1.3 Vector Equations

Theory: Vector Equations

Sections 1.1–1.2 looked at a linear system row by row (equations). Sections 1.3–1.4 look at the same system column by column (vectors). The punchline: solving a system, writing $\mathbf{b}$ as a linear combination of vectors, and solving $A\mathbf{x} = \mathbf{b}$ are the same problem, and all three are solved by row reducing the same augmented matrix.

Vectors in $\mathbb{R}^n$ (Lay §1.3)

A vector in $\mathbb{R}^n$ is an ordered list of $n$ real numbers, written as a column (an $n \times 1$ matrix):

$$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \mathbb{R}^n$$

Two vectors are equal iff their corresponding entries are equal, so $\begin{bmatrix} 7 \\ 4 \end{bmatrix} \ne \begin{bmatrix} 4 \\ 7 \end{bmatrix}$ (order matters). The zero vector $\mathbf{0}$ has every entry $0$.

Size convention: an $m \times n$ matrix has $m$ rows and $n$ columns (row first). Each column of an $m \times n$ matrix is a vector in $\mathbb{R}^m$.

Vector operations are done entry by entry:

$$\mathbf{u} + \mathbf{v} = \begin{bmatrix} u_1 + v_1 \\ \vdots \\ u_n + v_n \end{bmatrix}, \qquad c\,\mathbf{u} = \begin{bmatrix} c\,u_1 \\ \vdots \\ c\,u_n \end{bmatrix}$$

For example, with $\mathbf{u} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$, $\mathbf{v} = \begin{bmatrix} 2 \\ -5 \end{bmatrix}$: $\ 4\mathbf{u} - 3\mathbf{v} = \begin{bmatrix} 4 \\ -8 \end{bmatrix} + \begin{bmatrix} -6 \\ 15 \end{bmatrix} = \begin{bmatrix} -2 \\ 7 \end{bmatrix}$.

Geometry (in $\mathbb{R}^2$ and $\mathbb{R}^3$): a vector is an arrow from the origin to the point $(x_1, x_2)$.

• Parallelogram rule: $\mathbf{u} + \mathbf{v}$ is the fourth vertex of the parallelogram with vertices $\mathbf{0}, \mathbf{u}, \mathbf{v}$.
• $c\mathbf{u}$ stretches $\mathbf{u}$ by $|c|$; it points the opposite way if $c \lt 0$.

Algebraic properties of $\mathbb{R}^n$ (for all $\mathbf{u}, \mathbf{v}, \mathbf{w}$ and scalars $c, d$): $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$; $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$; $\mathbf{u} + \mathbf{0} = \mathbf{u}$; $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$; $c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}$; $(c + d)\mathbf{u} = c\mathbf{u} + d\mathbf{u}$; $c(d\mathbf{u}) = (cd)\mathbf{u}$; $1\mathbf{u} = \mathbf{u}$. In short: vectors behave just like numbers under $+$ and scaling.

Linear Combinations & Weights

Given $k$ vectors $\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathbb{R}^n$ and scalars $c_1, \dots, c_k$, the vector

$$\mathbf{y} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k$$

is a linear combination of $\mathbf{v}_1, \dots, \mathbf{v}_k$ with weights $c_1, \dots, c_k$. Weights can be any real numbers, including $0$ and negatives. For example, $\sqrt{3}\,\mathbf{v}_1 + \mathbf{v}_2$, $\tfrac{1}{2}\mathbf{v}_1$ and $\mathbf{0} = 0\mathbf{v}_1 + 0\mathbf{v}_2$ are all linear combinations of $\mathbf{v}_1, \mathbf{v}_2$.

Vector Equation $\iff$ Augmented Matrix

Question type: “Is $\mathbf{b}$ a linear combination of $\mathbf{a}_1, \dots, \mathbf{a}_n$?” This asks whether the vector equation

$$x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n = \mathbf{b}$$

has a solution. Writing it out entry by entry gives an ordinary linear system, whose augmented matrix simply has the vectors as its columns.

Key fact (Lay §1.3): The vector 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 $\left[\begin{array}{cccc|c} \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n & \mathbf{b} \end{array}\right].$ In particular, $\mathbf{b}$ is a linear combination of $\mathbf{a}_1, \dots, \mathbf{a}_n$ iff this system is consistent.

Mini-example (Lay §1.3, Example 1)

Is $\mathbf{b} = \begin{bmatrix} 7 \\ 4 \\ -3 \end{bmatrix}$ a linear combination of $\mathbf{a}_1 = \begin{bmatrix} 1 \\ -2 \\ -5 \end{bmatrix}$, $\mathbf{a}_2 = \begin{bmatrix} 2 \\ 5 \\ 6 \end{bmatrix}$? Row reduce $[\mathbf{a}_1\ \mathbf{a}_2\ \mathbf{b}]$ with $R_2 \leftarrow R_2 + 2R_1$, $R_3 \leftarrow R_3 + 5R_1$, then $R_3 \leftarrow R_3 - \tfrac{16}{9}R_2$:

$$\left[\begin{array}{cc|c} 1 & 2 & 7 \\ -2 & 5 & 4 \\ -5 & 6 & -3 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & 2 & 7 \\ 0 & 9 & 18 \\ 0 & 16 & 32 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & 2 & 7 \\ 0 & 9 & 18 \\ 0 & 0 & 0 \end{array}\right]$$

Consistent: $x_2 = 2$, $x_1 = 7 - 4 = 3$. So $\mathbf{b} = 3\mathbf{a}_1 + 2\mathbf{a}_2$ (check: $3(1,-2,-5) + 2(2,5,6) = (7,4,-3)$ ✓).

Recognising inconsistency: if the echelon form has a row $[\,0\ \cdots\ 0 \mid b\,]$ with $b \ne 0$, the augmented column contains a pivot, the system is inconsistent, and $\mathbf{b}$ is not a linear combination of the $\mathbf{a}_i$. You only need echelon form (not RREF) to decide this.

Span$\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$

The set of all linear combinations of $\mathbf{v}_1, \dots, \mathbf{v}_p$ is called the span:

$$\operatorname{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\} = \{c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p : c_1, \dots, c_p \in \mathbb{R}\}$$

So “Is $\mathbf{b} \in \operatorname{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$?” is exactly the question “Does $x_1\mathbf{v}_1 + \cdots + x_p\mathbf{v}_p = \mathbf{b}$ have a solution?”, i.e. “Is $[\mathbf{v}_1\ \cdots\ \mathbf{v}_p \mid \mathbf{b}]$ consistent?”

Every span contains $\mathbf{0}$ (all weights zero) and every $\mathbf{v}_i$ (weight $1$ on $\mathbf{v}_i$, $0$ elsewhere).

Geometric picture in $\mathbb{R}^3$:

Set	Picture
$\operatorname{Span}\{\mathbf{v}\}$, $\mathbf{v} \ne \mathbf{0}$	all multiples $c\mathbf{v}$: a line through the origin
$\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$, $\mathbf{v}$ not a multiple of $\mathbf{u}$	a plane through the origin containing $\mathbf{u}$ and $\mathbf{v}$
$\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$, $\mathbf{v} = c\mathbf{u}$	only the line through $\mathbf{u}$ (the second vector adds nothing new)

“Is $\mathbf{b}$ in the plane spanned by $\mathbf{a}_1$ and $\mathbf{a}_2$?” is a span question. If $[\mathbf{a}_1\ \mathbf{a}_2 \mid \mathbf{b}]$ is inconsistent, $\mathbf{b}$ sticks out of that plane.

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Solved Examples (Textbook Questions)

Exercise 1

(Adapted from Lay §1.3, Exercises 5 and 6)

Write a system of equations that is equivalent to the given vector equation.

(a) $\ x_1\begin{bmatrix} 6 \\ -1 \\ 5 \end{bmatrix} + x_2\begin{bmatrix} -3 \\ 4 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ -7 \\ -5 \end{bmatrix}$

(b) $\ x_1\begin{bmatrix} -2 \\ 3 \end{bmatrix} + x_2\begin{bmatrix} 8 \\ 5 \end{bmatrix} + x_3\begin{bmatrix} 1 \\ -6 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

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Solution

Scale each vector by its weight, add entry by entry, and set equal to the right-hand side.

(a)

$$\begin{bmatrix} 6x_1 \\ -x_1 \\ 5x_1 \end{bmatrix} + \begin{bmatrix} -3x_2 \\ 4x_2 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ -7 \\ -5 \end{bmatrix} \ \Longrightarrow\ \begin{bmatrix} 6x_1 - 3x_2 \\ -x_1 + 4x_2 \\ 5x_1 \end{bmatrix} = \begin{bmatrix} 1 \\ -7 \\ -5 \end{bmatrix}$$

Equating entries:

$$\begin{aligned} 6x_1 - 3x_2 &= 1 \\ -x_1 + 4x_2 &= -7 \\ 5x_1 &= -5 \end{aligned}$$

(b)

$$\begin{bmatrix} -2x_1 + 8x_2 + x_3 \\ 3x_1 + 5x_2 - 6x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \ \Longrightarrow\ \begin{aligned} -2x_1 + 8x_2 + x_3 &= 0 \\ 3x_1 + 5x_2 - 6x_3 &= 0 \end{aligned}$$

(In exams the intermediate vector step is usually not required, but it shows where each equation comes from: equation $i$ uses entry $i$ of every vector.)

Exercise 2

(Adapted from Lay §1.3, Exercises 9 and 10)

Write a vector equation that is equivalent to the given system.

(a) $\quad \begin{aligned} x_2 + 5x_3 &= 0 \\ 4x_1 + 6x_2 - x_3 &= 0 \\ -x_1 + 3x_2 - 8x_3 &= 0 \end{aligned}$

(b) $\quad \begin{aligned} 4x_1 + x_2 + 3x_3 &= 9 \\ x_1 - 7x_2 - 2x_3 &= 2 \\ 8x_1 + 6x_2 - 5x_3 &= 15 \end{aligned}$

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Solution

Read down each variable’s coefficients to build its column; missing variables get $0$.

(a) Note $x_1$ is missing from the first equation, so its coefficient there is $0$:

$$\begin{bmatrix} x_2 + 5x_3 \\ 4x_1 + 6x_2 - x_3 \\ -x_1 + 3x_2 - 8x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \ \Longrightarrow\ x_1\begin{bmatrix} 0 \\ 4 \\ -1 \end{bmatrix} + x_2\begin{bmatrix} 1 \\ 6 \\ 3 \end{bmatrix} + x_3\begin{bmatrix} 5 \\ -1 \\ -8 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

(b)

$$x_1\begin{bmatrix} 4 \\ 1 \\ 8 \end{bmatrix} + x_2\begin{bmatrix} 1 \\ -7 \\ 6 \end{bmatrix} + x_3\begin{bmatrix} 3 \\ -2 \\ -5 \end{bmatrix} = \begin{bmatrix} 9 \\ 2 \\ 15 \end{bmatrix}$$

Exercise 3

(Adapted from Lay §1.3, Exercises 11 and 12)

Determine if $\mathbf{b}$ is a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$.

(a) $\ \mathbf{a}_1 = \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix},\ \mathbf{a}_2 = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix},\ \mathbf{a}_3 = \begin{bmatrix} 5 \\ -6 \\ 8 \end{bmatrix},\ \mathbf{b} = \begin{bmatrix} 2 \\ -1 \\ 6 \end{bmatrix}$

(b) $\ \mathbf{a}_1 = \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix},\ \mathbf{a}_2 = \begin{bmatrix} 0 \\ 5 \\ 5 \end{bmatrix},\ \mathbf{a}_3 = \begin{bmatrix} 2 \\ 0 \\ 8 \end{bmatrix},\ \mathbf{b} = \begin{bmatrix} -5 \\ 11 \\ -7 \end{bmatrix}$

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Solution

“Is $\mathbf{b}$ a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$?” is the same as “Does $x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + x_3\mathbf{a}_3 = \mathbf{b}$ have a solution?”, i.e. “Is $M = [\mathbf{a}_1\ \mathbf{a}_2\ \mathbf{a}_3 \mid \mathbf{b}]$ consistent?” Row reduce until the pivots are visible.

(a) $R_2 \leftarrow R_2 + 2R_1$, then $R_3 \leftarrow R_3 - 2R_2$:

$$M = \left[\begin{array}{ccc|c} 1 & 0 & 5 & 2 \\ -2 & 1 & -6 & -1 \\ 0 & 2 & 8 & 6 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 2 & 8 & 6 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

No pivot in the augmented column, so the system is consistent: $\mathbf{b}$ is a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$.

(Bonus: $x_3$ is free, $x_1 = 2 - 5x_3$, $x_2 = 3 - 4x_3$. Taking $x_3 = 0$: $\mathbf{b} = 2\mathbf{a}_1 + 3\mathbf{a}_2$. Check: $(2, -4, 0) + (0, 3, 6) = (2, -1, 6)$ ✓. There are infinitely many choices of weights.)

(b) $R_2 \leftarrow R_2 + 2R_1$, $R_3 \leftarrow R_3 - 2R_1$, then $R_3 \leftarrow R_3 - R_2$:

$$M = \left[\begin{array}{ccc|c} 1 & 0 & 2 & -5 \\ -2 & 5 & 0 & 11 \\ 2 & 5 & 8 & -7 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 2 & -5 \\ 0 & 5 & 4 & 1 \\ 0 & 5 & 4 & 3 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 2 & -5 \\ 0 & 5 & 4 & 1 \\ 0 & 0 & 0 & 2 \end{array}\right]$$

The last row says $0 = 2$: the augmented column has a pivot, so the system is inconsistent. $\mathbf{b}$ is not a linear combination of $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$.

Exercise 4

(Adapted from Lay §1.3, Exercises 13 and 14)

Determine if $\mathbf{b}$ is a linear combination of the vectors formed from the columns of the matrix $A$.

(a) $\ A = \begin{bmatrix} 1 & -4 & 2 \\ 0 & 3 & 5 \\ -2 & 8 & -4 \end{bmatrix},\ \mathbf{b} = \begin{bmatrix} 3 \\ -7 \\ -3 \end{bmatrix}$

(b) $\ A = \begin{bmatrix} 1 & -2 & -6 \\ 0 & 3 & 7 \\ 1 & -2 & 5 \end{bmatrix},\ \mathbf{b} = \begin{bmatrix} 11 \\ -5 \\ 9 \end{bmatrix}$

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Solution

Call the columns $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$ and row reduce $[\mathbf{a}_1\ \mathbf{a}_2\ \mathbf{a}_3 \mid \mathbf{b}]$ to echelon form.

(a) One operation does it: $R_3 \leftarrow R_3 + 2R_1$:

$$\left[\begin{array}{ccc|c} 1 & -4 & 2 & 3 \\ 0 & 3 & 5 & -7 \\ -2 & 8 & -4 & -3 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -4 & 2 & 3 \\ 0 & 3 & 5 & -7 \\ 0 & 0 & 0 & 3 \end{array}\right]$$

Row 3 reads $0 = 3$, so the system is inconsistent: $\mathbf{b}$ is not a linear combination of the columns of $A$.

(b) $R_3 \leftarrow R_3 - R_1$:

$$\left[\begin{array}{ccc|c} 1 & -2 & -6 & 11 \\ 0 & 3 & 7 & -5 \\ 1 & -2 & 5 & 9 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -2 & -6 & 11 \\ 0 & 3 & 7 & -5 \\ 0 & 0 & 11 & -2 \end{array}\right]$$

This is echelon form with pivots in columns 1, 2, 3 and no pivot in the augmented column, so the system is consistent (in fact it has a unique solution). $\mathbf{b}$ is a linear combination of the columns of $A$.

The question only asks whether $\mathbf{b}$ is a combination, so stop at echelon form. You do not need the actual weights (here they are messy fractions).