A linear map T preserves the two operations a vector space has: T(\mathbf u + \mathbf v) = T(\mathbf u) + T(\mathbf v) and T(c\mathbf v) = cT(\mathbf v). Its kernel is everything it sends to zero, its image is everything it actually reaches, and the rank-nullity theorem ties the two together: \dim(\ker T) + \dim(\operatorname{im} T) = \dim V. Linear Algebra I proved this for matrices acting on \mathbb R^n; chapter 02 reproves it for an arbitrary linear map between arbitrary vector spaces, and the proof turns out to need nothing matrix-specific at all — the quotient space V/\ker T does the entire argument in one line.
Picture T as a machine. It shreds some directions completely (the kernel), and it stretches the survivors out into wherever it actually reaches. That reachable set, everything the machine can actually output, is the image. The rank-nullity theorem says the shredded pile and the surviving pile always add up to the same total size, the size of whatever went in. Feed in more directions than the machine can possibly preserve, and some are guaranteed to get shredded.
A linear map is a function T between two vector spaces that preserves both operations a vector space has: T(\mathbf u + \mathbf v) = T(\mathbf u) + T(\mathbf v) and T(c\mathbf v) = cT(\mathbf v). Its image is the set of every output it actually produces — everywhere it reaches inside the codomain. The rank-nullity theorem is really a conservation law: a linear map has a fixed "budget" of \dim V dimensions to work with, and every dimension that collapses into the kernel is a dimension unavailable to the image. This is why a linear map from a smaller space to a bigger one can never be onto (the image can't out-dimension its budget) and a linear map from a bigger space to a smaller one can never be one-to-one (something has to collapse). Both facts follow from rank-nullity alone, with no separate argument required.
Theorem
The Rank-Nullity Theorem
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · The rank-nullity theorem — first taught there.
For any m\times n matrix A, \operatorname{rank}(A) + \dim\left(\operatorname{Nul}A\right) = n.
Let k = \dim\left(\operatorname{Nul}A\right) and pick a basis \{\mathbf{v}_1,\dots,\mathbf{v}_k\} for \operatorname{Nul}A, the set of vectors in \mathbb{R}^n that A sends to \mathbf{0}. Since any basis for a subspace of \mathbb{R}^n extends to a basis of the whole space, choose vectors \mathbf{v}_{k+1},\dots,\mathbf{v}_n so that \{\mathbf{v}_1,\dots,\mathbf{v}_n\} is a basis for \mathbb{R}^n. The claim is that the n-k image vectors A\mathbf{v}_{k+1},\dots,A\mathbf{v}_n form a basis for \operatorname{Col}A, which pins \operatorname{rank}(A) = n-k and finishes the proof.
Spanning. Every \mathbf{b}\in\operatorname{Col}A has the form A\mathbf{x} for some \mathbf{x}\in\mathbb{R}^n. Write \mathbf{x} in the extended basis as \mathbf{x} = c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n. Applying A and using linearity, \mathbf{b} = A\mathbf{x} = c_1A\mathbf{v}_1 + \cdots + c_kA\mathbf{v}_k + c_{k+1}A\mathbf{v}_{k+1} + \cdots + c_nA\mathbf{v}_n. The first k terms vanish because \mathbf{v}_1,\dots,\mathbf{v}_k \in \operatorname{Nul}A, leaving \mathbf{b} = c_{k+1}A\mathbf{v}_{k+1} + \cdots + c_nA\mathbf{v}_n. So the n-k image vectors already span \operatorname{Col}A.
Independence. Suppose c_{k+1}A\mathbf{v}_{k+1} + \cdots + c_nA\mathbf{v}_n = \mathbf{0}. By linearity this is A(c_{k+1}\mathbf{v}_{k+1} + \cdots + c_n\mathbf{v}_n) = \mathbf{0}, which puts the vector \mathbf{w} = c_{k+1}\mathbf{v}_{k+1} + \cdots + c_n\mathbf{v}_n inside \operatorname{Nul}A. Since \{\mathbf{v}_1,\dots,\mathbf{v}_k\} is a basis for \operatorname{Nul}A, \mathbf{w} also equals some combination d_1\mathbf{v}_1 + \cdots + d_k\mathbf{v}_k. Subtracting the two expressions for \mathbf{w} gives d_1\mathbf{v}_1 + \cdots + d_k\mathbf{v}_k - c_{k+1}\mathbf{v}_{k+1} - \cdots - c_n\mathbf{v}_n = \mathbf{0}, a linear relation among all n vectors of the basis \{\mathbf{v}_1,\dots,\mathbf{v}_n\}. Basis vectors are linearly independent, so every coefficient here is 0 — in particular c_{k+1} = \cdots = c_n = 0. Only the trivial combination of A\mathbf{v}_{k+1},\dots,A\mathbf{v}_n gives \mathbf{0}, so these vectors are linearly independent.
Together, spanning and independence make \{A\mathbf{v}_{k+1},\dots, A\mathbf{v}_n\} a basis for \operatorname{Col}A, so \operatorname{rank}(A) = \dim\left(\operatorname{Col}A\right) = n-k. Moving k to the other side recovers \operatorname{rank}(A) + \dim\left(\operatorname{Nul}A\right) = n.
Sanity check: for the 3\times 3 identity matrix, \operatorname{Nul}A = \{\mathbf{0}\}, so k = 0 and \operatorname{rank}(A) = 3 = n, matching the fact that the identity is invertible and its columns already span \mathbb{R}^3. At the other extreme, the 3\times 3 zero matrix has \operatorname{Nul}A = \mathbb{R}^3, so k=3 and \operatorname{rank}(A) = 0 — no columns to span anything — and 0 + 3 = 3 still checks out.
Theorem
The Kernel Is a Subspace
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · Kernel — first taught there.
If T: V \to W is a linear transformation, then \ker T = \{\mathbf{v} \in V : T(\mathbf{v}) = \mathbf{0}\} is a subspace of V.
To show \ker T is a subspace of V, three things must hold: it contains the zero vector, it is closed under addition, and it is closed under scalar multiplication.
Contains the zero vector. Linearity forces T(\mathbf{0}) = \mathbf{0}: writing \mathbf{0} = 0\mathbf{v} for any \mathbf{v} \in V, T(\mathbf{0}) = T(0\mathbf{v}) = 0\, T(\mathbf{v}) = \mathbf{0}. So \mathbf{0} \in \ker T, and \ker T is nonempty.
Closed under addition. Let \mathbf{u}, \mathbf{v} \in \ker T, so T(\mathbf{u}) = \mathbf{0} and T(\mathbf{v}) = \mathbf{0}. Then T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) = \mathbf{0} + \mathbf{0} = \mathbf{0}, using additivity of T. Hence \mathbf{u} + \mathbf{v} \in \ker T.
Closed under scalar multiplication. Let \mathbf{v} \in \ker T and let c be any scalar. Then T(c\mathbf{v}) = c\, T(\mathbf{v}) = c\mathbf{0} = \mathbf{0}, using homogeneity of T. Hence c\mathbf{v} \in \ker T.
All three closure properties hold, so \ker T is a subspace of V.
Sanity check: if T is the zero transformation, T(\mathbf{v}) = \mathbf{0} for every \mathbf{v} \in V, so \ker T = V itself — the whole space, which is certainly a subspace of V. At the other extreme, if T is one-to-one, the only vector sent to \mathbf{0} is \mathbf{0} itself, so \ker T = \{\mathbf{0}\}, the trivial subspace. Both extremes are genuine subspaces, exactly as the theorem guarantees.
Watch the kernel appear as det collapses to zero
Drag toward a singular matrix — the kernel line appears in the left panel exactly when the image collapses to a line in the right panel.
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All four subspaces from one matrix, at once
Edit an entry of A and watch the row space, null space, column space, and left null space all update together — their dimensions always keep rank-nullity true on both sides.
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Worked example
Rank-nullity pins down the image dimension without computing it
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1rank-nullity applies to every linear map, so plug in what's already known
T:\mathbb R^2 \to \mathbb R^2 has matrix A = \begin{pmatrix}1&2\\2&4\end{pmatrix}. Row 2 is twice row 1, so \operatorname{rank}(A) = 1; find \dim(\ker T) without solving A\mathbf x = \mathbf 0 directly. \dim(\ker T) + \dim(\operatorname{im }T) = \dim(\mathbb R^2) = 2
rank-nullity applies to every linear map, so plug in what's already known
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2
\dim(\operatorname{im }T) = \operatorname{rank}(A) = 1, so \dim(\ker T) = 2 - 1 = 1.
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3\boxed{\dim(\ker T) = 1}
The kernel is a line through the origin — confirmed directly: A\mathbf x = \mathbf 0 reduces to the single equation x + 2y = 0, whose solution set is exactly one-dimensional.
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Where this shows up: an isomorphism is rank-nullity with no room to spare
Chapter 01's first isomorphism theorem for vector spaces is rank-nullity stated as an equation between quotient spaces: V/\ker T \cong \operatorname{im} T. A linear map is an isomorphism precisely when its kernel is as small as possible (just \{\mathbf 0\}) and its image is as large as possible (all of the codomain) — rank-nullity is what forces those two extremes to happen together whenever \dim V = \dim W.
Where this goes next
Duality and the transpose, chapter 03
"Row rank equals column rank" reads like a computational coincidence the first time it's proven by row-reducing a specific matrix. Chapter 03 proves it's not a coincidence at all: it's the statement \dim(\operatorname{im }T) = \dim(\operatorname{im }T^*) for the dual map T^*, and rank-nullity applied to T^* is exactly what makes that statement provable in one line instead of a computational check performed matrix by matrix.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
Is T(x,y,z) = (x - y + z,\ 2z), a map from \mathbb{R}^3 to \mathbb{R}^2, linear?
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Domain and codomain never need matching dimensions — a linear transformation can map \mathbb{R}^3 to \mathbb{R}^2, \mathbb{R}^5 to \mathbb{R}^{100}, or anything else, as long as additivity and homogeneity hold.
An output coordinate is allowed to ignore some input variables entirely (its coefficient is just 0) — that's still a linear combination.
Any number of input variables can combine linearly; the restriction is on how they combine (no products or powers between them), not how many there are.
Why
Both output coordinates are linear combinations of x, y, z with no products or powers, so T is linear despite mapping a 3-dimensional space to a 2-dimensional one.
Check additivity and homogeneity directly on T(x,y,z) = (x-y+z, 2z) rather than reasoning from domain/codomain size.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
T(x,y) = (4x - y,\ 2y) is linear. Write an expression for the first coordinate of T(x,y) in terms of x and y.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
The first coordinate is read directly off the formula: 4x - y — a linear combination of x and y, with no products or powers between them.
Correct answer
Also accepted:
Why
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
T(x,y) = (2x - y,\ -6x + 3y). If (x,y) is in \ker T and x = 4, what is y?
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Why
\ker T solves 2x - y = 0 and -6x+3y=0 (the second is -3\times the first, so it adds nothing new). From 2x - y = 0, y = 2x. At x=4, y = 8.
Correct answer
Also accepted:
Why
Use the surviving independent equation from the kernel condition (the other equation is a multiple of it) to relate y to x directly.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A = \begin{bmatrix} 1 & 4 \\ 2 & 8 \end{bmatrix}. What does \ker T look like for T(\mathbf x) = A\mathbf x?
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Check the rows: row 2 is exactly 2\times row 1, so row reduction leaves a free variable — that means more than just \mathbf x = \mathbf 0 solves A\mathbf x = \mathbf 0.
A isn't the zero matrix, so it doesn't crush every input to zero — only a whole line's worth, not the whole plane.
\mathbf x = \mathbf 0 always solves A\mathbf x = \mathbf 0 for any A, so the kernel is never empty — it contains at least the zero vector.
Why
Row 2 is 2\times row 1, so row reducing [A \mid \mathbf 0] leaves one pivot and one free variable: x_1 + 4x_2 = 0 \Rightarrow x_1 = -4x_2, giving \ker T = \operatorname{span}\{(-4,1)\}, a line.
Row reduce A and see how many free variables show up before matching that count to a shape (point, line, or plane).
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
T:\mathbb{R}^2\to\mathbb{R}^2, T(x,y)=(5x-2y,\ 10x-4y). If (4,\,m) is in \operatorname{im}(T), what is m?
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Why
A's columns are (5,10) and (-2,-4), both multiples of (1,2), so \operatorname{im}(T)=\operatorname{span}\{(1,2)\} — every reachable output has second coordinate exactly 2 times the first. With first coordinate 4, the second coordinate must be m=2(4)=8.
Correct answer
Also accepted:
Why
Determine whether the two columns of the standard matrix are parallel, and if so, find the fixed ratio between the two coordinates of every output.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
T:\mathbb{R}^2\to\mathbb{R}^2 sends every input to \mathbf{0} (the zero transformation). What is \operatorname{im}(T)?
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The zero transformation reaches exactly one output no matter the input — that's the opposite of covering the whole plane.
The zero transformation still has a perfectly good standard matrix — the 2\times 2 zero matrix — it's just the most degenerate linear map, not an undefined one.
The kernel of the zero map is all of \mathbb{R}^2 (every input maps to \mathbf{0}) — that happens to be a different set from the image here, even though both facts are true simultaneously; don't conflate the two subspaces.
Why
Every input produces the single output \mathbf{0}, so the image is the smallest possible subspace, \{\mathbf{0}\} — dimension 0.
Think about what the zero transformation actually outputs for every possible input before describing its image.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A 4 \times 6 matrix has rank 2, given directly. What is its nullity?
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That's the rank restated, not n - \operatorname{rank}(A). Nullity is n - \operatorname{rank}(A) = 6 - 2.
That's n itself, before subtracting the rank. Nullity is n - \operatorname{rank}(A), not n.
Rank-nullity says rank + nullity = n, not n + \operatorname{rank}(A). Solve 2 + \text{nullity} = 6.
Why
Rank-nullity gives \operatorname{rank}(A) + \operatorname{nullity}(A) = n, the number of columns. Here n = 6 and rank = 2, so nullity = 6 - 2 = 4. The row count, 4, never enters the identity at all.
From Linear Algebra — Vectors, Matrices, and the Geometry of Data
A 5 \times 3 matrix has nullity 1, given directly, and you must work backward. What is its rank?
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That's the nullity restated. Solve \operatorname{rank}(A) + 1 = n for rank, using n = 3.
4 would come from using the row count 5 somewhere -- rank-nullity only involves n, the column count.
That's n itself, before subtracting the nullity. Rank is n - \operatorname{nullity}(A) = 3 - 1.
Why
n = 3 here (the column count), and nullity = 1, so rank = n - \operatorname{nullity}(A) = 3 - 1 = 2.