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Rank, Null Space, and the Four Fundamental Subspaces

  • Understand the rank of a matrix and its connection to linear independence.
  • Define the null space (kernel) of a matrix and compute it.
  • Identify all four fundamental subspaces of a matrix: column space, null space, row space, and left null space.
  • Apply the Rank-Nullity Theorem to relate rank and nullity.
  • See how these subspaces reveal the structure of a linear transformation.

Rank of a Matrix

The rank of a matrix \( A \) is the dimension of its column space \( C(A) \) — equivalently, the maximum number of linearly independent columns (or rows) in \( A \). In practice, you find it by row-reducing \( A \) to echelon form and counting the pivot columns:

\[ \text{rank}(A) = \text{number of pivots in the row echelon form of } A \]

A matrix has full rank when its rank equals the smaller of its number of rows or columns. Recall from Gaussian Elimination and Understanding Determinants: a square matrix is invertible exactly when it has full rank, i.e. \( \det(A) \neq 0 \).

Null Space (Kernel)

The null space (or kernel) of \( A \) is the set of all vectors that \( A \) sends to zero:

\[ N(A) = \{ \mathbf{x} \mid A\mathbf{x} = \mathbf{0} \} \]

You compute it by solving the homogeneous system \( A\mathbf{x} = \mathbf{0} \) with row reduction. The free variables in the reduced system parametrize the null space; each free variable contributes one basis vector.

The Four Fundamental Subspaces

Every \( m \times n \) matrix \( A \) has four fundamental subspaces, paired up between \( \mathbb{R}^n \) and \( \mathbb{R}^m \):

  • Column space \( C(A) \subseteq \mathbb{R}^m \): all linear combinations of the columns of \( A \). Dimension \( r = \text{rank}(A) \).
  • Null space \( N(A) \subseteq \mathbb{R}^n \): solutions to \( A\mathbf{x} = \mathbf{0} \). Dimension \( n - r \).
  • Row space \( C(A^T) \subseteq \mathbb{R}^n \): all linear combinations of the rows of \( A \). Dimension \( r \).
  • Left null space \( N(A^T) \subseteq \mathbb{R}^m \): solutions to \( A^T\mathbf{y} = \mathbf{0} \). Dimension \( m - r \).

These four subspaces aren't just related by dimension — they're related geometrically: the row space and null space are orthogonal complements of each other in \( \mathbb{R}^n \), and the column space and left null space are orthogonal complements in \( \mathbb{R}^m \). We'll make that precise next time when we look at orthogonality and projections.

The Rank-Nullity Theorem

For any \( m \times n \) matrix \( A \), the dimensions of the column space and null space always add up to the number of columns:

\[ \text{rank}(A) + \dim N(A) = n \]

This is the Rank-Nullity Theorem. Intuitively: every column of \( A \) either contributes a new pivot direction (adding to the rank) or a free direction (adding to the null space) — there's nowhere else for a dimension to go.

Examples

Example 1: Find the rank and null space of:

\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 2 \end{bmatrix} \]

Step 1: Row reduce \( A \):

\[ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 2 \end{bmatrix} \xrightarrow{R_2 - 2R_1,\ R_3 - R_1} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -1 \end{bmatrix} \xrightarrow{\text{swap } R_2, R_3} \begin{bmatrix} 1 & 2 & 3 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \]

There are 2 pivots, so \( \text{rank}(A) = 2 \). By the Rank-Nullity Theorem, \( \dim N(A) = 3 - 2 = 1 \).

Step 2: Solve \( A\mathbf{x} = \mathbf{0} \) using the reduced form. Setting \( x_3 = t \) as the free variable: \( x_2 = -t \), \( x_1 = -2x_2 - 3x_3 = 2t - 3t = -t \). So:

\[ N(A) = \left\{ t \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} \;\middle|\; t \in \mathbb{R} \right\} \]

Exercises

  • Question 1: Find the rank of \( A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \).
  • Question 2: Find a basis for the null space of \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} \).
  • Question 3: If \( A \) is a \( 5 \times 7 \) matrix with rank 4, what is \( \dim N(A) \)?

  • Answer 1: Row 2 is \( 2 \times \) Row 1, so there is only 1 pivot: \( \text{rank}(A) = 1 \).
  • Answer 2: Setting \( x_3 = t \): \( x_1 = -2t \), \( x_2 = -3t \), so a basis is \( \left\{ \begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix} \right\} \).
  • Answer 3: By Rank-Nullity, \( \dim N(A) = 7 - 4 = 3 \).

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