1.3 Vector equations

Big idea

The same system of equations gets a second reading here. Instead of rows, look at columns. A system becomes a question about mixing a few fixed vectors to hit a target vector, and the numbers you mix them with are the unknowns.

Two operations generate everything: adding two vectors and scaling a vector by a number. Apply them to a fixed list of vectors in every possible way and you get a set called the span of that list. The central question of the section is whether a given vector belongs to a given span.

That question is not new. Writing the mixture out coordinate by coordinate produces a linear system whose augmented matrix has the given vectors as its columns and the target as its last column. So membership in a span is consistency of a system, and you already own the method for deciding consistency. The value here is the translation, because the geometric side (lines, planes, arrows) makes results memorable that would otherwise be rows of arithmetic.

Decoder

A vector in this chapter is a column of numbers, and $\mathbf{R}^n$ is the set of all columns with $n$ entries. Written in a sentence they get squeezed into a row with commas, but they are columns; the column shape is what makes the matrix connection work.

A vector equation has a solution exactly when the system with matching augmented matrix has one, and the two solution sets are the same list of numbers.

Unpacked: the vector equation $x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n = \mathbf{b}$ and the linear system whose augmented matrix has columns $\mathbf{a}_1, \dots, \mathbf{a}_n, \mathbf{b}$ are the same statement written twice. One line of the vector equation is one equation of the system. Nothing is proved by the sentence; it is a dictionary entry.

Three phrasings mean the same thing and get used interchangeably from here on: $\mathbf{b}$ is a linear combination of $\mathbf{a}_1, \dots, \mathbf{a}_n$; $\mathbf{b}$ belongs to the span of those vectors; the system with that augmented matrix is consistent.

Definitions and results

Vectors in $\mathbf{R}^n$. A vector is an ordered column of $n$ real numbers. Two vectors are equal when all matching entries are equal, so order matters and length matters. The zero vector $\mathbf{0}$ has every entry $0$.

Sum and scalar multiple. Add entry by entry. Multiply by a scalar $c$ by multiplying every entry by $c$. Both results live in the same $\mathbf{R}^n$.

Algebra of the two operations. Addition is commutative and associative. The zero vector is neutral, and $-\mathbf{u}$ undoes $\mathbf{u}$. Scalars distribute over vector sums, vector sums of scalars distribute over a vector, scalars multiply associatively, and $1\mathbf{u} = \mathbf{u}$. These are the rules you already use for numbers, and they let you rearrange vector expressions without thinking about entries.

Geometry in $\mathbf{R}^2$ and $\mathbf{R}^3$. Picture a vector as an arrow from the origin to the point with those coordinates. The sum is the fourth corner of the parallelogram built on the two arrows. A scalar multiple stretches the arrow by the size of the scalar and flips it when the scalar is negative. The set of all multiples of one nonzero vector is a line through the origin.

Linear combination. Given vectors $\mathbf{v}_1, \dots, \mathbf{v}_p$ and scalars $c_1, \dots, c_p$, the vector $c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p$ is a linear combination of them with those weights. Weights may be zero or negative.

Vector equation and matrix translation. The equation $x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n = \mathbf{b}$ has the same solution set as the linear system whose augmented matrix is the one with columns $\mathbf{a}_1, \dots, \mathbf{a}_n, \mathbf{b}$. In particular $\mathbf{b}$ is a linear combination of the $\mathbf{a}_i$ exactly when that system is consistent.

Span. The span of $\mathbf{v}_1, \dots, \mathbf{v}_p$, written $\text{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$, is the set of all their linear combinations. It always contains $\mathbf{0}$ (take every weight zero) and each $\mathbf{v}_i$ (take one weight $1$ and the rest zero). It is closed under sums and scalar multiples: combine two members and you land back inside.

Shape of a span. In $\mathbf{R}^3$, the span of one nonzero vector is a line through the origin. The span of two vectors is a plane through the origin when neither is a multiple of the other, and just a line when one is a multiple of the other. Asking whether $\mathbf{b}$ lies in a span asks whether the point sits on that line or plane.

Worked examples

Combining two vectors

Let

$$ \mathbf{u} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}, \qquad \mathbf{v} = \begin{bmatrix} -1 \\ 3 \end{bmatrix} $$

Compute $3\mathbf{u} - 2\mathbf{v}$. Scale first: $3\mathbf{u}$ has entries $6$ and $-3$, and $2\mathbf{v}$ has entries $-2$ and $6$. Subtract entry by entry:

$$ 3\mathbf{u} - 2\mathbf{v} = \begin{bmatrix} 6 - (-2) \\ -3 - 6 \end{bmatrix} = \begin{bmatrix} 8 \\ -9 \end{bmatrix} $$

Check by rebuilding: $\begin{bmatrix} 8 \\ -9 \end{bmatrix} + 2\mathbf{v}$ has entries $8 - 2 = 6$ and $-9 + 6 = -3$, which is $3\mathbf{u}$.

Deciding membership in a span

Let

$$ \mathbf{a}_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \qquad \mathbf{a}_2 = \begin{bmatrix} -2 \\ 1 \\ 3 \end{bmatrix}, \qquad \mathbf{b} = \begin{bmatrix} 3 \\ -1 \\ -1 \end{bmatrix} $$

Is $\mathbf{b}$ in $\text{Span}\{\mathbf{a}_1, \mathbf{a}_2\}$? Write the vector equation $x_1\mathbf{a}_1 + x_2\mathbf{a}_2 = \mathbf{b}$ and read it one coordinate at a time:

$$ x_1 - 2x_2 = 3 $$

$$ x_2 = -1 $$

$$ 2x_1 + 3x_2 = -1 $$

The middle line gives $x_2 = -1$. The first gives $x_1 = 3 + 2(-1) = 1$. The third must now hold on its own, and it does: $2(1) + 3(-1) = -1$. The system is consistent, so the answer is yes, with weights $1$ and $-1$.

Check the combination directly: $\mathbf{a}_1 - \mathbf{a}_2$ has entries $1 - (-2) = 3$, $0 - 1 = -1$, and $2 - 3 = -1$. That is $\mathbf{b}$.

A vector outside the span

Keep the same $\mathbf{a}_1$ and $\mathbf{a}_2$ and ask about

$$ \mathbf{c} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$

The second coordinate forces $x_2 = 1$. The first then forces $x_1 = 1 + 2(1) = 3$. The third coordinate demands $2(3) + 3(1) = 1$, but that left side is $9$. The system is inconsistent, so $\mathbf{c}$ is not in the span.

Geometrically, $\mathbf{a}_1$ and $\mathbf{a}_2$ are not multiples of each other, so their span is a plane through the origin in $\mathbf{R}^3$. The point $\mathbf{b}$ lies on that plane and the point $\mathbf{c}$ sits off it. A span in $\mathbf{R}^3$ built from two vectors is a plane, never all of space, so most vectors will miss it.

Practice

Start with the arithmetic of the two operations. These drills ask for sums, differences, and scalar multiples of small vectors.

Next, the weights. You are given vectors and a target and asked to produce the combination that hits it, or to evaluate a combination with stated weights.

Last, membership and shape. You decide whether a vector lies in a span and describe the span you are looking at.

Videos

The first video builds the arrow picture and the coordinate picture of a vector and shows why they are the same object. The second introduces span directly. Watch for the moment where the weights become sliders: that is the mental image behind every span question in this chapter.

Quiz

Six items across vector arithmetic, linear combinations, and span.