LINEAR-ALGEBRA · Unit 5 · Video 4 · Interactive Practice

The Column Space of a Matrix

IKey Formulas

Formula	Name	Description
$A\mathbf{x} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n$	Column view of $A\mathbf{x}$	Matrix-vector product as linear combination of columns
$C(A) = \{\mathbf{b} : \mathbf{b} = A\mathbf{x} \text{ for some } \mathbf{x}\}$	Column space	Set of all reachable outputs of $A$
$A\mathbf{x} = \mathbf{b}$ is solvable $\iff$ $\mathbf{b} \in C(A)$	Solvability theorem	Solution exists iff $\mathbf{b}$ is in the column space
$C(A) \subseteq \mathbb{R}^m$ for $A$ of size $m \times n$	Ambient space	Column space lives in $\mathbb{R}^m$ (the row count)

IIVisualization 1 — Ax as a Combination of Columns

Every output $A\mathbf{x} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2$ is the columns scaled and added tip-to-tail.

x₁ 1 x₂ 1

💡 The column space is a subspace: it always contains $\mathbf{0}$ (set both weights to zero).

IIIVisualization 2 — When the Span Collapses

Independent columns fill the plane; align $\mathbf{a}_2$ with $\mathbf{a}_1$ and $C(A)$ collapses to a line.

a₂ x 3 a₂ y 1

💡 Challenge: move $\mathbf{a}_2$ until $\det[\,\mathbf{a}_1\ \ \mathbf{a}_2\,] = 0$ — the instant the plane becomes a line.

IVVisualization 3 — Is b Reachable?

$A\mathbf{x} = \mathbf{b}$ is solvable exactly when $\mathbf{b}$ lies in the column space $C(A)$ .

b₁ 3 b₂ 2

💡 These columns are dependent, so $C(A)$ is only a line — most targets $\mathbf{b}$ are unreachable.

VQuiz Questions

Question 1

Which of the following best describes the column space $C(A)$ of a matrix $A$ ?

✅ Correct! The column space is exactly the span of the columns — all possible linear combinations.

❌ Not quite.

Show solution

Solution: The set of all linear combinations of the columns of $A$ .

The column space is defined as: $C(A) = \{x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \cdots + x_n\mathbf{a}_n \mid x_i \in \mathbb{R}\} = \{A\mathbf{x} \mid \mathbf{x} \in \mathbb{R}^n\}$

Why the others are wrong:

"The set of columns" — That's only $n$ individual vectors. $C(A)$ is the span of those vectors, which includes infinitely many linear combinations.
"Solutions to $A\mathbf{x} = \mathbf{0}$ " — That's the null space $N(A)$ , a completely different subspace.
"All vectors in $\mathbb{R}^n$ " — The column space lives in $\mathbb{R}^m$ (not $\mathbb{R}^n$ ), and it's usually a proper subspace, not all of $\mathbb{R}^m$ .

Question 2

True or False: The column space of a $4 \times 3$ matrix is a subspace of $\mathbb{R}^3$ .

✅ Correct! The columns have 4 entries, so they live in R⁴. The column space is a subspace of R⁴, not R³.

❌ Not quite. Think carefully: how many entries does each column vector have?

Show solution

Answer: False

A $4 \times 3$ matrix has 4 rows and 3 columns. Each column is a vector with 4 entries, so each column lives in $\mathbb{R}^4$ .

Since $C(A)$ is the span of these columns, we have $C(A) \subseteq \mathbb{R}^4$ , not $\mathbb{R}^3$ .

Key rule: For an $m \times n$ matrix, the column space lives in $\mathbb{R}^m$ (determined by the number of rows). The number of columns $n$ limits the dimension of $C(A)$ to at most $n$ , but the ambient space is always $\mathbb{R}^m$ .

Question 3

For $A = \begin{bmatrix} 1 & 3 \\ 2 & 3 \\ 4 & 1 \end{bmatrix}$ , which of the following vectors $\mathbf{b}$ makes $A\mathbf{x} = \mathbf{b}$ solvable?

Hint: try to write each $\mathbf{b}$ as $x_1(1,2,4) + x_2(3,3,1)$ .

✅ Correct! With x = (1, 1) we get 1·(1,2,4) + 1·(3,3,1) = (4,5,5).

❌ Not quite. Try plugging each option into the system and check whether the third equation causes a contradiction.

Show solution

Answer: b = (4, 5, 5)

We need $\mathbf{b} = x_1(1,2,4) + x_2(3,3,1)$ . Try $x_1 = 1,\; x_2 = 1$ :

$1 \cdot \begin{pmatrix}1\\2\\4\end{pmatrix} + 1 \cdot \begin{pmatrix}3\\3\\1\end{pmatrix} = \begin{pmatrix}1+3\\2+3\\4+1\end{pmatrix} = \begin{pmatrix}4\\5\\5\end{pmatrix} \;\checkmark$

For the other choices, the system leads to a contradiction. For example, $\mathbf{b} = (0,0,1)$ :

From eq. 1 and eq. 2: $x_1 = 0,\; x_2 = 0$ .
Eq. 3: $4(0) + 1(0) = 0 \neq 1$ . Contradiction — no solution.

Geometrically, $(4,5,5)$ lies on the column space plane while the others do not.

Question 4

A $3 \times 2$ matrix has two nonzero columns that are scalar multiples of each other ( $\mathbf{a}_2 = k\,\mathbf{a}_1$ for some $k \neq 0$ ). What is $C(A)$ ?

✅ Correct! Dependent columns collapse the column space — every combination points along one direction.

❌ Not quite.

Show solution

Answer: A line through the origin in R³

If $\mathbf{a}_2 = k\,\mathbf{a}_1$ , every linear combination simplifies:

$x_1\mathbf{a}_1 + x_2\mathbf{a}_2 = x_1\mathbf{a}_1 + x_2\,k\,\mathbf{a}_1 = (x_1 + kx_2)\,\mathbf{a}_1$

No matter what $x_1, x_2$ you choose, the result is always a scalar multiple of $\mathbf{a}_1$ . That traces out a line through the origin.

Why the others are wrong:

Plane — A plane needs two independent columns. Dependent columns collapse one dimension.
All of R³ — Even two independent columns in R³ only span a plane, never all of R³.
Just the zero vector — Since $\mathbf{a}_1 \neq \mathbf{0}$ , the span includes the full line, not just the origin.

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