Is a Vector in the Span?

A linear combination of vectors $v_1,\ldots,v_k$ (each in $\mathbb{R}^n$) is any sum $x_1v_1 + x_2v_2 + \cdots + x_kv_k,$ where the weights (coefficients) $x_1,\ldots,x_k$ are real numbers. The span, written $\text{span}\{v_1,\ldots,v_k\}$, is the set of all such combinations.

So the question "is $b$ in $\text{span}\{v_1,\ldots,v_k\}$?" really asks: can I choose weights $x_1,\ldots,x_k$ so that $x_1v_1 + x_2v_2 + \cdots + x_kv_k = b\ ?$

• If yes, then $b$ is in the span, and those weights tell you exactly how to build $b$ out of the given vectors.
• If no choice of weights works, then $b$ is not in the span.

Geometrically, $\text{span}\{v\}$ (one nonzero vector) is a line through the origin, $\text{span}\{u,v\}$ (two non-parallel vectors) is a plane through the origin, and so on. Asking whether $b$ is in the span is asking whether $b$ sits on that line/plane.

The vector equation $x_1v_1 + \cdots + x_kv_k = b$ is a linear system in disguise. Stack the vectors as the columns of a matrix $A = \begin{bmatrix} v_1 & v_2 & \cdots & v_k \end{bmatrix},$ so $A$ is $n\times k$ (here $n$ is the number of entries in each vector and $k$ is the number of vectors). The equation becomes the matrix equation $Ax = b, \qquad x = \begin{bmatrix} x_1 \\ \vdots \\ x_k \end{bmatrix}.$

To solve it, build the augmented matrix by placing $b$ as one extra column on the right: $[A \mid b] = \left[\begin{array}{ccc|c} v_1 & \cdots & v_k & b \end{array}\right].$

Each original vector is a column on the left of the bar; $b$ is the single column to the right. Row reducing this matrix solves the system — and the answer to "is $b$ in the span?" is read directly off the result.

A system is consistent if it has at least one solution and inconsistent if it has none. Span membership is exactly consistency: $b \in \text{span}\{v_1,\ldots,v_k\} \iff Ax=b \text{ is consistent.}$

To check, row reduce $[A\mid b]$ to echelon form and inspect the last column:

• If some row has the form $\begin{bmatrix} 0 & 0 & \cdots & 0 & \mid & c \end{bmatrix}$ with $c \neq 0$ — a pivot in the augmented column — that row asserts $0 = c$, a contradiction. The system is inconsistent, so $b$ is not in the span.
• If no such row appears, the system is consistent, so $b$ is in the span.

Equivalently, in rank language: the system is consistent iff $\operatorname{rank}(A) = \operatorname{rank}([A\mid b])$ — that is, adjoining $b$ created no new pivot.

When the system is consistent, finishing the row reduction hands you the weights.

• Unique solution — a pivot in every coefficient column, i.e. no free variables. There is exactly one weight vector $x$. This occurs precisely when $v_1,\ldots,v_k$ are linearly independent. For example, reducing to $\begin{bmatrix} 1 & 0 & \mid & 2 \\ 0 & 1 & \mid & 1 \end{bmatrix}$ gives $x_1=2,\ x_2=1$.
• Infinitely many solutions — at least one free variable (a coefficient column with no pivot). Then $b$ is still in the span, but it can be written as a combination in infinitely many ways. This happens when the spanning vectors are linearly dependent.

To read off a solution, set any free variables to a convenient value (often $0$) and back-substitute. Always sanity-check: plug the weights back in and confirm $x_1v_1 + \cdots + x_kv_k = b$.