Linear Algebra · N.VM.C.11

Multiplying a matrix by a vector

A matrix sits next to a vector. What single vector does their product stand for, and what does it tell you about which right sides an equation can reach?

You need

Matrix multiplicationPrecalculus Linear combinationLinear Algebra

Unlocks

Linear transformationLinear Algebra Homogeneous systems and solutions in parametric vector formLinear Algebra

Where you are 9 in Matrices as linear transformations

Linear algebra6

Matrices as linear transformations9

▴ 2 above Image Linear transformation

Multiplying a matrix by a vector

Kernel Rank and nullity ▾ Show all 9 (2 more)

When you reach for it

Three earlier lessons meet here. Linear combination scales vectors and adds them, and asks which weights build a given vector. Matrix multiplication gives each entry of a product as a row of the first factor times a column of the second. Writing a system as a matrix equation packs a system into one equation A\mathbf{x} = \mathbf{b}.

A matrix times a vector can be read row by row, as the multiplication rule computes it, or column by column, as a combination of the columns of A with the entries of \mathbf{x} as weights. The two readings always agree, and the column reading is the one that explains things: it is why A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}, and why A\mathbf{x} = \mathbf{b} has a solution exactly when \mathbf{b} is built from the columns. Reach for it whenever fixed ingredients are mixed in chosen amounts.

The procedure

Two readings of one product

Let A be an m \times n matrix with columns \mathbf{a}_1, \dots, \mathbf{a}_n, each in \mathbb{R}^m, and let \mathbf{x} = (x_1, \dots, x_n) be a column vector in \mathbb{R}^n. Then

A\mathbf{x} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \dots + x_n\mathbf{a}_n.

This is the column reading: entry j of \mathbf{x} is the weight on column j. It needs one weight per column, so \mathbf{x} must have n entries, and the result lives where the columns live, in \mathbb{R}^m.

The row reading is the multiplication rule with a one-column second factor: entry i of A\mathbf{x} is row i of A times \mathbf{x},

(A\mathbf{x})_i = a_{i1}x_1 + a_{i2}x_2 + \dots + a_{in}x_n.

They agree because they add the same products. Entry i of column j is a_{ij}, so entry i of x_1\mathbf{a}_1 + \dots + x_n\mathbf{a}_n is x_1a_{i1} + \dots + x_na_{in}, which is the row sum term for term (multiplication of numbers is commutative). The row reading groups the products by row, the column reading by column.

The standard vector \mathbf{e}_j puts weight 1 on column j and 0 elsewhere, so

A\mathbf{e}_j = \mathbf{a}_j:

multiplying by \mathbf{e}_j picks out column j.

Linearity

For every \mathbf{u} and \mathbf{v} in \mathbb{R}^n and every scalar c,

A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}, \qquad A(c\mathbf{u}) = c\,A\mathbf{u}.

Why. By the column reading, A(\mathbf{u} + \mathbf{v}) = (u_1 + v_1)\mathbf{a}_1 + \dots + (u_n + v_n)\mathbf{a}_n. Distribute each weight over its column, then reorder and regroup (the commutative and associative properties of vector addition) so the u terms come first:

\begin{aligned} A(\mathbf{u} + \mathbf{v}) &= (u_1\mathbf{a}_1 + \dots + u_n\mathbf{a}_n) \\ &\quad + (v_1\mathbf{a}_1 + \dots + v_n\mathbf{a}_n) \\ &= A\mathbf{u} + A\mathbf{v}. \end{aligned}

Likewise A(c\mathbf{u}) = (cu_1)\mathbf{a}_1 + \dots + (cu_n)\mathbf{a}_n, which is c(u_1\mathbf{a}_1 + \dots + u_n\mathbf{a}_n) by merging products of scalars and distributing c. Together the rules say A carries any combination to the same combination of outputs,

\begin{aligned} &A(c_1\mathbf{u}_1 + \dots + c_k\mathbf{u}_k) \\ &\quad = c_1A\mathbf{u}_1 + \dots + c_kA\mathbf{u}_k, \end{aligned}

and taking c = 0 gives A\mathbf{0} = \mathbf{0}.

Which right sides are reachable

As \mathbf{x} runs over \mathbb{R}^n, the column reading says A\mathbf{x} runs over every combination of the columns, which is their span. So:

A\mathbf{x} = \mathbf{b} is consistent exactly when \mathbf{b} is in \operatorname{span}\{\mathbf{a}_1, \dots, \mathbf{a}_n\}, and each solution is a list of weights that builds \mathbf{b} from the columns.

Row reduction of [\,A \mid \mathbf{b}\,] decides it: the system is inconsistent exactly when a row reduces to [\,0 \ \cdots \ 0 \mid d\,] with d \neq 0, and carrying a general \mathbf{b} through turns such rows into equations on \mathbf{b}. And A\mathbf{x} = \mathbf{b} is consistent for every \mathbf{b} in \mathbb{R}^m exactly when the columns span \mathbb{R}^m, which happens exactly when an echelon form of A has a pivot in every row, as Span shows.

Worked examples

Example 1

Both readings, side by side

Compute A\mathbf{x} for

A = \begin{bmatrix} 3 & -2 & 1 \\ 1 & 0 & 4 \end{bmatrix}, \qquad \mathbf{x} = \begin{bmatrix} 2 \\ 5 \\ -1 \end{bmatrix},

first row by row, then as a combination of columns.

Solution

A is 2 \times 3 and \mathbf{x} has 3 entries, one per column, so the product is defined and lies in \mathbb{R}^2.

Row by row. Row 1 times \mathbf{x} is 3(2) + (-2)(5) + 1(-1) = 6 - 10 - 1 = -5. Row 2 times \mathbf{x} is 1(2) + 0(5) + 4(-1) = 2 + 0 - 4 = -2.

By columns. The weights 2, 5, -1 go on the three columns:

\begin{aligned} A\mathbf{x} &= 2\begin{bmatrix} 3 \\ 1 \end{bmatrix} + 5\begin{bmatrix} -2 \\ 0 \end{bmatrix} - 1\begin{bmatrix} 1 \\ 4 \end{bmatrix} \\ &= \begin{bmatrix} 6 \\ 2 \end{bmatrix} + \begin{bmatrix} -10 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 \\ -4 \end{bmatrix} \\ &= \begin{bmatrix} -5 \\ -2 \end{bmatrix}. \end{aligned}

Both give A\mathbf{x} = (-5, -2), from the same six products.

Example 2

A product drawn in the plane

Let A = \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix}. Find A\mathbf{x} for \mathbf{x} = (2, 1) and read it on a picture, then find A\mathbf{e}_1 and A\mathbf{e}_2.

Solution

The columns are \mathbf{a}_1 = (3, 1) and \mathbf{a}_2 = (-1, 2). By the column reading,

A\mathbf{x} = 2\begin{bmatrix} 3 \\ 1 \end{bmatrix} + 1\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}.

The row reading agrees: 3(2) - 1(1) = 5 and 1(2) + 2(1) = 4. In the figure the product is a walk along the columns.

Figure 1With columns \mathbf{a}_1 = (3, 1) and \mathbf{a}_2 = (-1, 2) and \mathbf{x} = (2, 1), the product A\mathbf{x} = 2\mathbf{a}_1 + \mathbf{a}_2 = (5, 4) is where two steps along the first column and one along the second end up.

The standard vectors put all the weight on one column, so A\mathbf{e}_1 = \mathbf{a}_1 = (3, 1) and A\mathbf{e}_2 = \mathbf{a}_2 = (-1, 2): one step along one column.

Example 3

Fertilizer in the spreader

Delali, Elikem and Senyo feed their fields from three fertilizer blends. A sack of blend 1 carries (4, 1, 0) kilograms of (nitrogen, phosphorus, potassium), blend 2 carries (0, 3, 2) and blend 3 carries (1, 1, 5). Put the blends in the columns of

D = \begin{bmatrix} 4 & 0 & 1 \\ 1 & 3 & 1 \\ 0 & 2 & 5 \end{bmatrix}.

(a) Delali spreads \mathbf{x} = (2, 1.5, 1) sacks of the three blends. How much of each nutrient goes on her field? (b) Elikem spreads \mathbf{y} = (1, 2, 0) sacks on the same field. Find the total nutrients two ways. (c) Senyo spreads three times Delali’s recipe. Find his nutrients without a new product.

Solution

(a) Sacks times kilograms per sack is kilograms, so blend j adds x_j times column j:

\begin{aligned} D\mathbf{x} &= 2\begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} + 1.5\begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix} \\ &\quad + 1\begin{bmatrix} 1 \\ 1 \\ 5 \end{bmatrix} = \begin{bmatrix} 9 \\ 7.5 \\ 8 \end{bmatrix}. \end{aligned}

That is 9 kilograms of nitrogen, 7.5 of phosphorus and 8 of potassium. Row 1, (4, 0, 1), lists the nitrogen in a sack of each blend, so row 1 times \mathbf{x} counts the nitrogen.

(b) Elikem’s sacks carry D\mathbf{y} = 1(4, 1, 0) + 2(0, 3, 2) = (4, 7, 4). On one field the nutrients add: D\mathbf{x} + D\mathbf{y} = (13, 14.5, 12). The other way, the combined recipe is \mathbf{x} + \mathbf{y} = (3, 3.5, 1) sacks, and

\begin{aligned} D(\mathbf{x} + \mathbf{y}) &= 3\begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} + 3.5\begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix} \\ &\quad + \begin{bmatrix} 1 \\ 1 \\ 5 \end{bmatrix} = \begin{bmatrix} 13 \\ 14.5 \\ 12 \end{bmatrix}. \end{aligned}

The agreement is the first linearity rule, D(\mathbf{x} + \mathbf{y}) = D\mathbf{x} + D\mathbf{y}.

Example 4

A matrix recovered from two outputs

A 2 \times 2 matrix A satisfies A(1, 1) = (3, 5) and A(1, -1) = (1, -1). Find A(4, 2) and then A itself.

Solution

Write the input as a combination of the known inputs, and linearity does the rest. Weights with c_1(1, 1) + c_2(1, -1) = (4, 2) satisfy c_1 + c_2 = 4 and c_1 - c_2 = 2, so c_1 = 3, c_2 = 1. Then

\begin{aligned} A\begin{bmatrix} 4 \\ 2 \end{bmatrix} &= 3A\begin{bmatrix} 1 \\ 1 \end{bmatrix} + A\begin{bmatrix} 1 \\ -1 \end{bmatrix} \\ &= \begin{bmatrix} 9 \\ 15 \end{bmatrix} + \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 10 \\ 14 \end{bmatrix}. \end{aligned}

The columns of A are A\mathbf{e}_1 and A\mathbf{e}_2, so the same move finds them. Since \displaystyle \mathbf{e}_1 = \frac{1}{2}(1, 1) + \frac{1}{2}(1, -1) and \displaystyle \mathbf{e}_2 = \frac{1}{2}(1, 1) - \frac{1}{2}(1, -1),

\begin{aligned} A\mathbf{e}_1 &= \tfrac{1}{2}(3, 5) + \tfrac{1}{2}(1, -1) = (2, 2), \\ A\mathbf{e}_2 &= \tfrac{1}{2}(3, 5) - \tfrac{1}{2}(1, -1) = (1, 3). \end{aligned}

So A = \begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix}. Check: A(1, 1) = (3, 5), A(1, -1) = (1, -1), and A(4, 2) = (8 + 2, 8 + 6) = (10, 14).

Example 5

Reachable and unreachable right sides

Let

A = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 5 & -1 \\ -1 & 1 & -3 \end{bmatrix}.

Find the condition on \mathbf{b} = (b_1, b_2, b_3) for A\mathbf{x} = \mathbf{b} to be consistent, and decide the cases \mathbf{b} = (3, 7, 0) and \mathbf{b} = (1, 0, 1), solving the consistent one.

Solution

Row reduce [\,A \mid \mathbf{b}\,], carrying a general right side. Subtract 2 times row 1 from row 2, and add row 1 to row 3:

\left[\begin{array}{ccc|c} 1 & 2 & 0 & b_1 \\ 0 & 1 & -1 & b_2 - 2b_1 \\ 0 & 3 & -3 & b_3 + b_1 \end{array}\right].

Subtracting 3 times row 2 from row 3 leaves [\,0 \ \ 0 \ \ 0 \mid 7b_1 - 3b_2 + b_3\,]. So the system is consistent exactly when

7b_1 - 3b_2 + b_3 = 0.

The columns span only this plane, not all of \mathbb{R}^3.

For (3, 7, 0): 21 - 21 + 0 = 0, so it is reachable. The reduced rows give x_2 - x_3 = 7 - 6 = 1 and x_1 + 2x_2 = 3; with x_3 = t free,

\mathbf{x} = (1 - 2t,\ 1 + t,\ t).

With t = 0, \mathbf{a}_1 + \mathbf{a}_2 = (1, 2, -1) + (2, 5, 1) = (3, 7, 0), as it should. For (1, 0, 1): 7 - 0 + 1 = 8 \neq 0, so no \mathbf{x} exists. The free weight comes from \mathbf{a}_3 = 2\mathbf{a}_1 - \mathbf{a}_2, which says A(2, -1, -1) = \mathbf{0}.

Example 6

Every right side reachable

Show that A\mathbf{x} = \mathbf{b} is consistent for every \mathbf{b} in \mathbb{R}^2 when A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 6 & 1 \end{bmatrix}, and find one solution for \mathbf{b} = (1, 7).

Solution

Subtracting 3 times row 1 from row 2 gives the echelon form \begin{bmatrix} 1 & 2 & -1 \\ 0 & 0 & 4 \end{bmatrix}, with a pivot in each row. No row can become [\,0 \ \ 0 \ \ 0 \mid d\,] with d \neq 0, so the columns span \mathbb{R}^2 and every \mathbf{b} is reachable.

For \mathbf{b} = (1, 7) the reduced right side is 7 - 3 = 4, so x_3 = 1, and row 1 gives x_1 + 2x_2 = 1 + x_3 = 2. Taking x_2 = 0, \mathbf{x} = (2, 0, 1). Check by columns: 2(1, 3) + 1(-1, 1) = (1, 7). Other values of x_2 give other solutions.

Boundaries

Sizes must match. A\mathbf{x} is defined only when \mathbf{x} has one entry per column of A. An m \times n matrix takes vectors in \mathbb{R}^n to vectors in \mathbb{R}^m, two different spaces when m \neq n.
Linearity is for products, not products plus a constant. The rule \mathbf{x} \mapsto A\mathbf{x} + \mathbf{c} with \mathbf{c} \neq \mathbf{0} sends \mathbf{0} to \mathbf{c}, so it breaks both rules.
Consistency is not uniqueness. \mathbf{b} in the span of the columns gives at least one solution; there is exactly one only when A\mathbf{x} = \mathbf{0} has no solution but \mathbf{0} (Examples 5 and 6 have many).

Where this goes next

Every matrix gives a Linear transformation, \mathbf{x} \mapsto A\mathbf{x}, and A\mathbf{e}_j = \mathbf{a}_j is how Finding the matrix of a transformation builds a matrix from a geometric move. All the A\mathbf{x} together form the image; the \mathbf{x} with A\mathbf{x} = \mathbf{0}, met in Example 5, form the kernel. Two solutions of A\mathbf{x} = \mathbf{b} differ by a solution of A\mathbf{x} = \mathbf{0}, the start of Homogeneous systems and solutions in parametric vector form. The pivot count that decides spanning is the rank, and one matrix applied after another is Composing transformations as a matrix product.

Exercises

Both readings

In Exercises 1 to 3, compute the product row by row, then check it as a combination of columns.

\begin{bmatrix} 2 & 0 & -3 \\ 1 & 4 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix}
\begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ -3 \end{bmatrix}
\begin{bmatrix} 3 & 1 & 0 & -2 \\ -1 & 2 & 4 & 1 \end{bmatrix}\mathbf{x} with \mathbf{x} = (1, 0, -1, 2)
Find \begin{bmatrix} 2 & 7 \\ -3 & 5 \end{bmatrix}\mathbf{e}_2 without multiplying, and say why.
Write 3(1, -2) - (4, 0) + 2(0, 5) as A\mathbf{x} for a matrix A and vector \mathbf{x}, and compute it.
Let A = \begin{bmatrix} 1 & 0 & 2 \\ 4 & -1 & 3 \end{bmatrix}. Decide whether A\mathbf{x} is defined for (a) \mathbf{x} = (2, 1) and (b) \mathbf{x} = (1, 1, 1), and compute it where it is.

Linearity

Suppose A\mathbf{u} = (2, -1, 4) and A\mathbf{v} = (0, 3, -2). Find A(\mathbf{u} + \mathbf{v}), A(3\mathbf{u}) and A(\mathbf{u} - 2\mathbf{v}).
A 2 \times 2 matrix A satisfies A(1, 2) = (5, 1) and A(0, 1) = (2, -1). Find A(1, 0) and then A.
With the blends of Example 3, Senyo spreads 1 sack of blend 1, none of blend 2 and 2 sacks of blend 3. Find the nutrients he spreads, then the nutrients for a field needing three times as much, using linearity.
With the same blends, Elikem wants exactly (5, 8, 9) kilograms of (nitrogen, phosphorus, potassium). How many sacks of each blend does he spread?
A matrix A satisfies A(2, 1) = (4, 0, 6). Find A(-6, -3) without knowing A.

Reachable right sides

Let A = \begin{bmatrix} 1 & -2 \\ -3 & 6 \end{bmatrix}. Decide whether A\mathbf{x} = \mathbf{b} is consistent for (a) \mathbf{b} = (3, -9) and (b) \mathbf{b} = (1, 1), and describe the span of the columns.
Let A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}. Find the condition on \mathbf{b} for A\mathbf{x} = \mathbf{b} to be consistent, decide the cases (2, 3, 5) and (1, 1, 1), and give a solution where one exists.
Let A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 2 & 3 \end{bmatrix}. Find h so that A\mathbf{x} = (1, h, 4) is consistent, and give the solution.
Let A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & k \end{bmatrix}. For which k is A\mathbf{x} = \mathbf{b} consistent for every \mathbf{b} in \mathbb{R}^3?
Let B = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 3 & 1 \end{bmatrix}. Find every \mathbf{x} with B\mathbf{x} = \mathbf{0}. Then compute B(1, 1, 1) and use linearity to give two other vectors with the same product.

Reasoning

Prove A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} and A(c\mathbf{u}) = c\,A\mathbf{u} from the row reading instead of the column reading.
Suppose A\mathbf{p} = \mathbf{b} and A\mathbf{q} = \mathbf{b}. Show that A(\mathbf{p} - \mathbf{q}) = \mathbf{0}. Conversely, show that if A\mathbf{z} = \mathbf{0} then \mathbf{p} + \mathbf{z} also solves A\mathbf{x} = \mathbf{b}.
Prove that the set of \mathbf{b} for which A\mathbf{x} = \mathbf{b} is consistent contains \mathbf{0} and is closed under addition and under scaling.
Give a counterexample to the claim “if A\mathbf{x} = A\mathbf{y}, then \mathbf{x} = \mathbf{y}.”
Explain why, for a 3 \times 2 matrix A, the equation A\mathbf{x} = \mathbf{b} cannot be consistent for every \mathbf{b} in \mathbb{R}^3.

Answers

(-4, -5). Rows: 2 + 0 - 6 = -4 and 1 - 8 + 2 = -5. Columns: (2, 1) - 2(0, 4) + 2(-3, 1) = (-4, -5).
(7, -6, 11). Rows: 4 + 3, 0 - 6, 20 - 9. Columns: 4(1, 0, 5) - 3(-1, 2, 3) = (7, -6, 11).
(-1, -3). Rows: 3 + 0 + 0 - 4 = -1 and -1 + 0 - 4 + 2 = -3. Columns: (3, -1) - (0, 4) + 2(-2, 1) = (-1, -3).
(7, 5), column 2: \mathbf{e}_2 puts weight 1 on column 2 and 0 on column 1.
A = \begin{bmatrix} 1 & 4 & 0 \\ -2 & 0 & 5 \end{bmatrix}, \mathbf{x} = (3, -1, 2): the vectors are the columns and the scalars the weights. A\mathbf{x} = (3 - 4 + 0, -6 + 0 + 10) = (-1, 4).
(a) Not defined: A has 3 columns and \mathbf{x} has 2 entries. (b) (3, 6), from 1 + 0 + 2 and 4 - 1 + 3.
A(\mathbf{u} + \mathbf{v}) = (2, 2, 2), A(3\mathbf{u}) = (6, -3, 12), A(\mathbf{u} - 2\mathbf{v}) = (2, -1, 4) - (0, 6, -4) = (2, -7, 8), all by linearity.
(1, 0) = (1, 2) - 2(0, 1), so A(1, 0) = (5, 1) - 2(2, -1) = (1, 3). The columns are A\mathbf{e}_1 = (1, 3) and A\mathbf{e}_2 = (2, -1), so A = \begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix}. Check: A(1, 2) = (1 + 4, 3 - 2) = (5, 1).
D(1, 0, 2) = (4, 1, 0) + 2(1, 1, 5) = (6, 3, 10) kilograms. Tripled: D(3, 0, 6) = 3D(1, 0, 2) = (18, 9, 30) kilograms.
1 sack of blend 1, 2 of blend 2 and 1 of blend 3: (4, 1, 0) + 2(0, 3, 2) + (1, 1, 5) = (5, 8, 9). It is the only recipe: reducing D gives a pivot in every column.
(-6, -3) = -3(2, 1), so A(-6, -3) = -3(4, 0, 6) = (-12, 0, -18).
(a) Consistent: (3, -9) = 3(1, -3), so \mathbf{x} = (3, 0) is one solution. (b) Inconsistent. Column 2 is -2 times column 1, so the span of the columns is the line through \mathbf{0} along (1, -3), and (1, 1) is not a multiple of (1, -3).
Consistent exactly when b_3 = b_1 + b_2: row 3 minus row 1 minus row 2 leaves [\,0 \ \ 0 \ \ 0 \mid b_3 - b_1 - b_2\,]. (2, 3, 5) is reachable, with \mathbf{x} = (2, 3, 0) for one; (1, 1, 1) is not, since 1 \neq 2.
h = -2, with \mathbf{x} = (5, -2). Rows 1 and 3 give x_1 + 2x_2 = 1 and 2x_1 + 3x_2 = 4, so x_2 = -2, x_1 = 5, and row 2 forces h = x_2.
Every k except -1. Row 3 minus row 1 is (0, -1, k), and adding row 2 gives (0, 0, k + 1): a pivot in every row exactly when k + 1 \neq 0.
\mathbf{x} = t(-5, 3, 1) for any t: row 2 minus 2 times row 1 gives -x_2 + 3x_3 = 0, so x_2 = 3x_3 and x_1 = -2x_2 + x_3 = -5x_3. B(1, 1, 1) = (2, 6), and by linearity (-4, 4, 2) and (6, -2, 0) (t = 1 and t = -1) also give (2, 6).
Entry i of A(\mathbf{u} + \mathbf{v}) is \sum_j a_{ij}(u_j + v_j) = \sum_j a_{ij}u_j + \sum_j a_{ij}v_j by the distributive property, which is entry i of A\mathbf{u} + A\mathbf{v}. Entry i of A(c\mathbf{u}) is \sum_j a_{ij}cu_j = c\sum_j a_{ij}u_j. Equal in every entry, so the vectors are equal.
A(\mathbf{p} - \mathbf{q}) = A\mathbf{p} - A\mathbf{q} = \mathbf{b} - \mathbf{b} = \mathbf{0}, by linearity with weights 1, -1. And A(\mathbf{p} + \mathbf{z}) = A\mathbf{p} + A\mathbf{z} = \mathbf{b} + \mathbf{0} = \mathbf{b}.
\mathbf{0} = A\mathbf{0}. If \mathbf{b} = A\mathbf{x} and \mathbf{b}' = A\mathbf{x}', then \mathbf{b} + \mathbf{b}' = A(\mathbf{x} + \mathbf{x}') and c\mathbf{b} = A(c\mathbf{x}), so both are reachable.
A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \mathbf{x} = (1, 0), \mathbf{y} = (0, 1): A\mathbf{x} = A\mathbf{y} = (1, 1) but \mathbf{x} \neq \mathbf{y}.
The reachable \mathbf{b} form the span of the two columns. An echelon form of A has at most one pivot per column, so at most two, and A has three rows; some row has no pivot, so the columns do not span \mathbb{R}^3 and some \mathbf{b} is unreachable.