LINEAR-ALGEBRA · Unit 6 · Video 2 · Interactive Practice

Column Space of a Matrix

IKey Formulas

Formula	Name	Description
$Ax = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n$	Matrix-vector product	$Ax$ is a linear combination of $A$ 's columns
$C(A) = \{Ax : x \in \mathbb{R}^n\}$	Column space	Set of all possible output vectors
$\text{rank}(A) = \dim C(A)$	Rank	Equals the number of pivot (independent) columns
$Ax = b$ solvable $\iff b \in C(A)$	Solvability	$b$ must lie in the column space

IIVisualization 1 — The Redundant Column

The third column $\mathbf{a}_3 = \mathbf{a}_1 + \mathbf{a}_2$ adds no new reach — its weight redistributes onto the first two.

x₁ 1 x₂ 0 x₃ 1

💡 For the video's $A$ with $\mathbf{a}_1=(1,2,3,4)$ , $\mathbf{a}_2=(1,1,1,1)$ , $\mathbf{a}_3=(2,3,4,5)=\mathbf{a}_1+\mathbf{a}_2$ , replacing $x_3\mathbf{a}_3$ by $x_3\mathbf{a}_1+x_3\mathbf{a}_2$ gives weights $x_1'=x_1+x_3$ , $x_2'=x_2+x_3$ . The axes are coordinates inside $C(A)$ , measured against the pivot columns $\mathbf{a}_1,\mathbf{a}_2$ .

IIIVisualization 2 — Seeing the Column Space

Two independent columns fill the whole plane; slide $\mathbf{a}_2$ parallel to $\mathbf{a}_1$ and $C(A)$ collapses to a line.

a₂ x 1 a₂ y 2

💡 A 2-D stand-in for the video's plane in $\mathbb{R}^4$ . Challenge: move $\mathbf{a}_2$ until $\det[\,\mathbf{a}_1\ \ \mathbf{a}_2\,]=0$ — the instant the plane becomes a line, and $\mathbf{a}_3=\mathbf{a}_1+\mathbf{a}_2$ collapses onto it too.

IVVisualization 3 — Is b Reachable?

$A\mathbf{x}=\mathbf{b}$ is solvable exactly when $\mathbf{b}$ lands on the column-space line — predict, then reveal.

target b

💡 These columns are dependent ( $\mathbf{a}_2 = 2\mathbf{a}_1$ ), so $C(A)$ is only a line and most targets are unreachable — the same reasoning that left $\mathbf{b}=(0,0,0,1)$ outside the video's rank-2 column space.

VQuiz Questions

Question 1

When you compute the matrix-vector product $Ax$ , the result is best described as:

✅ Correct! Each entry of x weights one column of A, and the output is their weighted sum.

❌ Not quite. Think about what each entry of x does — it scales one column of A.

Show solution

Solution:

The matrix-vector product $Ax$ equals: $Ax = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n$

Each entry $x_i$ acts as a weight on column $\mathbf{a}_i$ . The result is a linear combination of the columns of $A$ , with $x$ providing the weights. This is the fundamental column-space interpretation of matrix multiplication.

Question 2

The video's matrix $A$ is $4 \times 3$ with columns $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$ where $\mathbf{a}_3 = \mathbf{a}_1 + \mathbf{a}_2$ . What is $\dim C(A)$ ?

✅ Correct! Only 2 of the 3 columns are independent, so the column space is 2-dimensional.

❌ Common mistake! Having 3 columns does not mean dim = 3. Column 3 is dependent — it's the freeloader!

❌ Not quite. Count the linearly independent columns of A.

Show solution

Solution: dim C(A) = 2

$A$ has 3 columns, but $\mathbf{a}_3 = \mathbf{a}_1 + \mathbf{a}_2$ means column 3 is linearly dependent. Only columns 1 and 2 are independent — these are the pivot columns.

$\dim C(A) = \text{rank}(A) = \text{number of pivot columns} = \boxed{2}$

Key lesson: The number of columns does NOT determine the dimension. Only independent columns count!

Question 3

True or False: If $A$ is a $5 \times 4$ matrix with $\text{rank}(A) = 3$ , then $Ax = b$ is solvable for every $\mathbf{b} \in \mathbb{R}^5$ .

✅ Correct! Rank 3 < 5 rows, so C(A) doesn't fill all of R⁵. Most b vectors are unreachable.

❌ Think again. Compare rank to the number of rows. Does the column space fill all of R⁵?

Show solution

Solution: False

For $Ax = b$ to be solvable for every $\mathbf{b} \in \mathbb{R}^5$ , the column space must fill all of $\mathbb{R}^5$ : $\text{rank}(A) = m = 5 \quad \text{(full row rank)}$

But $\text{rank} = 3 < 5 = m$ . The column space $C(A)$ is only a 3-dimensional subspace of $\mathbb{R}^5$ . Most vectors in $\mathbb{R}^5$ lie outside this subspace and are unreachable.

There are $5 - 3 = 2$ constraints that $\mathbf{b}$ must satisfy for the system to be solvable.

Question 4

For the matrix $A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix}$ , which vector is NOT in the column space $C(A)$ ?

✅ Correct! As shown in the video, (0,0,0,1) leads to a contradiction — it floats off the column space plane.

❌ Not quite. Try writing each vector as x₁(1,2,3,4) + x₂(1,1,1,1) and check for contradictions.

Show solution

Solution: (0, 0, 0, 1)

Since rank = 2 with pivot columns $\mathbf{a}_1$ and $\mathbf{a}_2$ , we need $\mathbf{b} = x_1(1,2,3,4) + x_2(1,1,1,1)$ .

Check (0, 0, 0, 1):

Row 1: $x_1 + x_2 = 0 \implies x_2 = -x_1$
Row 2: $2x_1 + x_2 = 0 \implies x_1 = 0$ , so $x_2 = 0$
Row 4: $4(0) + 1(0) = 0 \neq 1$ — Contradiction! ❌

The other three are all in $C(A)$ :

$(1,2,3,4) = 1\cdot\mathbf{a}_1 + 0\cdot\mathbf{a}_2$ ✓
$(2,3,4,5) = 1\cdot\mathbf{a}_1 + 1\cdot\mathbf{a}_2$ ✓
$(0,-1,-2,-3) = (-1)\cdot\mathbf{a}_1 + 1\cdot\mathbf{a}_2$ ✓

Solved: 0 / 4