Vector Equations and the Matrix Equation Ax = b

A linear system can be read through the columns of its coefficient matrix rather than its rows. Two questions come with this view: whether a target vector is reachable as a combination of given vectors, and whether a fixed set of vectors reaches everything. Both restate the consistency question, and both are answered by pivots.

Vectors in $R^{n}$

A matrix with a single column is a column vector, or simply a vector. For now a vector is an ordered list of numbers written vertically; the general definition comes with vector spaces. The set of all vectors with $n$ real entries is $R^{n}$ ,

u = u_{1} u_{2} ⋮ u_{n} \in R^{n} .

Two vectors are equal exactly when their corresponding entries agree, so order matters: $(4, 7) \neq = (7, 4)$ . Two operations are defined entrywise.

A scalar is written in lightface to distinguish it from a boldface vector. In $R^{2}$ the operations have a picture: identify the vector $(a, b)$ with the point $(a, b)$ in the plane, or with the arrow from the origin to that point. Addition is then the parallelogram rule, and scalar multiplication stretches or reverses the arrow.

Addition of two vectors completes the parallelogram on 0, u, v; a scalar multiple c u stretches u by |c| and reverses it when c is negative.

The eight familiar rules — commutativity, associativity, distributivity, and so on — hold entry by entry because they hold for real numbers. They are the axioms that later define a vector space; here they are theorems about lists.

Linear combinations and span

Scaling and adding, combined, produce the central construction of the subject.

Deciding whether a specific $b$ is such a combination reduces to a linear system.

The set of all reachable combinations gets a name.

Every span contains $0$ (take all weights zero) and every scalar multiple of each $v_{i}$ .¹

The geometry of a span

In $R^{3}$ a span has a shape that grows with the number of independent directions it holds.

One nonzero vector. $Span {v}$ is the set of all scalar multiples of $v$ : a line through the origin.
Two vectors, neither a multiple of the other. $Span {u, v}$ is the plane through the origin containing both.

In R^3 the span of one vector is a line through 0; the span of two non-parallel vectors is the plane through 0 that contains both.

A membership test remains a row reduction.

b lies in Span of a_1, a_2 exactly when it sits in their plane through 0; a target off that plane makes the system inconsistent.

The matrix equation $A x = b$

Packaging the vectors as columns of a matrix compresses the notation and defines the matrix-vector product.

So $A x$ is by definition a linear combination of the columns of $A$ , with the entries of $x$ as weights. This definition ties the three formulations together.²

One problem, three equivalent forms; all are solved by row reducing the same augmented matrix.

A worked case: writing the combination $3 v_{1} - 5 v_{2} + 7 v_{3}$ as $A x$ just places the $v_{i}$ into the columns of $A$ and the weights into $x = (3, - 5, 7)$ .

Computing $A x$ : the row-vector rule

The column definition explains what $A x$ is; for hand computation the row rule is faster.

For example,

2 - 1 6 35 - 2 4 - 3 8 x_{1} x_{2} x_{3} = 2 x_{1} + 3 x_{2} + 4 x_{3} - x_{1} + 5 x_{2} - 3 x_{3} 6 x_{1} - 2 x_{2} + 8 x_{3} .

The identity matrix $I_{n}$ (ones on the diagonal, zeros elsewhere) satisfies $I_{n} x = x$ for every $x$ , because each row picks off one entry. Two algebraic properties follow directly from the definition and will be used constantly.

Both follow by writing each side as a linear combination of columns and using the entrywise rules of $R^{n}$ . These two identities underlie linear transformations, where $x \mapsto A x$ becomes the object of study.

Existence: one $b$ versus every $b$

Two existence questions sit side by side, and they are not the same.

Is $A x = b$ consistent for one given $b$ ? By the definition of $A x$ , this holds if and only if $b$ is a linear combination of the columns of $A$ — the span membership test, decided by row reducing $[A b]$ .

Is $A x = b$ consistent for every $b$ in $R^{m}$ ? This is stronger, and it does not depend on any particular $b$ . It depends only on the pivot pattern of $A$ itself.

Statements (a), (b), (c) are the same fact worded three ways; (d) connects it to the pivot pattern. If $A$ has a pivot in every row, then no echelon form of $[A b]$ can carry a row $[0 \dots 0 ∣ 1]$ — every row already spends its pivot on a column of $A$ — so the system is consistent whatever $b$ is. If some row of the echelon form of $A$ is all zeros, a $b$ can be chosen to put a $1$ there, forcing inconsistency.³

A description of the reachable $b$

When the columns do not span $R^{m}$ , the set of reachable $b$ is itself describable.

Question	Depends on	Test
$A x = b$ solvable for this $b$ ?	$A$ and $b$	reduce $[A b]$ ; no pivot in last column
$A x = b$ solvable for every $b$ ?	$A$ alone	reduce $A$ ; a pivot in every row
Which $b$ are reachable?	$A$	span of the columns of $A$

Lay, Linear Algebra and Its Applications, §1.3 — Vector Equations: vectors in $R^{n}$ , the parallelogram rule, linear combinations, span, and the equivalence between a vector equation and the linear system with the vectors as columns. ↩
Lay, §1.4 — The Matrix Equation $A x = b$ : the definition of $A x$ as a linear combination of columns, the row-vector rule, and the three equivalent formulations of a linear system. ↩
Lay, §1.4, Theorem 4 and its proof: the equivalence of the four spanning statements, resting on the fact that $A$ has a pivot in every row iff no echelon form of $[A b]$ has an inconsistent last row. ↩

Vectors in Rn

Linear combinations and span

The geometry of a span

The matrix equation Ax=b

Computing Ax: the row-vector rule

Existence: one b versus every b

A description of the reachable b

Footnotes

Vectors in $R^{n}$

The matrix equation $A x = b$

Computing $A x$ : the row-vector rule

Existence: one $b$ versus every $b$

A description of the reachable $b$