Rank, Nullity, and the Fundamental Theorem of Linear Maps

A linear map $T:V\to W$ does only two things to the space it acts on: it crushes some directions to nothing, and it carries the rest through faithfully. The crushed directions form the kernel $\ker T=\{v: Tv=0\}$ ; the directions that survive fill out the image $\operatorname{im}T=\{Tv: v\in V\}$ . Both are honest subspaces — the kernel inside the domain $V$ , the image inside the codomain $W$ — and that alone is a consequence of linearity. The dimension of the kernel is the nullity, the dimension of the image is the rank, and the whole subject of this article is a single, exact accounting identity between them. The Fundamental Theorem of Linear Maps — the rank-nullity theorem — says these two numbers are not free to roam: their sum is pinned at $\dim V$ . The domain has a fixed budget of dimensions, and every one is spent exactly once, either annihilated into the kernel or transmitted into the image. The statement is a conservation law, and like the conservation laws of physics it converts a hard structural question into simple bookkeeping. It needs only that $V$ be finite-dimensional; $W$ may be as large as you like, even infinite-dimensional. That one identity does an astonishing amount of work. It decides when a linear system has solutions and how many; it proves that a square operator is injective exactly when it is surjective exactly when it is invertible; it measures the degeneracy of an energy level as the nullity of $H-E\,\mathbb{1}$ . It is the first structural theorem of linear algebra that a working physicist uses every day, almost always without naming it.

Key takeaways

A linear map $T:V\to W$ splits the domain into a part it kills (the kernel, dimension = nullity) and a part it transmits (the image, dimension = rank).
The Fundamental Theorem of Linear Maps is the conservation law $\dim V=\operatorname{rank}T+\operatorname{null}T$ — every domain dimension is spent exactly once.
The proof extends a basis of the kernel to a basis of $V$ and shows the images of the added vectors form a basis of $\operatorname{im}T$ ; only $\dim V<\infty$ is needed.
In matrix form $\operatorname{rank}A+\dim(\operatorname{null}A)=n$ for $A\in\mathbb{F}^{m\times n}$ , and remarkably column rank always equals row rank.
Immediate physics payoffs: a square operator is injective iff surjective iff invertible; a system $A\mathbf{x}=\mathbf{b}$ is solvable iff $\operatorname{rank}[A\,|\,\mathbf{b}]=\operatorname{rank}A$ ; the geometric multiplicity of an eigenvalue $\lambda$ is $n-\operatorname{rank}(A-\lambda\mathbb{1})$ .
Worked cleanly: a $3\times4$ matrix of rank $2$ has nullity $2$ (so $2+2=4$ ), and the defective matrix with eigenvalue $2$ of algebraic multiplicity $2$ has geometric multiplicity only $1$ .
Key idea: Every dimension of the domain is accounted for exactly once: $\dim V=\operatorname{rank}T+\operatorname{null}T$ . The map either kills a direction (kernel) or transmits it to the image — never both, never neither.

Where this lives today

Rank and nullity have quietly become the load-bearing numbers of computational science. Every time an engineer runs a singular value decomposition, the count of singular values above the noise floor is a numerical rank, and the near-null space it exposes is what principal component analysis, image compression, and recommender systems all exploit — a data matrix of nominal rank $1000$ that is "really" rank $20$ is the premise of low-rank approximation and compressed sensing. In control engineering, Kalman's controllability and observability tests are literally rank conditions: a system is controllable exactly when a certain matrix has full rank, i.e. nullity zero. In first-principles materials physics, phonon calculations enforce the acoustic sum rule precisely so the dynamical matrix has the correct three-dimensional null space of rigid translations. And in inverse problems — X-ray tomography, seismic imaging, gravitational-wave parameter estimation — a rank-deficient forward operator is the signature of information you did not measure, its nullity naming exactly the parameter combinations the data can never constrain. The theorem that began as bookkeeping for Victorian determinants is now the diagnostic that tells a scientist which questions the data can answer.

The physical picture

Think of $T$ as a pipeline that swallows an $n$ -dimensional input and delivers whatever comes out the far end. Some input directions are squeezed flat and emerge as zero — that information is gone, and those directions make up the kernel. Every direction that is not flattened has to appear at the output as a genuinely new, independent direction of the image. There is no third fate, so the input dimension is conserved: flattened plus transmitted equals the total. The first figure draws this ledger with real numbers — a map from a $5$ -dimensional domain whose kernel is $2$ -dimensional must have a $3$ -dimensional image, because $5=3+2$ and there is nowhere else for the dimensions to go.

Figure

A linear map from a 5-dimensional domain V to W. A 2-dimensional kernel (red) is crushed to the single zero vector, while the complementary 3 dimensions (blue) are carried faithfully onto a 3-dimensional image. The ledger closes exactly: dim V = rank T + nullity T, that is 5 = 3 + 2.

The cleanest concrete instance is a projection. Drop every vector of $\mathbb{R}^3$ straight down onto the horizontal plane: the shadow is its image, and the vertical direction you project along is squashed to zero. The image is the whole plane, so the rank is $2$ ; the kernel is the single vertical line, so the nullity is $1$ ; and $2+1=3$ , the domain dimension. The second figure makes the trade-off vivid: you cannot enlarge the image without shrinking the kernel, because they are two halves of one fixed budget.

Figure

Orthogonal projection of ℝ³ onto the horizontal plane. A vector v drops to its shadow P v in the plane (the image, rank 2), while the vertical direction it is projected along is sent to zero (the kernel, nullity 1). The budget closes as 2 + 1 = 3.

Choosing a basis adapted to $T$ makes the theorem visible at a glance: take basis vectors inside the kernel, then add just enough more to fill out $V$ . The kernel vectors map to zero and the added ones map to an independent set spanning the image, so the added vectors number $\dim V$ minus the nullity and their images number the rank. This is also why solving a square system is all-or-nothing: if no direction is crushed, none is missing from the image either, so existence and uniqueness arrive together.

The core equations

Everything begins with two sets attached to the map: the vectors it sends to zero, and the vectors it can produce. Linearity guarantees both are subspaces, so it makes sense to speak of their dimensions.

\ker T=\{v\in V: Tv=0\},\qquad \operatorname{im}T=\{Tv: v\in V\} \qquad (1)

where: $T$ — a linear map $V\to W$ obeying $T(\alpha u+\beta v)=\alpha\,Tu+\beta\,Tv$ ; $\ker T$ — the kernel or null space, a subspace of the domain $V$ ; $\operatorname{im}T$ — the image or range, a subspace of the codomain $W$ . Reading: the kernel collects what is destroyed, the image what is reachable. Both are subspaces by linearity — if $Tu=Tv=0$ then $T(\alpha u+\beta v)=0$ , and images are closed the same way. The two numbers that matter are the dimensions of these subspaces. They are intrinsic to the map, unchanged by any choice of basis or by any similarity transform, which is exactly why they carry physical meaning.

\operatorname{rank}T=\dim(\operatorname{im}T),\qquad \operatorname{null}T=\dim(\ker T) \qquad (2)

where: $\operatorname{rank}T$ — the number of independent output directions the map produces; $\operatorname{null}T$ — the number of independent input directions it destroys. Reading: rank measures how much gets through, nullity how much is lost. The theorem itself ties them together. It is a single equation, and it is exact for any finite-dimensional domain regardless of how large the codomain is.

\dim V=\operatorname{rank}T+\operatorname{null}T \qquad (3)

where: $\dim V$ — the dimension of the finite-dimensional domain. Reading: domain dimension = transmitted + annihilated. This is the conservation law; the derivation below turns "spend each dimension once" into a rigorous basis count. Fixing bases turns $T$ into a matrix $A$ , and the theorem becomes a statement about columns. The number of columns is the domain dimension; the pivots count the rank; the leftover free columns count the nullity.

\operatorname{rank}A+\dim(\operatorname{null}A)=n,\qquad A\in\mathbb{F}^{m\times n} \qquad (4)

where: $A$ — the $m\times n$ matrix of $T:\mathbb{F}^n\to\mathbb{F}^m$ ; $n$ — number of columns, equal to $\dim V$ ; $\operatorname{rank}A$ — number of pivot columns, equal to $\dim(\text{column space})$ . Reading: pivots + free columns = total columns, so $\dim(\operatorname{null}A)=n-\operatorname{rank}A$ — each non-pivot column is one free parameter of $A\mathbf{x}=\mathbf{0}$ . A companion fact makes the word "rank" unambiguous: the number of independent columns equals the number of independent rows. Row reduction produces the same pivot count whether you read it down the columns or across the rows.

\operatorname{rank}A=\operatorname{rank}A^{\mathsf T}\quad(\text{column rank}=\text{row rank}) \qquad (5)

where: $A^{\mathsf T}$ — the transpose, whose column space is the row space of $A$ . Reading: the $r$ pivots of the echelon form count both independent rows and independent columns, so the two ranks coincide — a nontrivial theorem, not a definition. Finally, rank decides solvability. Appending the right-hand side $\mathbf{b}$ as an extra column must not create a new independent direction, or there is no solution at all.

A\mathbf{x}=\mathbf{b}\ \text{is solvable}\iff \mathbf{b}\in\operatorname{im}A\iff \operatorname{rank}[A\,|\,\mathbf{b}]=\operatorname{rank}A \qquad (6)

where: $[A\,|\,\mathbf{b}]$ — the augmented matrix. Reading: consistency is a rank condition, and when it holds the solution set is an affine copy of $\ker A$ of dimension $n-\operatorname{rank}A$ — one point when the nullity is zero, a whole flat when it is positive.

Derivation AProof of the fundamental theorem of linear maps

The goal is to prove $\dim V=\operatorname{rank}T+\operatorname{null}T$ from scratch, using nothing but the definition of a basis and the linearity of $T$ . The strategy is the standard and beautiful one: start with a basis of the kernel, extend it to a basis of the whole domain, and then show that the images of exactly the added vectors form a basis of the image. Counting the vectors in each basis then delivers the identity with no arithmetic to spare. Throughout, let $T:V\to W$ with $\dim V=n$ finite and $\operatorname{null}T=k$ .

Step 1. Choose a basis $u_1,\dots,u_k$ of the kernel, so that by definition $\operatorname{null}T=k$ ; if the kernel is trivial we take $k=0$ and start from the empty set. Why: the kernel is where the map does its destroying, and pinning down a basis for it lets us separate the destroyed directions cleanly from everything else.

Step 2. Extend that independent set to a basis $\{u_1,\dots,u_k,\,v_1,\dots,v_{n-k}\}$ of $V$ ; the basis-extension theorem guarantees this is always possible, and the number of added vectors is forced to be $n-k$ because a basis of an $n$ -dimensional space has exactly $n$ elements. Why: the added $v_j$ are the candidate carriers of the image — they are precisely the directions the map does not kill.

Step 3. Show the images $Tv_1,\dots,Tv_{n-k}$ span $\operatorname{im}T$ . Any output is $Tv$ for some $v=\sum_i a_i u_i+\sum_j b_j v_j$ , and since $Tu_i=0$ the kernel part drops out, leaving $Tv=\sum_j b_j\,Tv_j$ . Why: the kernel basis vectors contribute nothing to any output, so whatever the image contains is already built from the images of the added vectors alone.

Step 4. Show those images are linearly independent. Suppose $\sum_j c_j\,Tv_j=0$ ; then $T\!\left(\sum_j c_j v_j\right)=0$ , so $\sum_j c_j v_j\in\ker T$ and can be written $\sum_i d_i u_i$ . Rearranged, $\sum_j c_j v_j-\sum_i d_i u_i=0$ , a linear relation among the basis of $V$ , which forces every coefficient — in particular every $c_j$ — to vanish. Why: independence is what upgrades "spanning set" to "basis", and it rests entirely on the $u_i,v_j$ together being a basis of $V$ .

Step 5. Conclude. The $Tv_j$ form a basis of the image, so $\operatorname{rank}T=\dim(\operatorname{im}T)=n-k$ , and adding back the nullity gives $\operatorname{rank}T+\operatorname{null}T=(n-k)+k=n=\dim V$ . Why: this is the whole theorem, and it used only $\dim V<\infty$ — the codomain was never assumed finite-dimensional, which is why it survives into the operator theory of quantum mechanics.

Go deeper · Derivation B · 1 minInjective equals surjective equals invertible for a square operator

Derivation BInjective equals surjective equals invertible for a square operator

The goal is to prove one of the most-used facts in applied mathematics: when a linear operator maps a finite-dimensional space to itself, being injective, being surjective, and being invertible are three names for the same condition, each equivalent to the nullity being zero. This equivalence fails badly in infinite dimensions, so the proof leans entirely on the rank-nullity theorem, which is what makes finiteness decisive. Let $T:V\to V$ be linear with $\dim V=n<\infty$ .

Step 1. Suppose $T$ is injective, i.e. $\operatorname{null}T=0$ . Rank-nullity gives $\operatorname{rank}T=n-0=n$ , so $\operatorname{im}T$ is an $n$ -dimensional subspace of the $n$ -dimensional space $V$ , hence all of $V$ : $T$ is surjective. Why: a subspace whose dimension equals that of the whole space cannot be a proper subspace, so full rank forces the image to fill the codomain.

Step 2. Conversely suppose $T$ is surjective, i.e. $\operatorname{rank}T=n$ . Rank-nullity gives $\operatorname{null}T=n-n=0$ , so the kernel is trivial and $T$ is injective. Why: the same conservation law run backwards — filling the image completely leaves no dimensions over to be destroyed.

Step 3. Each property implies the other, so together they make $T$ a bijection, and a bijective linear map has a linear inverse $T^{-1}$ . In matrix terms this is exactly the condition $\det A\neq0$ . Why: existence and uniqueness of solutions to $T\mathbf{x}=\mathbf{b}$ are governed by surjectivity and injectivity respectively, and here they stand or fall together.

Step 4. Note where finiteness is essential: the right-shift operator $S(x_1,x_2,\dots)=(0,x_1,x_2,\dots)$ on $\ell^2$ is injective but not surjective, and the left shift is surjective but not injective. Why: with infinitely many dimensions the "budget" is inexhaustible, so a map can gain or lose room without contradiction — the equivalence is a purely finite-dimensional gift of rank-nullity.

Go deeper · Derivation C · 1 minCounting the solutions of a linear system

Derivation CCounting the solutions of a linear system

The goal is to extract, from rank-nullity alone, the complete decision tree for the equation $A\mathbf{x}=\mathbf{b}$ : when a solution exists, when it is unique, and when it is not, how large the family of solutions is. The answer is entirely dimensional — existence is a rank condition on the augmented matrix, and the size of the solution set is the nullity — so a physicist can read off the structure of a linear problem without ever computing a solution. Let $A\in\mathbb{F}^{m\times n}$ and write $r=\operatorname{rank}A$ .

Step 1. Existence. A solution exists if and only if $\mathbf{b}$ lies in the image of $A$ , equivalently $\operatorname{rank}[A\,|\,\mathbf{b}]=\operatorname{rank}A=r$ ; if appending $\mathbf{b}$ raises the rank, the system is inconsistent and has no solution. Why: $A\mathbf{x}$ is by definition a combination of the columns of $A$ , so it can equal $\mathbf{b}$ only when $\mathbf{b}$ adds no new column direction.

Step 2. Structure. When consistent, the general solution is $\mathbf{x}=\mathbf{x}_p+\mathbf{x}_h$ , one particular solution $\mathbf{x}_p$ with $A\mathbf{x}_p=\mathbf{b}$ plus any homogeneous solution $\mathbf{x}_h$ with $A\mathbf{x}_h=\mathbf{0}$ . Why: the difference of any two solutions lies in the kernel, so the solution set is a single translate of $\ker A$ — an affine flat, not a subspace unless $\mathbf{b}=\mathbf{0}$ .

Step 3. Size. By rank-nullity the homogeneous solutions fill $\ker A$ , of dimension $n-r$ , so the solution set is an affine space of dimension $n-r$ . Why: the nullity directly counts the free parameters, so it is literally the number of independent knobs left after the equation is imposed.

Step 4. Special cases. If $r=n$ (full column rank) the kernel is trivial and there is at most one solution; if in addition $r=m=n$ the matrix is square and invertible, and there is exactly one solution $\mathbf{x}=A^{-1}\mathbf{b}$ for every $\mathbf{b}$ . Why: these are the invertible and overdetermined-consistent limits, and they are exactly the cases the square-operator equivalence above singles out.

Limiting cases

The single identity specializes into sharply different regimes depending on the shape of the map and how much of its domain survives:

Case	Result	Meaning
Injective map	$\ker T=\{0\}\iff \operatorname{null}T=0\iff \operatorname{rank}T=\dim V$	Injectivity is exactly a trivial kernel; the rank then equals the full domain dimension, which requires $\dim W\geq\dim V$ .
Surjective map	$\operatorname{im}T=W\iff \operatorname{rank}T=\dim W$	Onto means the image fills the codomain; combined with rank-nullity this forces $\operatorname{null}T=\dim V-\dim W\geq0$ , so $\dim V\geq\dim W$ .
Square operator $V\to V$	$\operatorname{null}T=0\iff \operatorname{rank}T=\dim V\iff T\text{ invertible}$	When $\dim V=\dim W$ , injective, surjective, and bijective coincide — the workhorse behind invertibility tests and nonzero determinants.
Rank deficiency / degeneracy	$\operatorname{rank}T=r<\dim V\Rightarrow \operatorname{null}T=\dim V-r>0$	A nontrivial kernel means non-unique solutions and, for $A-\lambda\mathbb{1}$ , a degenerate (multi-dimensional) eigenspace.
Zero map	$\operatorname{rank}T=0,\ \operatorname{null}T=\dim V$	The entire domain is crushed to $\{0\}$ ; the kernel is everything — the extreme opposite of an isomorphism.

Common trapRank plus nullity equals the dimension of the domain, not the codomain

The most common slip is to write $\operatorname{rank}T+\operatorname{null}T=\dim W$ . It is $\dim V$ — the domain — every time. The reason is structural: the kernel lives inside $V$ and the image inside $W$ , two different spaces, and it is the domain whose dimensions get partitioned into "killed" and "transmitted". The codomain $W$ only has to be large enough to hold the image; any extra room in $W$ outside $\operatorname{im}T$ is never reached and plays no role in the count. A related trap: the kernel and image are not orthogonal complements and generally do not even live in the same space, so $\ker T\oplus\operatorname{im}T$ is meaningless in general. That picture only appears once you introduce an inner product and the adjoint $T^{\dagger}$ , where $\operatorname{im}T=(\ker T^{\dagger})^{\perp}$ — a different theorem about the four fundamental subspaces.

Worked examples

1 · Rank, nullity, and the null space of a matrix (fully worked)

Find the rank, the nullity, and an explicit basis of the null space of

A=\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 4 & 6 & 8\\ 0 & 1 & 1 & 1 \end{pmatrix},\qquad A\in\mathbb{R}^{3\times4}.

Step 1. Row reduce. The second row is exactly twice the first, so $R_2\to R_2-2R_1=(0,0,0,0)$ vanishes. Swapping the zero row to the bottom leaves two nonzero rows, $(1,2,3,4)$ with a pivot in column 1 and $(0,1,1,1)$ with a pivot in column 2. There are two pivots, so $\operatorname{rank}A=2$ .

Step 2. Apply the theorem. The domain is $\mathbb{R}^4$ , so $n=4$ and rank-nullity gives $\dim(\operatorname{null}A)=n-\operatorname{rank}A=4-2=\mathbf{2}$ . Columns 3 and 4 are the non-pivot (free) columns, matching the two free parameters we are about to find.

Step 3. Solve $A\mathbf{x}=\mathbf{0}$ . With free variables $x_3=s$ and $x_4=t$ , the pivot row $(0,1,1,1)$ gives $x_2=-s-t$ , and the pivot row $(1,2,3,4)$ gives $x_1=-2x_2-3x_3-4x_4=-s-2t$ . Hence

\mathbf{x}=s\begin{pmatrix}-1\\-1\\1\\0\end{pmatrix}+t\begin{pmatrix}-2\\-1\\0\\1\end{pmatrix}.

Substituting back confirms $A(-1,-1,1,0)^{\mathsf T}=\mathbf{0}$ and $A(-2,-1,0,1)^{\mathsf T}=\mathbf{0}$ , so these two vectors are a basis of the null space, of dimension $2$ .

Step 4. Cross-check. Column rank $=2$ (columns 1 and 2 are independent), row rank $=2$ (rows 1 and 3 are independent, row 2 dependent); they agree, illustrating equation (5). The ledger closes: $\operatorname{rank}A+\dim(\operatorname{null}A)=2+2=4=n$ . Rank 2, nullity 2, and 2 + 2 = 4.

2 · Geometric multiplicity and a defective matrix (fully worked)

For $B=\begin{pmatrix}2&1&0\\0&2&0\\0&0&3\end{pmatrix}$ find the geometric multiplicity of the eigenvalue $\lambda=2$ and decide whether $B$ is diagonalizable.

Step 1. The characteristic polynomial is $\det(B-\lambda\mathbb{1})=(2-\lambda)^2(3-\lambda)$ , so $\lambda=2$ has algebraic multiplicity 2 and $\lambda=3$ has algebraic multiplicity 1.

Step 2. Form $B-2\mathbb{1}=\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}$ . Its nonzero rows $(0,1,0)$ and $(0,0,1)$ are independent, so $\operatorname{rank}(B-2\mathbb{1})=2$ .

Step 3. The geometric multiplicity is the nullity of $B-2\mathbb{1}$ , and rank-nullity on the $3\times3$ operator gives $\dim\ker(B-2\mathbb{1})=3-\operatorname{rank}(B-2\mathbb{1})=3-2=\mathbf{1}$ . Solving $(B-2\mathbb{1})\mathbf{x}=\mathbf{0}$ gives $x_2=0,\ x_3=0,\ x_1$ free, so the single eigenvector is $(1,0,0)^{\mathsf T}$ .

Step 4. Compare. The geometric multiplicity $1$ is strictly less than the algebraic multiplicity $2$ , so $B$ has too few eigenvectors to form a basis: it is defective and not diagonalizable, and its structure is captured instead by the Jordan canonical form. The whole diagnosis was a nullity computation. Geometric multiplicity 1 < algebraic multiplicity 2, so B is defective.

3 · Counting molecular vibrational modes

How many independent vibrational modes does a nonlinear water molecule (H₂O, $N=3$ atoms) have, and how many for linear carbon dioxide (CO₂, $N=3$ )?

Step 1. A molecule of $N$ atoms has $3N$ Cartesian coordinates, so its configuration space is $3N$ -dimensional. For both molecules $3N=9$ .

Step 2. The zero-frequency modes — rigid translations and rotations of the whole molecule — form the kernel of the mass-weighted dynamical (Hessian) matrix, since a rigid displacement costs no restoring energy. A nonlinear molecule has $3$ translations and $3$ rotations, nullity $6$ ; a linear molecule has $3$ translations and only $2$ rotations, nullity $5$ .

Step 3. By rank-nullity the vibrational modes are the rank of the dynamical matrix, $3N$ minus the nullity: water gives $9-6=\mathbf{3}$ vibrational modes, and CO₂ gives $9-5=\mathbf{4}$ . H₂O: 3 modes; CO₂: 4 modes — the standard 3N − 6 and 3N − 5 counts.

Physics of the portrait

The rank-nullity theorem carries no single discoverer's name because it grew up alongside the language of matrices themselves. Its vocabulary, though, is almost entirely the invention of one flamboyant Victorian.

Origins · 1 minJames Joseph Sylvester and how this was discovered

James Joseph Sylvester

1814–1897 · matrix algebra and invariant theory; coined the words "matrix", "minor", and "nullity"

Sylvester was one of the great word-coiners of mathematics, and much of the language of this article is his. Barred for decades from a professorship in England because he was Jewish, he worked as an actuary and a lawyer before holding chairs at Virginia, Johns Hopkins — where he founded the American Journal of Mathematics in 1878 — and finally Oxford. In an 1850 paper he introduced the term "matrix" for a rectangular array from which determinants (his "minors") could be cut, and in his 1884 lectures he introduced "nullity" for the dimension of what we now call the kernel, along with his "law of nullity" bounding the nullity of a product. His collaborator Arthur Cayley built the abstract algebra of matrices around these ideas, while Georg Frobenius gave the rank its rigorous modern theory in 1878, establishing the equality of row and column rank. The clean statement in terms of abstract linear maps — $\dim V=\operatorname{rank}T+\operatorname{null}T$ — belongs to the twentieth-century vector-space viewpoint of Peano, Noether, and van der Waerden, but the two nouns in its name are Sylvester's gift.

History

1844

— Grassmann's Ausdehnungslehre founds the ideas of linear independence and dimension

1850

— Sylvester coins the term "matrix" for a rectangular array of numbers

1858

— Cayley's memoir builds the abstract algebra of matrices, including the inverse

1878

— Frobenius gives the rank of a matrix its rigorous theory and proves row rank equals column rank

1884

— Sylvester introduces "nullity" and states his law of nullity for products

1930

— van der Waerden's Moderne Algebra frames rank plus nullity for abstract linear maps

Going further

Texts. Axler, Linear Algebra Done Right, 3rd ed., Ch. 3 (the Fundamental Theorem of Linear Maps) · Stone & Goldbart, Mathematics for Physics, Ch. 1 (Vector Spaces and Linear Operators) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 2–3 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 8 · Strang, Introduction to Linear Algebra, Ch. 3–4 (the four fundamental subspaces) · Hassani, Mathematical Physics, Ch. 2 · Nakahara, Geometry, Topology and Physics, §2 (linear maps and their invariants). Key results to keep in mind: the basis-extension proof yields $\operatorname{rank}=n-k$ using only $\dim V<\infty$ ; the worked $3\times4$ matrix has rank $2$ , nullity $2$ , and the stated null-space basis; the defective matrix $B$ has geometric multiplicity $1$ versus algebraic multiplicity $2$ ; geometric multiplicity $=n-\operatorname{rank}(A-\lambda\mathbb{1})$ ; and the molecular mode counts are $3N-6$ (nonlinear) and $3N-5$ (linear). No nonstandard numerical values are used anywhere in the article.

Builds on. Vector Spaces and Subspaces · Linear Independence, Basis, and Dimension · Linear Operators and Matrix Representations. Leads to. Eigenvalues, Eigenvectors, and the Characteristic Equation · Diagonalization and the Spectral Decomposition · Jordan Canonical Form and Defective Matrices · Adjoint and Hermitian Operators.

Ask the tutor: Ask me to row-reduce any matrix you like and read off its rank, nullity, null-space basis, and exactly how many solutions $A\mathbf{x}=\mathbf{b}$ has.

Why does the theorem need the domain to be finite-dimensional but not the codomain?

The proof counts basis vectors of $V$ , splitting them into a kernel basis and the added vectors whose images span the range. That count only makes sense if $V$ has finitely many basis vectors to distribute — the whole argument is a finite partition of the domain's dimensions. The codomain $W$ never gets counted; it only has to be large enough to contain the image. This is why the theorem survives into quantum mechanics, where operators act on finite-dimensional spin spaces mapping into infinite-dimensional function spaces. In genuinely infinite-dimensional domains the identity fails, and one needs the Fredholm index instead.

What is the difference between geometric and algebraic multiplicity?

The algebraic multiplicity of an eigenvalue $\lambda$ is how many times $(\lambda-x)$ divides the characteristic polynomial; the geometric multiplicity is $\dim\ker(A-\lambda\mathbb{1})$ , the number of independent eigenvectors, computed by rank-nullity as $n-\operatorname{rank}(A-\lambda\mathbb{1})$ . Geometric multiplicity is always at least $1$ and never exceeds the algebraic multiplicity. When they are equal for every eigenvalue the matrix is diagonalizable; when the geometric falls short — as in Worked Example 2, where a double eigenvalue supplies only one eigenvector — the matrix is defective and requires the Jordan canonical form. In physics the geometric multiplicity is the degeneracy of an energy level, so this gap is precisely where the tidy spectral picture breaks down.

Is the rank of a matrix the same as the number of its nonzero rows?

Only after you row-reduce it. In a general matrix the rows may be linearly dependent even though none is literally zero — the second row of Worked Example 1, $(2,4,6,8)$ , is nonzero yet contributes nothing to the rank because it is twice the first. The rank is the number of pivots in the row echelon form, equivalently the number of independent rows, equivalently the number of independent columns. Counting visibly nonzero rows in the original matrix overcounts whenever there is hidden dependence. The safe recipe is always the same: reduce to echelon form and count the pivots, which simultaneously gives you the rank and, by subtraction from the number of columns, the nullity.

How does rank-nullity relate to solving A x = b?

It supplies the complete answer to both existence and uniqueness. Existence is the rank condition of equation (6): a solution exists iff appending $\mathbf{b}$ leaves the rank unchanged, i.e. $\mathbf{b}$ is in the column space. Uniqueness is the nullity: the solution set, when nonempty, is a translate of the null space, so it has dimension $n-\operatorname{rank}A$ . Zero nullity means at most one solution; positive nullity means an infinite affine family. This is why full column rank guarantees uniqueness and full row rank guarantees existence for every $\mathbf{b}$ , and why a square matrix with both — nonzero determinant — gives exactly one solution always. The theorem turns "how many solutions?" into two dimension counts.

Why is column rank always equal to row rank?

It is a genuine theorem, not a definition, and it can feel surprising because rows and columns look so different. One clean argument: row-reduce $A$ to echelon form. Row operations never change the row space, so the number of nonzero rows (the pivots) is the row rank; but those same pivots sit in specific columns, and the corresponding columns of the original $A$ are independent and span its column space, making the pivot count the column rank too. So a single integer, the number of pivots, is both ranks at once — equivalently $\operatorname{rank}A=\operatorname{rank}A^{\mathsf T}$ . The consequence is that "rank" is unambiguous, which is quietly essential to the whole theory.

Do the kernel and image ever have any overlap?

They live in different spaces — the kernel in the domain $V$ , the image in the codomain $W$ — so in general asking whether they overlap is a category error. The confusion arises for an operator $T:V\to V$ mapping a space to itself, where both are subspaces of the same $V$ and can genuinely intersect. A pure projection has $\ker T$ and $\operatorname{im}T$ complementary, meeting only at $0$ ; but a nilpotent shift $(x,y,z)\mapsto(y,z,0)$ has vectors that are both in the image and in the kernel. So for self-maps, rank plus nullity still equals $\dim V$ , yet the two subspaces need not be complementary — that holds only for special operators such as projections and, with an inner product, normal operators.