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If $A$ is $m\times n$, then $Y$ is $n\times m$. The equation $AY=I_m$ makes every standard basis vector of ${\mathbb{R}}^m$ a linear combination of the columns of $A$: take the corresponding column of $Y$ as coefficients. Hence the column space is all of ${\mathbb{R}}^m$, giving $\mathrm{rank}(A)=m$. In particular, $n\ge m$ and $A$ has full row rank.

Choose $r$ independent columns of $A$ that form a basis of its column space, and put them in $B\in {\mathbb{R}}^{m\times r}$. For each column $a_j$ of $A$, let $c_j$ be its coordinate vector in that basis; put these coordinate vectors in $C\in {\mathbb{R}}^{r\times n}$. The column equations $a_j=B{c}_j$ give $A=BC$.

By construction $B$ has full column rank $r$. The columns of $C$ corresponding to the selected basis columns are the standard vectors of ${\mathbb{R}}^r$, so $C$ has full row rank $r$. This proves the requested rank factorization. When $r=0$, $A=0$ and the corresponding empty-column and empty-row factors have the required shapes and rank zero.

Let $r=\mathrm{rank}(A)$. If $r=0$, the matrix is zero and the identity is immediate. Otherwise choose the rank factorization $A=BC$ from part b. The column space of $B$ is the image of $A$.

An idempotent matrix acts as the identity on its image: if $u=Av$, then $Au=A^{2}v=Av=u$. Applying this to each column of $B$ gives $AB=B$. Substituting $A=BC$ yields $B(CB-I_r)=0$. Since $B$ has full column rank, the represented map is injective, so each column of $CB-I_r$ is zero. Thus $CB=I_r$.

Now the rectangular cyclic property of the trace givesThis proves that the trace of an idempotent equals its rank using the supplied factorization method. Orthogonality of the projection is unnecessary: the argument applies to every real idempotent matrix.

Gaussian elimination gives the row operations $R_2\leftarrow R_2-3R_1$, $R_3\leftarrow R_3-R_1$, $R_4\leftarrow R_4-2R_1$, producing the lower three rows $(0,1,4,1)$, $(0,1,4,1)$, $(0,-3,2,-1)$. Subtract the new row two from row three and add three times row two to row four; the remaining nonzero row is $(0,0,14,2)$. Move it above the zero row and finish elimination to obtainThe pivot columns are one, two and three. Row operations preserve the linear dependence relations among columns, so the corresponding columns of the original matrix form a basis for the column space, which is the image of the linear map:In particular, the matrix rank is three. Columns of the reduced matrix itself would generally give a different subspace, so the original columns must be used.

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.

A steady state ${\mathbf{c}}_{\ast }$ satisfies $0=A{\mathbf{c}}_{\ast }-\mathbf{b}$, henceThe sign is positive on the right after rearranging the supplied equation. A solution exists exactly when $\mathbf{b}$ lies in the column space of $A$, which above was found to be the zero-sum vector subspace. Equivalently, multiply by the left null space vector $1^T$:This condition is necessary and sufficient for an algebraic steady state, since both the column space and that zero-sum vector subspace have dimension $3$. It expresses global conservation of mass: in the closed box model, total tracer production must balance total tracer removal. With the paper's convention $-\mathbf{b}$, a positive $b_i$ is a sink and a negative $b_i$ is a source. If the sum is nonzero, $dM/dt=-{\sum }_i{b}_i$ and the inventory changes linearly in time, preventing a steady state.

When the condition holds, all algebraic steady states form the affine solution space of a linear equation ${\mathbf{c}}_p+\alpha 1$, for any particular solution $A{\mathbf{c}}_p=\mathbf{b}$. The initial inventory selects one member through $1^T{\mathbf{c}}_{\ast }=1^T\mathbf{c}(0)$. For a physically admissible solution, its four concentrations must additionally be nonnegative; the compatibility condition alone imposes no such inequalities.

A pivot column of a matrix is a column whose position contains a leading nonzero matrix entry in its row echelon form. The corresponding columns of the original matrix form a basis of its column space. The column positions survive row operations, although the column vectors themselves generally change.^[ref]

A reduced QR decomposition of an $m\times k$ matrix over the real numbers with full column rank, $m\ge k$, writes $A=QR$, where $Q$ is $m\times k$ with orthonormal columns and $R$ is a $k\times k$ upper triangular matrix with positive diagonal. Applying the Gram-Schmidt process to the columns $v_j$ of $A$ constructs $Q=(q_1\cdots q_k)$ and records the coefficientsThis last column identity is exactly the matrix multiplication $A=QR$. The upper triangular matrix records that the $j$th input uses only the first $j$ new directions, while its positive diagonal records the norms of the nonzero orthogonal projection residuals. Since $Q^{T}Q=I$, one also has $R=Q^{T}A$; $Q^T$ denotes the transpose and $I$ the identity matrix. Over the complex numbers, use the conjugate transpose $Q^{\ast }$ in these identities.

For the three-dimensional Gram-Schmidt process example, the exact matrices areFor example, the last column reconstructs asThe diagonal entries are approximately $1.414214,1.224745,1.154701$. Each is a norm of the new direction before it was scaled to length one.

For a linear least-squares problem, the orthogonal projection of $b$ onto the column space of $A$ is $Q{Q}^{T}b$. Hence the minimizing coefficients solve the linear system $Rx=Q^{T}b$ by back substitution. Let $A_2=(v_1{v}_2)$ consist of the first two columns of the example; its reduced factors are $Q_2=(q_1{q}_2)$ and the leading $2\times 2$ block $R_2$ of $R$. For $b=(0,0,1{)}^T$, the linear system $R_{2}x=Q_2^{T}b$ isThus $A_{2}x=(1/3,-1/3,2/3{)}^T$, agreeing with the orthogonal projection computed above, and the minimum Euclidean norm of $A_{2}x-b$ is $1/\sqrt{3}$. The QR decomposition separates building an orthonormal basis from recovering coefficients in the original basis.^[ref]

Choose a random sampling matrix $X$, form $Y=AX$, and find an orthonormal basis $Q$ for the sampled column space by a thin QR decomposition. The low-rank approximation $\hat{A}=Q{Q}^{T}A$ keeps the component of $A$ in that space. Since $Q{Q}^{T}AX=AX$, it reproduces the sampled action exactly. If $Y$ has deficient matrix rank, one can use only a basis of its column space; alternatively an orthonormal completion must be specified.

For a matrix of matrix rank $r$, choose an ordered basis of its column space as the columns of $B$. Express every column of $A$ in that basis and put its coordinates in $C$. Then $A=BC$, $B$ has full column rank and $C$ has full row rank. Choosing independent original columns makes the corresponding columns of $C$ the standard basis vectors, directly proving its rank.

The basis choice is not unique: for any invertible $r\times r$ matrix $S$, the factors $BS$ and $S^{-1}C$ have the same product. Once $B$ is fixed, its coordinate matrix is unique. For rank zero, the empty factors of the displayed shapes represent the zero matrix.

For row reduction with a parameterized right-hand side at $\tau =5$, the nonzero reduced equations areColumn 3 has no pivot element, so $x_3$ is a free variable in a linear system. Taking $x_3=t$ gives every solution:There is no further restriction on $t$. Independent substitution into the original linear system givesConversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the kernel $\ker A=\mathrm{span}\{v\}$. Its dimension is one. The rank-nullity theorem appears here as $2+1=3$: two pivot columns and one freely chosen coordinate account for the three unknowns.^[ref]

In general a consistent linear system has solution set $x_0+\ker A$. Subtracting $A{x}_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the affine subspace, and the kernel supplies its directions.^[ref] Over the real numbers this example is an entire line; over the rational numbers the same formula uses rational $t$. A free variable in a linear system over a finite field instead has only finitely many choices.

The pivot columns identify a different object: the original column space. Here they selectwhich form its basis, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0{)}^T$ and $(0,1,0,0{)}^T$; their linear span is a different plane. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0{)}^T$ does not. Since $E$ is an invertible matrix, column linear relations survive $EA=R$ even though the output vectors move. This is why choosing original columns at the reduced pivot positions works.

Consider the linear map $A:{\mathbb{R}}^3\to {\mathbb{R}}^2$ represented by the rectangular matrixTo compute its kernel, solve the homogeneous linear system $x+2y=0$, $y+z=0$. Taking $y=t$ gives $x=-2t$, $z=-t$, soThus $v$ represents a null direction of a linear map: decreasing the first input by two, increasing the second by one, and decreasing the third by one changes neither output. Solving the homogeneous linear system is the concrete computation behind a kernel; the column space instead decides which outputs are attainable.^[ref]

Now prescribe the nonzero output $b=(3,2{)}^T$. The linear system $Au=b$ becomes $x+2y=3$, $y+z=2$. Again take $y=t$:Direct substitution gives $A{u}_0=(3,2{)}^T$, and $A(u_0+tv)=A{u}_0+tAv=b$. Conversely, the two linear equations force exactly this form, so no solutions have been omitted. This fiber is an affine subspace of dimension one in the three-dimensional input vector space. It is parallel to the kernel, but does not pass through the origin. For instance, $u_0=(3,0,2{)}^T$ and $u_1=(1,1,1{)}^T$ both give $b$, while their difference $u_1-u_0=v$ gives zero. Adding $u_0$ and $u_1$ gives output $2b$, which is another way to see that this nonzero-output fiber is not a vector subspace.

The first and third columns of $A$ are $(1,0{)}^T$ and $(0,1{)}^T$, a basis of the output vector space. Hence $A$ has matrix rank two and reaches every output. The rank-nullity theorem predicts nullity $3-2=1$ before we solve any linear system. It provides a useful completeness check: one nonzero kernel vector already forms a basis here, so a second independent null direction of a linear map would signal a calculation error. In general, matrix rank $r$ for an $m\times n$ matrix leaves $n-r$ independent kernel directions, rather than $m-r$.^[ref]

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

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 $A$.

The row space of a matrix is the linear span of its rows. It is the column space of the matrix transpose.

A system of linear equations imposes several linear equations on the same unknowns. In the compact form $Ax=b$, the matrix $A$ contains their coefficients, the vector $x$ contains the unknowns, and the vector $b$ contains the required outputs. Over the real numbers, the following examples exhibit the three possible outcomes: one solution, infinitely many solutions, or no solution.

For $x+y=3$ and $x-y=1$, adding the two linear equations gives $2x=4$, hence $x=2$ and $y=1$. Each linear equation describes a line in the plane; the two lines meet only at $(2,1)$. Substitution checks both requirements: $2+1=3$ and $2-1=1$. The two measurements together determine both unknowns.

Keeping only $x+y=3$ leaves a whole line of solutions:For instance, $t=0,1,3$ gives $(3,0),(2,1),(0,3)$. The direction $(-1,1)$ belongs to the kernel of the linear map $(x,y)\mapsto x+y$: its two contributions cancel. Moving in that direction changes the unknowns without changing their measured sum. The solution line is an affine space, translated away from the origin.

For $x+y=3$ and $2x+2y=7$, the second linear equation instead demands $x+y=3.5$. The parallel lines never meet. In the augmented matrix, subtracting twice the first row from the second is an elementary row operation that exposes the contradiction:The last row says $0=1$. A zero row of coefficients is harmless only when its appended entry is also zero.

The column viewpoint unifies these pictures. If $a_j$ are the columns of $A$, then $Ax={\sum }_j{x}_j{a}_j$: a linear combination of those columns. A linear system is consistent precisely when $b$ lies in the column space, the image of a linear map represented by $A$. Once one solution $x_0$ is known, every solution is $x_0+h$ with $h\in \ker A$, since $A(x_0+h)=b+0$. Conversely, subtracting any two solutions gives an element of the kernel. Thus the kernel measures freedom, while the image of a linear map decides existence. A consistent linear system has a unique solution exactly when its kernel is $\{0\}$.^[ref]

Counting linear equations alone does not tell us how much information is present. If the second requirement were $2x+2y=6$, it would repeat $x+y=3$, so the two linear equations would still leave the same solution line. The matrix rank counts independent requirements. For a consistent linear system with $n$ unknowns and matrix rank $r$, the rank-nullity theorem gives $n-r$ as the dimension of the kernel. This is the number of independent directions in which a solution can move. In the single-measurement example, one measured combination leaves one direction of freedom; in the sum-and-difference example, neither unknown can change without changing an output.