A square system of equations can be deceptive. Three equations in three unknowns may look determined, but if the third equation is obtained by adding the first two, then only two independent constraints are present. Rank is the invariant that separates apparent size from effective linear information.

[example: Redundant Equations] Over $\mathbb{R}$, consider \begin{align*} x_1+x_2=2. \end{align*} \begin{align*} x_2+x_3=3. \end{align*} \begin{align*} x_1+2x_2+x_3=5. \end{align*} Adding the left-hand sides of the first two equations gives \begin{align*} (x_1+x_2)+(x_2+x_3)=x_1+2x_2+x_3, \end{align*} and adding the right-hand sides gives \begin{align*} 2+3=5. \end{align*} Thus the third equation imposes no new condition beyond the first two. To solve, set $x_2=t$ with $t \in \mathbb{R}$. From $x_1+x_2=2$ we get \begin{align*} x_1+t=2, \end{align*} so $x_1=2-t$. From $x_2+x_3=3$ we get \begin{align*} t+x_3=3, \end{align*} so $x_3=3-t$. Hence every solution has the form \begin{align*} (x_1,x_2,x_3)=(2-t,t,3-t). \end{align*} Conversely, substituting this expression into the three equations gives \begin{align*} (2-t)+t=2, \end{align*} \begin{align*} t+(3-t)=3, \end{align*} and \begin{align*} (2-t)+2t+(3-t)=5. \end{align*} So the solution set is the affine line $\{(2-t,t,3-t):t\in\mathbb{R}\}$. The coefficient columns are $(1,0,1)$, $(1,1,2)$, and $(0,1,1)$. They satisfy \begin{align*} (1,1,2)=(1,0,1)+(0,1,1), \end{align*} so the column space is contained in the span of $(1,0,1)$ and $(0,1,1)$. If \begin{align*} \alpha(1,0,1)+\beta(0,1,1)=(0,0,0), \end{align*} then comparing coordinates gives $\alpha=0$, $\beta=0$, and $\alpha+\beta=0$. Thus these two columns are independent, the column space has dimension $2$, and the coefficient matrix has rank $2$. [/example]

The example reveals the central problem of the subject: count independence, not presentation. Rank answers the same question for matrices, linear maps, systems of equations, finitely generated modules, and tensor decompositions. Each setting has its own technical language, but the guiding idea is stable: rank measures how many independent simple directions are needed to describe the object.

## Definition

A matrix can be read as a list of output vectors. If a right-hand side $b$ is to be produced by a system $Ax=b$, then $b$ must be built as a linear combination of the columns. The primary numerical question is therefore: how many independent output directions do the columns actually generate?

[definition: Matrix Rank] Let $k$ be a field, and let $A \in k^{m \times n}$ have columns $a_1, \dots, a_n \in k^m$. The rank of $A$ is \begin{align*} \operatorname{rank}(A) := \dim_k \operatorname{span}_k\{a_1, \dots, a_n\}. \end{align*} [/definition]

This definition measures the output content of the matrix without being distracted by how many columns were written down. To use rank in systems, though, we need to refer not only to its dimension but to the actual set of attainable right-hand sides. Naming that set lets us ask whether a given vector $b$ can occur as $Ax$.

[definition: Column Space] Let $k$ be a field, and let $A \in k^{m \times n}$ have columns $a_1, \dots, a_n \in k^m$. The column space of $A$ is the subspace \begin{align*} \operatorname{Col}(A) := \operatorname{span}_k\{a_1, \dots, a_n\} \subset k^m. \end{align*} [/definition]

With this notation, the rank is $\operatorname{rank}(A)=\dim_k \operatorname{Col}(A)$. Column space records outputs, but row reduction works by changing equations. To connect computation with geometry, we also need the subspace generated by the equations themselves.

[definition: Row Space] Let $k$ be a field, and let $A \in k^{m \times n}$ have rows $r_1, \dots, r_m \in k^n$. The row space of $A$ is the subspace \begin{align*} \operatorname{Row}(A) := \operatorname{span}_k\{r_1, \dots, r_m\} \subset k^n. \end{align*} [/definition]

A problem remains: the same [linear map](/page/Linear%20Map) can be represented by many matrices after changing bases. To avoid making rank depend on a coordinate choice, we need an intrinsic definition using the image of the map itself.

[definition: Rank of a Linear Map] Let $k$ be a field, let $V$ and $W$ be finite-dimensional $k$-vector spaces, and let $T: V \to W$ be a $k$-linear map. The rank of $T$ is \begin{align*} \operatorname{rank}(T) := \dim_k \operatorname{im}(T). \end{align*} [/definition]

This leaves a common decision problem: sometimes the exact rank matters less than whether it is as large as the dimensions allow. That question motivates the language of full rank.

[definition: Full Rank] Let $k$ be a field, and let $A \in k^{m \times n}$. The matrix $A$ has full rank if \begin{align*} \operatorname{rank}(A)=\min\{m,n\}. \end{align*} [/definition]

Full rank turns independence into a yes-or-no test, but rectangular matrices force us to separate two different ways of being maximal. When a map has more output coordinates than input coordinates, it cannot reach every output vector; the useful question is instead whether it loses any input information. This is the situation measured by full column rank: every column must contribute a new independent direction, so distinct input vectors remain distinguishable after applying $A$.

[definition: Full Column Rank] Let $k$ be a field, and let $A \in k^{m \times n}$. The matrix $A$ has full column rank if \begin{align*} \operatorname{rank}(A)=n. \end{align*} [/definition]

Full column rank says that no input direction is lost. When there are at least as many columns as rows, the opposite question becomes more important: can the columns generate the whole target space, so that every possible right-hand side is reachable?

[definition: Full Row Rank] Let $k$ be a field, and let $A \in k^{m \times n}$. The matrix $A$ has full row rank if \begin{align*} \operatorname{rank}(A)=m. \end{align*} [/definition]

In a rectangular matrix, full column rank separates having no lost input directions from full row rank, which separates reaching every output direction.

[example: A Rank-Two Matrix Without Matrix Notation] Let $A: \mathbb{Q}^4 \to \mathbb{Q}^3$ be the linear map whose columns are \begin{align*} a_1=(1,0,1),\quad a_2=(2,1,3),\quad a_3=(0,1,1),\quad a_4=(1,3,4). \end{align*} We compute the span of the columns. First, \begin{align*} 2a_1+a_3=2(1,0,1)+(0,1,1)=(2,0,2)+(0,1,1)=(2,1,3)=a_2. \end{align*} Also, \begin{align*} a_1+3a_3=(1,0,1)+3(0,1,1)=(1,0,1)+(0,3,3)=(1,3,4)=a_4. \end{align*} Thus each of $a_1,a_2,a_3,a_4$ lies in $\operatorname{span}_{\mathbb{Q}}\{a_1,a_3\}$, so \begin{align*} \operatorname{span}_{\mathbb{Q}}\{a_1,a_2,a_3,a_4\}=\operatorname{span}_{\mathbb{Q}}\{a_1,a_3\}. \end{align*} It remains to check that $a_1$ and $a_3$ are independent. Suppose \begin{align*} \alpha a_1+\beta a_3=(0,0,0) \end{align*} with $\alpha,\beta \in \mathbb{Q}$. Substituting $a_1=(1,0,1)$ and $a_3=(0,1,1)$ gives \begin{align*} \alpha(1,0,1)+\beta(0,1,1)=(\alpha,\beta,\alpha+\beta)=(0,0,0). \end{align*} Comparing the first coordinate gives $\alpha=0$, and comparing the second coordinate gives $\beta=0$. Hence $a_1$ and $a_3$ form an independent spanning set for the column space. Therefore the column space has dimension $2$, so $\operatorname{rank}(A)=2$. [/example]