf{A}\). They are dependent: with transposed rows \(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3\), the relation is \(\mathbf{r}_2-\mathbf{r}_1-\mathbf{r}_3=\mathbf{0}\). In entries, this is \((2,4,1,3)-(1,2,0,1)-(1,2,1,2)=(0,0,0,0)\). The correction is to use the two nonzero rows of an echelon form. Those rows span the same row space and, by the theorem, are independent.
Another useful check is the number of coordinates in a row-space vector: it must equal the number of columns in the matrix. The row space of a \(3\times4\) matrix lies in \(\mathbb{R}^4\), not \(\mathbb{R}^3\). Keeping the rows as transposed column vectors makes this dimension and the dot-product equations explicit.
Check Your Understanding
Use row operations and the definition of row space to justify each answer.
- Find a basis for the row space of \(\begin{bmatrix}1&0&2\\2&0&4\\0&1&1\end{bmatrix}\).
- Explain why elementary row operations preserve the row space, including why the reverse inclusion holds.
- For the matrix in question 1, write the homogeneous coordinate equations and identify any redundant equation.
- Can two matrices with different entries have equal row spaces? Give a reason based on row operations.
- What must be checked before claiming that a spanning list of rows is a basis?
