The Four Fundamental Subspaces

⚡ Key Takeaways

The Four Fundamental Subspaces are the column space, row space, null space, and left null space of an $m \times n$ matrix $A$ . Together they decompose the domain $\mathbb{R}^n$ and codomain $\mathbb{R}^m$ into orthogonal complementary pieces, encoding everything about solvability of $Ax = b$ and the geometry of the linear map. Applies wherever linear systems, least squares, or rank arguments appear. Crystallized as a teaching framework by Gilbert Strang (MIT, late 20th century).

💡 The Core Insight

A matrix is not just a grid of numbers — it is a linear transformation. That transformation tears $\mathbb{R}^n$ into two perpendicular halves (row space and null space) and $\mathbb{R}^m$ into two perpendicular halves (column space and left null space). The "real action" of $A$ is a one-to-one correspondence between the row space and the column space, both of dimension $r$ . Everything in the null space is annihilated; everything in the left null space is unreachable. Solvability, uniqueness, and least-squares all follow from where $b$ sits relative to these halves.

DefinitionThe Four Subspaces of an m×n Matrix

Let $A \in \mathbb{R}^{m \times n}$ with rank $r$ . Define:

1

\quad C(A) = \{\, Ax : x \in \mathbb{R}^n \,\} \subseteq \mathbb{R}^m

Derivation hint: the column space is the set of all linear combinations of the columns of $A$ , equivalently all attainable outputs $b$ .

2

\quad N(A) = \{\, x \in \mathbb{R}^n : Ax = 0 \,\} \subseteq \mathbb{R}^n

Derivation hint: the null space (kernel) is the solution set of the homogeneous system $Ax=0$ .

3

\quad C(A^T) = \{\, A^T y : y \in \mathbb{R}^m \,\} \subseteq \mathbb{R}^n

Derivation hint: the row space is the column space of $A^T$ , i.e. all linear combinations of the rows of $A$ viewed as vectors in $\mathbb{R}^n$ .

4

\quad N(A^T) = \{\, y \in \mathbb{R}^m : A^T y = 0 \,\} \subseteq \mathbb{R}^m

Derivation hint: the left null space contains all $y$ with $y^T A = 0$ , hence the name "left." Here $m$ = number of rows (output dimension), $n$ = number of columns (input dimension), $r$ = rank = number of pivots = dimension of $C(A)$ .

TheoremFundamental Theorem of Linear Algebra

For any $A \in \mathbb{R}^{m \times n}$ of rank $r$ :

Part I (Dimensions):

5

\quad \dim C(A) = \dim C(A^T) = r, \quad \dim N(A) = n - r, \quad \dim N(A^T) = m - r

Derivation hint: column rank equals row rank ( $r$ ); the rank–nullity theorem gives the null-space dimensions in each space.

Part II (Orthogonality): Inside $\mathbb{R}^n$ , the row space and null space are orthogonal complements; inside $\mathbb{R}^m$ , the column space and left null space are orthogonal complements:

6

\quad C(A^T) \perp N(A), \quad C(A^T) \oplus N(A) = \mathbb{R}^n

Derivation hint: $Ax=0$ means each row of $A$ is orthogonal to $x$ ; the dimensions $r + (n-r) = n$ force a direct-sum decomposition.

7

\quad C(A) \perp N(A^T), \quad C(A) \oplus N(A^T) = \mathbb{R}^m

Derivation hint: apply [6] to $A^T$ in place of $A$ .

ProofEquality of Row Rank and Column Rank

Step 1. Reduce $A$ to its reduced row echelon form $R = EA$ , where $E$ is an invertible product of elementary matrices. Row operations are invertible linear combinations of rows, so they preserve the row space: $C(A^T) = C(R^T)$ .

Step 2. The nonzero rows of $R$ are linearly independent (each has a leading 1 in a column where all other rows are zero). There are exactly $r$ pivots, so $\dim C(A^T) = r$ .

Step 3. The $r$ pivot columns of $R$ are the standard basis vectors $e_1,\dots,e_r$ (truncated), hence independent; the non-pivot columns are combinations of them. Therefore $\dim C(R) = r$ .

Step 4. Since $E$ is invertible, multiplication by $E$ is a bijection between $C(A)$ and $C(R)$ , preserving dimension: $\dim C(A) = \dim C(R) = r$ .

Conclusion. $\dim C(A) = r = \dim C(A^T)$ , so row rank equals column rank. $\blacksquare$

∎

ProofOrthogonality of Row Space and Null Space

Step 1. Let $x \in N(A)$ , so $Ax = 0$ . Write $A$ in terms of its rows $r_1^T, \dots, r_m^T$ . Then $Ax = 0$ means $r_i^T x = 0$ for every $i$ .

Step 2. Thus $x$ is orthogonal to every row $r_i$ , hence orthogonal to every linear combination of rows, i.e. to all of $C(A^T)$ . This proves $N(A) \subseteq C(A^T)^\perp$ .

Step 3. Count dimensions. By rank–nullity, $\dim N(A) = n - r$ . By [5], $\dim C(A^T)^\perp = n - \dim C(A^T) = n - r$ .

Step 4. A subspace contained in another of equal finite dimension must equal it: $N(A) = C(A^T)^\perp$ . Since their only common vector is $0$ and their dimensions sum to $n$ , we get the direct sum $C(A^T) \oplus N(A) = \mathbb{R}^n$ . $\blacksquare$

∎

mathematician · Gilbert Strang | b. 1934

While the underlying theorems (rank–nullity, the closed range orthogonality relations) are 19th- and early-20th-century results, the unifying "four subspaces" picture — with its now-iconic diagram of two perpendicular planes mapping into two perpendicular planes — is the pedagogical creation of Gilbert Strang of MIT. Through his textbook Introduction to Linear Algebra and his widely viewed MIT OpenCourseWare lectures (18.06), Strang made the simultaneous visualization of all four spaces the organizing principle of an entire course. He coined the phrase "the Fundamental Theorem of Linear Algebra" to package the dimension and orthogonality facts into a single memorable statement.

🧠 Intuition

Picture the matrix $A$ as a machine that takes input vectors $x$ from $\mathbb{R}^n$ and produces outputs $b$ in $\mathbb{R}^m$ . Split the input space $\mathbb{R}^n$ into two perpendicular pieces. The null space is the "blind spot": vectors there are mapped to zero and lost forever. The row space is everything perpendicular to the blind spot — and the machine treats it perfectly, mapping it one-to-one onto the column space. Now split the output space $\mathbb{R}^m$ . The column space is the set of all outputs the machine can actually produce — its "reachable" zone. The left null space is the perpendicular "unreachable" zone: no input ever lands there. So solving $Ax = b$ asks: Is $b$ in the reachable zone? If yes, there is a solution. Is the blind spot trivial? If yes, the solution is unique. The row space and column space are "twins" of dimension $r$ , linked by an invertible correspondence; this is why $A$ behaves like an invertible $r \times r$ matrix when restricted to its own essential parts.

Figure

The Four Fundamental Subspaces: domain ℝⁿ and codomain ℝᵐ

ExampleA 3×4 Matrix — Find All Four Subspaces

Given:

A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 1 & 2 & 1 & 4 \end{bmatrix}

Step 1 — Row reduce. Subtract row 1 from row 3: $(1,2,1,4) - (1,2,0,1) = (0,0,1,3)$ . Then subtract row 2 from the new row 3: $(0,0,1,3)-(0,0,1,3)=(0,0,0,0)$ . The RREF is

R = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}.

Pivots in columns 1 and 3, so $r = 2$ . Here $m=3, n=4$ .

Step 2 — Dimensions. $\dim C(A)=\dim C(A^T)=2$ ; $\dim N(A)=n-r=2$ ; $\dim N(A^T)=m-r=1$ .

Step 3 — Column space. Take the pivot columns of the original $A$ :

C(A) = \operatorname{span}\left\{ \begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix}\right\}.

Step 4 — Row space. The nonzero rows of $R$ :

C(A^T) = \operatorname{span}\left\{ (1,2,0,1),\ (0,0,1,3)\right\}.

Step 5 — Null space. Free variables $x_2 = s,\ x_4 = t$ . From $R$ : $x_1 = -2s - t$ , $x_3 = -3t$ . Special solutions:

N(A) = \operatorname{span}\left\{ \begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\-3\\1\end{bmatrix}\right\}.

Step 6 — Left null space. Solve $A^T y = 0$ , i.e. find $y$ with $y_1 + y_3 = 0$ , $2y_1 + 2y_3 = 0$ , $y_2 + y_3 = 0$ , $y_1 + 3y_2 + 4y_3 = 0$ . From the recorded elimination, the combination that zeroed row 3 was $\text{row}_3 - \text{row}_1 - \text{row}_2 = 0$ , giving:

\boxed{\,N(A^T) = \operatorname{span}\{(-1,-1,1)\}\,}

Check: $(-1)(1,2,0,1) + (-1)(0,0,1,3) + (1)(1,2,1,4) = (0,0,0,0)$ . ✓ And $(-1,-1,1)$ is orthogonal to both columns of $C(A)$ : $(-1)(1)+(-1)(0)+(1)(1)=0$ and $(-1)(0)+(-1)(1)+(1)(1)=0$ . ✓

ExampleSolvability via the Subspaces

Given: Using the same $A$ above, decide whether $Ax = b$ is solvable for $b_1 = (1, 1, 2)^T$ and $b_2 = (1, 1, 1)^T$ , and count solutions.

Step 1 — Criterion. $Ax=b$ is solvable iff $b \in C(A)$ , equivalently iff $b \perp N(A^T)$ , i.e. $b^T(-1,-1,1)=0$ .

Step 2 — Test $b_1$ . $(-1)(1)+(-1)(1)+(1)(2) = -1-1+2 = 0$ . So $b_1 \in C(A)$ → solvable. Because $\dim N(A) = 2 > 0$ , there are infinitely many solutions (a 2-parameter family: any particular solution plus $N(A)$ ).

Step 3 — Test $b_2$ . $(-1)(1)+(-1)(1)+(1)(1) = -1-1+1 = -1 \ne 0$ . So $b_2 \notin C(A)$ → no solution.

\boxed{\,Ax=b_1:\ \infty\text{ solutions};\qquad Ax=b_2:\ \text{inconsistent}\,}

Interpretation: $b_1$ lies in the reachable column space; $b_2$ has a nonzero projection onto the unreachable left null space, so no input can produce it exactly. (A least-squares "best fit" would still exist via Orthogonal Projection.)

⚠️ Common mistake

The column space is NOT spanned by the pivot columns of the RREF. Row operations change the column space (they preserve the row space and null space, but not $C(A)$ ). To get a basis for $C(A)$ , identify which columns are pivot columns, then take those columns from the original matrix $A$ , not from $R$ . By contrast, the row space is read directly off the nonzero rows of $R$ , since row operations preserve it.

⚠️ Common mistake

Orthogonality is between the right pairs, in the right space. The row space and null space are orthogonal complements in $\mathbb{R}^n$ (the input side). The column space and left null space are orthogonal complements in $\mathbb{R}^m$ (the output side). The column space and null space generally live in different spaces ( $\mathbb{R}^m$ vs $\mathbb{R}^n$ ) and need not even have the same dimension — comparing them directly is meaningless unless $m=n$ .

CounterexampleEqual Rank ≠ Equal Subspaces

Consider

A = \begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}, \qquad B = \begin{bmatrix}0 & 0\\0 & 1\end{bmatrix}.

Both are $2\times 2$ of rank $1$ , so all four corresponding dimensions match. Yet $C(A)=\operatorname{span}\{(1,0)\}$ while $C(B)=\operatorname{span}\{(0,1)\}$ — different subspaces. Dimension data alone does not determine the subspaces; rank tells you the sizes of the four spaces, not their orientations.

connection

The four subspaces are the geometric scaffolding for much of linear algebra. The orthogonal decompositions [6] and [7] are exactly what make Orthogonal Projection and Least Squares Approximation work: the least-squares solution projects $b$ onto $C(A)$ , killing its $N(A^T)$ component. The Singular Value Decomposition makes the entire framework explicit — its right singular vectors split into orthonormal bases for $C(A^T)$ and $N(A)$ , and its left singular vectors split into orthonormal bases for $C(A)$ and $N(A^T)$ , with the singular values quantifying the row-space-to-column-space isomorphism. The dimension count is the Rank–Nullity Theorem applied twice (to $A$ and $A^T$ ), and the abstract version replaces "transpose" with the Adjoint Operator in inner-product spaces. The Pseudoinverse $A^+$ is precisely the inverse of the row-space-to-column-space map, extended by zero on $N(A^T)$ .

🔬 Where this appears

⏳ History

- 1750s — Cramer & Euler: Early systematic study of linear systems and the conditions under which solutions exist or fail (foreshadowing rank).

- 1844 — Hermann Grassmann: Ausdehnungslehre introduces abstract linear (vector) spaces, subspaces, and dimension.

- 1850s–1860s — Sylvester & Cayley: The word "matrix" (Sylvester, 1850) and matrix algebra (Cayley, 1858) give the subject its modern symbolic apparatus.

- 1878 — Frobenius: Develops the theory of rank rigorously; the equality of row rank and column rank becomes a standard result.

- 1900s — Hilbert & Schmidt: Orthogonality, projections, and the geometry of solution spaces are extended to infinite dimensions (precursor to closed-range / Fredholm theory).

- 1907–1910 — Erhard Schmidt: Formalizes orthonormalization and orthogonal decomposition of inner-product spaces.

- 1936 — Eckart & Young: The SVD's low-rank approximation theorem ties singular values to the four subspaces' geometry.

- 1955 — Penrose: Defines the generalized (pseudo)inverse, making explicit the row-space-to-column-space isomorphism.

- 1976–present — Gilbert Strang: Packages the dimension and orthogonality results as "the Fundamental Theorem of Linear Algebra" and centers undergraduate teaching on the four-subspaces diagram.

open-problem

The clean orthogonal-complement story relies on the standard inner product and a finite-dimensional real (or complex) space. Several subtleties arise at the edges:

Other inner products / non-orthonormal bases: "Orthogonal complement" depends on the chosen inner product. Change the inner product (e.g. a weighted least squares with metric $W$ ) and the complement of the row space changes, even though the null space does not. Reconciling the algebraic null space with the geometric complement requires care.
Infinite dimensions: For operators on Hilbert spaces, $C(A)$ need not be closed, so $\overline{C(A)} = N(A^T)^\perp$ holds for the closure of the range, not the range itself. The four-subspace picture survives only with topological hedges (the Fredholm alternative, closed-range theorem).
Finite fields: Over $\mathbb{F}_q$ the dimension counts and rank–nullity carry over verbatim, but "orthogonality" can be degenerate — a nonzero vector can be self-orthogonal — so the orthogonal-complement direct sum [6]/[7] can fail. This is studied in self-dual codes.
Numerical rank: In floating-point computation, rank is ambiguous: tiny singular values blur the boundary between "in the column space" and "in the left null space." Choosing a threshold is genuinely problem-dependent, and naive Gaussian elimination can report the wrong rank for ill-conditioned matrices.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also