Matrices as Linear Maps: the Column Picture

Part 1 showed that a matrix is a table of where the basis lands, so applying it is a linear combination of its columns. Take that formula seriously and every question about a matrix map becomes a question about its columns: which outputs are reachable, how many are truly independent, and what gets sent to zero on the way. Those answers are the column space, the rank, and the four fundamental subspaces.

The product is a combination of columns

Write $A$ by its columns $A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \end{bmatrix}$ with each $a_j \in \mathbb{R}^m$. Then the matrix-vector product weights those columns by the entries of $x$:

$$A x = x_1 a_1 + x_2 a_2 + \cdots + x_n a_n .$$

This is the whole game. Multiplying by $A$ is not $m$ separate dot products, it is one vector equation in the output space $\mathbb{R}^m$. The row picture asks where the hyperplanes meet; the column picture asks which combination of columns hits a target. Drag the weights and watch $A x$ trace out the reachable set:

A x = x₁ c₁ + x₂ c₂interactive

c₁ₓ = 1c₂ₓ = 2c₁ᵧ = 3c₂ᵧ = 4

x₁ = 1x₂ = 1

x₁ c₁

(1, 3)

x₂ c₂

(2, 4)

A x

(3, 7)

det A

-2

rank / dim null

2 / 0

Columns independent: the span is all of ℝ², so every target is reachable.

Column space and rank

The set of every output $A x$, as $x$ ranges over all of $\mathbb{R}^n$, is by the formula above exactly the span of the columns. This is the column space,

$$C(A) = \{\, A x : x \in \mathbb{R}^n \,\} = \operatorname{span}\{a_1, \dots, a_n\} \subseteq \mathbb{R}^m .$$

It is a subspace of the output space, and $A x = b$ has a solution precisely when $b$ lies in it. Some columns may be redundant, so its dimension is not $n$ but the number of independent columns. That number is the rank,

$$r = \operatorname{rank}(A) = \dim C(A) .$$

A remarkable fact is that the column rank equals the row rank, so $r$ also counts the independent rows.¹ With $r$ pinned down, everything else follows by counting.

A worked example

Take the invertible case

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \qquad x = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$

The columns are $c_1 = (1, 3)$ and $c_2 = (2, 4)$, and the product is their sum weighted by $x$:

$$A x = 1 \begin{bmatrix} 1 \\ 3 \end{bmatrix} + 1 \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}.$$

Here $\det A = 1\cdot 4 - 2 \cdot 3 = -2 \neq 0$, so the two columns are independent, $r = 2$, and the column space is all of $\mathbb{R}^2$. Every target is reachable, and by the counting rule below nothing but $0$ is sent to $0$.

Rank deficiency: when a column is redundant

Now nudge the second column until it is a multiple of the first:

$$A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}, \qquad c_2 = 2\,c_1 .$$

The determinant is $1\cdot 4 - 2\cdot 2 = 0$, so $r = 1$. The column space is no longer the plane but the single line $\operatorname{span}\{(1,2)\}$, and $A x$ can never leave it no matter how the weights are chosen. Because a whole direction of input now maps to nothing, some nonzero $x$ solves $A x = 0$. Solving,

$$A \begin{bmatrix} 2 \\ -1 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 2 \end{bmatrix} - 1 \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix},$$

so the null space is $N(A) = \operatorname{span}\{(2, -1)\}$, a line in the input space. The one dimension the columns lost is exactly the one dimension the null space gained: $\dim N(A) + r = 1 + 1 = 2 = n$.

The four fundamental subspaces

An $m \times n$ matrix organizes four subspaces, two in the input space $\mathbb{R}^n$ and two in the output space $\mathbb{R}^m$:

$$\begin{array}{llll} \text{column space} & C(A) \subseteq \mathbb{R}^m & \dim = r & \text{outputs reachable}\\ \text{left null space} & N(A^{\top}) \subseteq \mathbb{R}^m & \dim = m - r & \text{outputs missed}\\ \text{row space} & C(A^{\top}) \subseteq \mathbb{R}^n & \dim = r & \text{inputs that matter}\\ \text{null space} & N(A) \subseteq \mathbb{R}^n & \dim = n - r & \text{inputs killed} \end{array}$$

They split each space into orthogonal complements. In the output, $\mathbb{R}^m = C(A) \oplus N(A^{\top})$: a target either lies in the column space or has a piece the map can never produce. In the input, $\mathbb{R}^n = C(A^{\top}) \oplus N(A)$: the row space carries the part of $x$ that survives, and the null space is the part that vanishes. The map is a clean one-to-one correspondence between the row space and the column space, both of dimension $r$.

Rank-nullity

Counting input dimensions gives the identity that ties it together. The $n$ input directions split into the $r$ that the map sends to independent outputs and the rest that it annihilates:

$$\underbrace{\dim N(A)}_{\text{killed}} + \underbrace{\operatorname{rank}(A)}_{\text{survives}} = n .$$

Read left to right, a bigger null space forces a smaller rank: every input direction the matrix crushes is one fewer independent output. Both worked cases obey it with $n = 2$, once as $0 + 2$ and once as $1 + 1$. This is the single most useful counting fact in linear algebra, and it is why $\det A = 0$, rank loss, a nontrivial null space, and non-invertibility are all the same statement.

Footnotes

1. Strang, Introduction to Linear Algebra (5th ed., 2016). ↩