Linear Algebra · Lesson 07

Column space and rank

The columns of a matrix span a subspace: everything the matrix can reach. Its dimension counts how many of the columns really pull in different directions, and it answers the question the lesson on systems opened, of which right-hand sides can be reached.

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rank (A) = dim Col (A)

Col (A)

span (a_{1}, \dots, a_{n})

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Intuition

The columns of a matrix span a subspace: everything the matrix can reach. Its dimension counts how many of the columns really pull in different directions, and it answers the question the lesson on systems opened, of which right-hand sides can be reached.

A workshop with many machines may still only produce a few kinds of thing, if several machines do the same job. What matters is how many different jobs are covered.

The column space and the rank

The column space of $A$ is the span of its columns, written $\operatorname{Col}A$ . It is a subspace of , because every span is. Its dimension is the rank of : the chapter on elimination defined the rank as the number of pivots, and the properties below show that the two agree. The rank never exceeds the number of columns and never exceeds the number of rows.

b \in Col (A)

Lesson complete

\mathbb{R}^m

R^{m}

A

\operatorname{Col}A=\operatorname{span}\{\mathbf{a}_1,\ldots,\mathbf{a}_n\}

What rank decides

$A\mathbf{x}=\mathbf{b}$ is solvable exactly when $\mathbf{b}\in\operatorname{Col}A$ .
$\operatorname{rank}A=\dim\operatorname{Col}A$ , which is at most the smaller of the two counts.
A square $n\times n$ matrix is invertible exactly when its rank is n.
Repeating a column adds nothing to the column space and so does not raise the rank.
Row operations keep every relation among the columns, because $A\mathbf{x}=\mathbf{0}$ and $R\mathbf{x}=\mathbf{0}$ have the same solutions for an echelon form $R$ of $A$ . In the reduced form the pivot columns are independent and reach every other column.
So the pivot columns of $A$ itself, not of its echelon form, are a basis of $\operatorname{Col}A$ , and $\dim\operatorname{Col}A$ is the number of pivots.
The row space $\operatorname{Row}A$ , the span of the rows, is not changed by row operations, and the non-zero rows of an echelon form are a basis of it. So $\dim\operatorname{Row}A=\operatorname{rank}A$ , and $\operatorname{rank}A^{T}=\operatorname{rank}A$ .

Three columns, and every one of them lies along a single line through the origin. The column space is that line, so the rank is 1 and not 3: a column the others already reach adds nothing to what the matrix can produce.

Rank decides invertibility

A\ (n\times n)\ \text{invertible}\iff\operatorname{rank}A=n

Invertibility asks that every right-hand side is reached and reached once. Reaching every one means the column space is all of the space, which for a subspace of dimension n inside an n-dimensional space means the rank is n. Reaching it only once means the columns are independent, and n spanning vectors in an n-dimensional space are independent exactly when the rank is n as well.

Proof steps

\text{every }\mathbf{b}\ \text{reached}\iff\operatorname{Col}A=\mathbb{R}^n

Reachability is membership of the column space, so reaching everything means the column space is everything.

\operatorname{Col}A=\mathbb{R}^n\iff\operatorname{rank}A=n

A subspace of $\mathbb{R}^n$ fills it exactly when its dimension is n.

n\ \text{columns},\ \operatorname{rank}n\implies\text{independent}

The n columns span a space of dimension n, so none of them is spare and they are independent.

\text{independent}\implies\text{one solution only}

Independence is exactly what makes the weights unique, proved in the lesson on bases.

\text{invertible}\iff\operatorname{rank}A=n

Both halves of invertibility have been matched to the rank being full, in both directions.

Applications

Practice

The column space is what is reachable

$A\mathbf{x}\in\operatorname{Col}A$ for every $\mathbf{x}$

And every vector of $\operatorname{Col}A$ arises as some $A\mathbf{x}$ .

Try it

What is the column space of $A$ ?