Basis, Dimension and the Rank Theorem

Quick answer

A basis of a subspace is a linearly independent set that spans it, and every vector in the subspace has exactly one set of coordinates relative to it. All bases of a subspace have the same number of vectors, and that number is its dimension. For a matrix, the dimension of the column space is the rank, the number of pivot columns, and the dimension of the null space is the number of free variables. Since every column is one or the other, rank plus nullity equals the number of columns.

What you'll learn

Decide whether a set is a basis
Find the coordinates of a vector relative to a basis
Find the dimension of a subspace
Use the rank theorem to relate rank and nullity

Enough vectors, and no more

A basis of a subspace $H$ is a set of vectors in $H$ that is

linearly independent, and
spans $H$ .

Spanning means every vector in $H$ can be built. Independence means none of the building blocks is wasted. The standard vectors $\mathbf{e}_1, \ldots, \mathbf{e}_n$ form a basis of ℝⁿ, but they are far from the only one: $(1, 1)$ and $(1, -1)$ form a basis of ℝ² as well.

Relative to a basis, each vector has coordinates: the coefficients that build it. They are unique. If two combinations of the basis gave the same vector, subtracting them would give a combination equal to $\mathbf{0}$ with a nonzero coefficient, which independence rules out.

Why every basis has the same size

The key fact is a counting argument. Suppose $H$ is spanned by $k$ vectors, and take any $k + 1$ vectors in $H$ . Each is a combination of the $k$ spanning vectors, so a dependence among them comes down to a homogeneous system with $k$ equations and $k + 1$ unknowns. More unknowns than equations forces a free variable, hence a nonzero solution, hence dependence. In a subspace spanned by $k$ vectors, any $k + 1$ vectors are dependent. So a basis can never have more vectors than another basis, and by symmetry the two sizes match.

That common size is the dimension of $H$ , written $\dim H$ . A line through the origin has dimension $1$ , a plane through the origin $2$ , ℝⁿ has dimension $n$ , and $\{\mathbf{0}\}$ has dimension $0$ .

Rank and the rank theorem

For a matrix $A$ , the previous lesson found bases for both of its subspaces. The basis for $\text{Col}\,A$ has one vector per pivot column, and the basis for $\text{Nul}\,A$ has one vector per free variable. So

\text{rank}\,A = \dim\text{Col}\,A = \text{number of pivot columns} \qquad \dim\text{Nul}\,A = \text{number of free variables}

Every column of $A$ is either a pivot column or a free-variable column, never both. Counting columns gives the rank theorem:

For an $m \times n$ matrix $A$ , $\ \text{rank}\,A + \dim\text{Nul}\,A = n$ .

The rank also cannot exceed the number of rows, since each pivot needs its own row, so $\text{rank}\,A \le \min(m, n)$ .

Worked examples

Common mistakes

Practice problems

Is $\{(1, 2),\ (2, 4)\}$ a basis of ℝ²?

Answer

No

Full solution

$(2, 4) = 2(1, 2)$ , so the set is dependent and spans only a line.
Is $\{(1, 0, 0),\ (0, 1, 0)\}$ a basis of ℝ³?

Answer

No

Full solution

It is independent but spans only the $xy$ -plane. A basis of ℝ³ needs three vectors.
Find the coordinates of $(4, 6)$ relative to $\{(1, 1),\ (1, -1)\}$ .

Answer

$(5, -1)$

Full solution

$c_1 + c_2 = 4$ and $c_1 - c_2 = 6$ give $c_1 = 5$ and $c_2 = -1$ . Check: $5(1, 1) - (1, -1) = (4, 6)$ .
Find $\dim\text{Span}\{(1, 2, 3),\ (2, 4, 6),\ (0, 1, 1)\}$ .

Answer

$2$

Full solution

The second vector is twice the first, and the third is not a multiple of the first. Row reducing the three columns gives two pivots.
A $4 \times 6$ matrix has rank $4$ . Find $\dim\text{Nul}\,A$ , and describe $\text{Col}\,A$ .

Answer

$\dim\text{Nul}\,A = 2$ ; $\text{Col}\,A =$ ℝ⁴

Full solution

The rank theorem gives $6 - 4 = 2$ . A $4$ -dimensional subspace of ℝ⁴ is all of ℝ⁴.
A $6 \times 4$ matrix has a null space of dimension $1$ . What is its rank?

Answer

$3$

Full solution

$\text{rank} = 4 - 1$ , counting the four columns.
An $n \times n$ matrix has $\det A \ne 0$ . What are its rank and nullity?

Answer

Rank $n$ , nullity $0$

Full solution

A nonzero determinant means $A$ is invertible, so it has $n$ pivots and only the trivial solution to $A\mathbf{x} = \mathbf{0}$ .
Find the rank and nullity of $\begin{bmatrix} 1 & 2 & 1 & 3\\ 2 & 4 & 3 & 7\\ 1 & 2 & 2 & 4 \end{bmatrix}$ .

Answer

Rank $2$ , nullity $2$

Full solution

Row reduction gives $\begin{bmatrix} 1 & 2 & 0 & 2\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 \end{bmatrix}$ : pivots in columns $1$ and $3$ , free variables $x_2$ and $x_4$ .
Can three vectors span ℝ⁴?

Answer

No

Full solution

Their span has dimension at most $3$ , and ℝ⁴ has dimension $4$ . The matrix with them as columns has at most three pivots, leaving a row without one.
A student says the column space of a $3 \times 5$ matrix has dimension $5$ , since the matrix has five columns. What went wrong?

Hint

Where does the column space live?

Answer

The dimension is the rank, the number of pivot columns, which is at most $3$ .

Full solution

The column space is a subspace of ℝ³, so its dimension cannot exceed $3$ . Five columns in ℝ³ are always dependent, and only the pivot columns count toward the dimension.

Frequently asked questions

What is a basis?

A set of vectors that is linearly independent and spans the subspace: enough vectors to reach every vector in it, and no redundant ones.

What is the dimension of a subspace?

The number of vectors in any basis for it. Every basis of the same subspace has the same size.

What is the rank of a matrix?

The dimension of its column space, which equals the number of pivot columns.

What is the rank theorem?

For an m × n matrix, rank A + dim Nul A = n. Every column either holds a pivot or belongs to a free variable.

When do n vectors form a basis of ℝⁿ?

Exactly when they are independent, which for n vectors in ℝⁿ is the same as spanning ℝⁿ. Checking either condition is enough.

What to learn next

Eigenvalues and EigenvectorsFind the directions a matrix only stretches: solve det(A − λI) = 0 for the eigenvalues, then the null space of A − λI for each eigenvector.

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