Fundamental Theorem of Linear Algebra¶

Theorem¶

Let $A$ be a real $m\times n$ matrix of rank $r$. Its four fundamental subspaces are

• the row space $\operatorname{Row}(A)\subseteq\mathbb{R}^n$ and the null space (kernel) $\operatorname{Null}(A)=\{\boldsymbol{x}\colon A\boldsymbol{x}=\boldsymbol{0}\}\subseteq\mathbb{R}^n$${\mathbb{R}}^n$;
• the column space (range) $\operatorname{Col}(A)=\{A\boldsymbol{x}\colon\boldsymbol{x}\in\mathbb{R}^n\}\subseteq\mathbb{R}^m$ and the null space of $A^{\top}$, $\operatorname{Null}(A^{\top})=\{\boldsymbol{y}\colon A^{\top}\boldsymbol{y}=\boldsymbol{0}\}\subseteq\mathbb{R}^m$${\mathbb{R}}^m$.

Then, with respect to the dot product,

1. (dimensions) $\dim\operatorname{Row}(A)=r$$r$, $\dim\operatorname{Null}(A)=n-r$$r$, $\dim\operatorname{Col}(A)=r$$r$, $\dim\operatorname{Null}(A^{\top})=m-r$$r$;
2. (orthogonality) $\operatorname{Null}(A)=\operatorname{Row}(A)^\perp$ and $\operatorname{Null}(A^{\top})=\operatorname{Col}(A)^\perp$$\mathrm{Col}(A{)}^{\perp }$.

Consequently $\mathbb{R}^n=\operatorname{Row}(A)\oplus\operatorname{Null}(A)$ and $\mathbb{R}^m=\operatorname{Col}(A)\oplus\operatorname{Null}(A^{\top})$$\mathrm{Null}(A^{\top })$, both direct sums of mutually orthogonal subspaces. Math II writes $\ker A$ and $\operatorname{range}A$ for $\operatorname{Null}(A)$ and $\operatorname{Col}(A)$. The course gives the fourth subspace, $\operatorname{Null}(A^{\top})$, no separate name; some textbooks call it the left null space.

Interpretation¶

Part 1 combines the Fundamental Theorem of Linear Maps (rank–nullity, $\dim\operatorname{Null}(A)+\dim\operatorname{Col}(A)=n$$n$) with the fact that row rank equals column rank, applied to both $A$ and its transpose.

In the figure the domain $\mathbb{R}^n$ is split into the row space (dimension $r$) and the null space (dimension $n-r$$r$), and the codomain $\mathbb{R}^m$ into the column space (dimension $r$) and the null space of $A^{\top}$ (dimension $m-r$$r$); the right-angle marks stand for the orthogonality in Part 2. The matrix $A$ carries the row space one-to-one onto the column space and sends the whole null space to $\boldsymbol{0}$. In the example below, $r=1$$1$, $n=2$$2$, $m=3$$3$, so the four dimensions are $1,1,1,2$.

Part 2 is almost a one-line proof. $A\boldsymbol{x}=\boldsymbol{0}$ says that every row of $A$ has dot product $0$ with $\boldsymbol{x}$, so $\boldsymbol{x}$ is orthogonal to all rows and hence to their span; this is exactly membership of the orthogonal complement of the row space. Applying the same argument to $A^{\top}$ gives the second identity.

The pairs live in different spaces: row space and null space in the domain $\mathbb{R}^n$, column space and $\operatorname{Null}(A^{\top})$ in the codomain $\mathbb{R}^m$. Row space and column space have the same dimension $r$. For a non-square matrix they lie in different spaces, and even for a square matrix they are in general neither equal nor orthogonal.

Key Results¶

• Since $\operatorname{Col}(A)=\operatorname{Null}(A^{\top})^\perp$$\mathrm{Null}(A^{\top }{)}^{\perp }$, $A\boldsymbol{x}=\boldsymbol{b}$ is solvable if and only if $\boldsymbol{y}^{\top}\boldsymbol{b}=0$ for every $\boldsymbol{y}$ with $A^{\top}\boldsymbol{y}=\boldsymbol{0}$$0$.
• Every $\boldsymbol{x}\in\mathbb{R}^n$ splits uniquely as $\boldsymbol{x}=\boldsymbol{x}_r+\boldsymbol{x}_n$ with $\boldsymbol{x}_r\in\operatorname{Row}(A)$$\mathrm{Row}(A)$, $\boldsymbol{x}_n\in\operatorname{Null}(A)$$\mathrm{Null}(A)$, and $A\boldsymbol{x}=A\boldsymbol{x}_r$$A{\mathbf{x}}_r$.
• $A$ restricted to the row space is a bijection onto the column space.
• Full column rank ($r=$$r=n$$n$) holds if and only if $\operatorname{Null}(A)=\{\boldsymbol{0}\}$$\{0\}$; full row rank ($r=$$r=m$$m$) holds if and only if $\operatorname{Null}(A^{\top})=\{\boldsymbol{0}\}$$\{0\}$, if and only if $A\boldsymbol{x}=\boldsymbol{b}$ is solvable for every $\boldsymbol{b}$.
• $\operatorname{Null}(A^{\top}A)=\operatorname{Null}(A)$$\mathrm{Null}(A)$, so $\operatorname{rank}(A^{\top}A)=\operatorname{rank}A$$\mathrm{rank}A$; this is what makes the normal equations of least squares work.

Examples¶

Let

so $m=3$$3$, $n=2$ and $r=1$$1$.

• Row space: $\operatorname{span}\{(1,2)^{\top}\}$, dimension $r=1$$1$.
• Null space: $x_1+2x_2=0$$0$, so $\operatorname{span}\{(-2,1)^{\top}\}$, dimension $n-r=1$$1$; indeed $(1,2)\cdot(-2,1)=0$$0$.
• Column space: $\operatorname{span}\{(1,2,3)^{\top}\}$, dimension $r=1$$1$.
• Null space of $A^{\top}$: $y_1+2y_2+3y_3=0$$0$, spanned by $\boldsymbol{y}_1=(-2,1,0)^{\top}$ and $\boldsymbol{y}_2=(-3,0,1)^{\top}$$(-3,0,1{)}^{\top }$, dimension $m-r=2$$2$; both are orthogonal to $(1,2,3)^{\top}$.

Solvability: for $\boldsymbol{b}=(1,0,0)^{\top}$$(1,0,0{)}^{\top }$, $\boldsymbol{y}_1^{\top}\boldsymbol{b}=-2\neq0$$0$, so $A\boldsymbol{x}=\boldsymbol{b}$ has no solution. For $\boldsymbol{b}=(2,4,6)^{\top}$$(2,4,6{)}^{\top }$, $\boldsymbol{y}_1^{\top}\boldsymbol{b}=\boldsymbol{y}_2^{\top}\boldsymbol{b}=0$$0$, and $\boldsymbol{x}=(2,0)^{\top}$ solves it.

Prerequisites¶

• Kernel
• Range of a Linear Map
• Fundamental Theorem of Linear Maps
• Row Space
• Orthogonal Complement

Related Concepts¶

• Orthogonal Projection
• Least Squares
• Singular Value Decomposition
• Moore-Penrose Inverse

Sources¶

• Math V: Advanced Linear Algebra · Week 5

Aliases: four fundamental subspaces theorem, theorem of the four subspaces Keywords: four fundamental subspaces, left null space, Col(A), Row(A), Null(A), Null(A) = Row(A)^⊥, Null(A^T) = Col(A)^⊥, r, n-r, r, m-r Symbols: element of, real numbers, subset

Last updated: 16 October 2026 22:28