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Over the underlying field $F$, the row space is the linear span of the row vectors of $A$, a vector subspace of $F^n$. The column space is the linear span of its column vectors, a vector subspace of $F^m$. The null space isa vector subspace of the domain. The column space is the image of a linear map represented by $A$; the null space is its kernel.

Gaussian elimination givesElementary row operations preserve the row space, so $\{(1,0,1),(0,1,2)\}$ is a basis. The two rows are linearly independent because their first two coordinates form the standard basis of ${\mathbb{R}}^2$.

Reducing the augmented matrix givesThere is no contradictory row. Hence the system is consistent, with affine solution spaceThe three planes intersect in this common line. Their normals span a two-dimensional row space, leaving a one-dimensional common direction, the null space. More explicitly, five times the first equation minus eleven times the second plus twice the third is the zero equation, including the right-hand side; the third plane therefore contains the intersection of the first two.

The rank of $A$ is the dimension of its column space, equivalently the dimension of its row space or the number of pivot columns in a row-echelon form. For an $n\times n$ matrix it lies between zero and $n$.

Since $L$ is an invertible matrix, $Av=0$ is equivalent to $Uv=0$. The last nontrivial row gives $v_3=v_4$; the second gives $v_2=v_4$; and the first gives $2v_1=v_2+v_4$, so $v_1=v_4$. A basis of the null space is thereforeIts single basis vector has Euclidean norm $2$, so the unit null space basis vector is $(1,1,1,1{)}^T/2$.

The pivot elements of $U$ lie in columns $1,2,3$. Take these columns from the original matrix $A$ to obtain a basis of its column space:The fourth column is minus the sum of these three columns, and the leading three-by-three matrix minor has nonzero determinant $-4$, confirming linear independence.

Elementary row operations preserve the row space. Thus the nonzero rows of $U$ give a basis of the row space of $A$:These basis vectors need not have unit Euclidean norm or be orthogonal. Their Euclidean norms are $\sqrt{6},\sqrt{2},2\sqrt{2}$, respectively; the three column basis vectors have Euclidean norms $2\sqrt{2},\sqrt{6},2\sqrt{2}$.

Regard $A$ as a linear map ${\mathbb{R}}^4\to {\mathbb{R}}^4$. By the rank-nullity theorem and matrix rank $3$,All three are vector subspaces of ${\mathbb{R}}^4$, with the row space identified with four-component row vectors in the usual way. The null space lies in the input vector space of concentrations; the column space lies in the output vector space of concentration-change rates. Both have four coordinates, although they play different physical roles.

In this particular matrix, the row sums and column sums both vanish. Put $1=(1,1,1,1{)}^T$. Then $A1=0$ and $1^{T}A=0$, and the two three-dimensional vector subspaces coincide:This equality follows here from both sum identities and the dimensions of a vector space; it does not make $A$ a symmetric matrix and is not true for every square matrix.

The reduced column echelon form of a matrix $A$ is defined by taking its transpose, forming the reduced row echelon form, and transposing back:It is obtained from $A$ by elementary column operations. Its nonzero columns precede its zero columns, the topmost nonzero entry of each nonzero column is $1$, these leading entries move strictly downward as the column number increases, and each leading entry is the only nonzero entry in its row.

Every elementary column operation preserves the column space, because it and its inverse express each new column as a linear combination of the old columns. Thus the nonzero columns of the reduced form give a basis of the original column space. In contrast, elementary row operations preserve the row space. There is no general identity $\mathrm{rcef}(A)=\mathrm{rref}(A{)}^T$: the definition requires reducing $A^T$, not $A$.

For an invertible matrix of size $n$, both reduced forms are $I_n$. A singular example showing the distinction is

A row echelon form is in reduced row echelon form when every pivot element is one and is the only nonzero entry in its column.^[ref] This extra reduction displays each pivot unknown directly in terms of the free variables in a linear system. The zero matrix also satisfies the definition: it has no pivot elements.

Every matrix has exactly one reduced row echelon form under elementary row operations, with its size and column order held fixed.^[ref] Ordinary row echelon form is not unique: multiplying a nonzero row by two over the rational numbers keeps the staircase but changes the matrix. The reduced conditions remove that freedom.

For a fixed number of columns, each row space has a unique reduced matrix after zero rows are removed; at a fixed full size, the number of appended zero rows is also determined. Thus reduced row echelon form represents a vector subspace independently of the starting basis. This is the uniqueness used by Gaussian binomial coefficients to count vector subspaces over a finite field.

The fixed column order matters. Elementary column operations can change the row space, the kernel and the resulting reduced row echelon form. For instance, swapping the columns of $(12)$ over the rational numbers gives $(21)$, whose reduced row echelon form is $(11/2)$ instead of $(12)$. When solving a linear system, a column interchange also requires tracking the corresponding interchange of unknowns.

The row space of a matrix is the span of its rows, and its column space is the span of its columns. Both have dimension equal to the matrix rank, though they lie in different ambient coordinate spaces for a rectangular matrix.

A matrix over a field $\mathbb{F}$ is in row echelon form when all zero rows follow the nonzero rows, each nonzero row's first nonzero entry lies strictly to the right of the preceding row's first nonzero entry, and all entries below each such leading entry vanish. These leading entries are its pivot elements. They need not be one, and entries above them need not vanish.^[ref]

The staircase turns a linear system into a sequence of simpler equations. For example,is in row echelon form, with pivot columns 1 and 2. Its second equation determines $x_2$ once $x_3$ is chosen; its first then determines $x_1$ by back substitution. It is not in reduced row echelon form, because the second pivot column contains a nonzero entry above its pivot element. Neither being square nor having a nonzero entry in every row is required.

Elementary row operations bring every matrix to row echelon form: choose a nonzero entry in the leftmost available nonzero column, interchange rows, clear entries below it, and repeat on the remaining rows and later columns. Skip columns with no remaining nonzero entry. Clearing above each pivot element and scaling it to one completes reduced row echelon form.^[ref]

The nonzero rows of an row echelon form are linearly independent: in a vanishing linear combination, inspecting the first pivot column forces the first coefficient to be zero, then the next forces the second, and so on. They give a basis of the original row space. The number of pivot elements is both the row rank and the column rank. To recover a basis of the original column space, select the corresponding original pivot columns; the reduced column vectors generally belong to a different column space.

Here the arithmetic is exact in the specified field. For floating-point arithmetic, deciding whether a computed entry counts as zero requires a numerical criterion; see Tiny-pivot roundoff in Gaussian elimination and condition number. An exact row echelon form classification alone does not settle numerical stability.

Two matrices of the same size over the same field are row-equivalent matrices if a sequence of elementary row operations transforms one into the other.^[ref] Every operation is reversible, so they have the same row space and the same column linear relations. Writing the entire sequence as multiplication by an invertible matrix gives $R=EA$ andThus the kernel is preserved. For a nonzero right-hand side, perform every operation on the whole augmented matrix $[A\mid b]$: changing the coefficients while keeping $b$ fixed generally changes the solutions. The column space instead transforms to $E\mathrm{col}A$, which need not equal $\mathrm{col}A$.

The row rank of a matrix is the dimension of the vector space spanned by its rows. Elementary row operations preserve this row space; multiplication on the right by an invertible matrix preserves all linear relations among the rows.

To see why row rank equals column rank, write a reduced row echelon form as $R=EA$, where $E$ is a product of elementary matrices and is an invertible matrix. Reversibility of the row operations gives equal row spaces for $R$ and $A$. Also $Rz=0$ if and only if $Az=0$, so the columns have exactly the same linear relations. If $R$ has $r$ nonzero rows, its pivot columns are the first $r$ standard basis vectors of ${\mathbb{F}}^m$ and span its column space. Its nonzero rows are linearly independent, since their matrix entries in the pivot positions form an identity matrix. Both dimensions therefore equal $r$, and both counts transfer back to $A$.

Here is a short proof that uses only the row space. For a vector subspace $W\subseteq {\mathbb{F}}^n$, let $d_j$ be the dimension of its projection map onto the first $j$ coordinates, with $d_0=0$. In any reduced row echelon form with row space $W$, $d_j$ counts the pivot columns among those first $j$ columns: later-pivot rows project to zero, while earlier-pivot rows remain linearly independent. Thus the positions where $d_j-d_{j-1}=1$ determine all pivot positions from $W$ alone.

If those positions are $p_1,\dots ,p_r$, the map sending $w\in W$ to $(w_{p_1},\dots ,w_{p_r})$ is an isomorphism $W\to {\mathbb{F}}^r$: the reduced nonzero rows map to the standard basis. Its $i$th row is therefore the unique vector of $W$ with pivot-coordinate values equal to the $i$th standard basis vector. Any other reduced row echelon form with that row space has identical nonzero rows in identical order. At a fixed matrix size, the remaining rows must be zero. Together with existence by Gaussian elimination, this proves uniqueness.