Linear Algebra - Lecture 12 - Fall, 2026

Overview

The lecture develops linear independence, bases, and matrix transformations through worked examples, then defines the kernel and image of a linear map and proves that each is a subspace. It culminates in the rank-nullity theorem: for a linear transformation from a finite-dimensional space V, dim(ker T) + dim(im T) = dim(V), with the proof built by combining preimages of an image basis with a basis for the kernel.

Key takeaways

A set of vectors is linearly dependent exactly when its associated homogeneous system has a nonzero solution; row reduction can reveal the coefficients of that relation.
For any m×n matrix A, the outputs Ax are precisely the linear combinations of A’s columns, so the image of the associated map is the column space.
A linear transformation is fully determined by its values on a basis: for Rn, those values become the columns of its representing matrix.
Both ker(T) and im(T) are subspaces: linearity preserves zero, vector addition, and scalar multiplication in the ways required for each set.
The rank-nullity theorem follows by taking preimages of an image basis and adjoining a kernel basis; the resulting vectors form a basis for the domain.
For a map from R3 to R2 whose image is all of R2, rank-nullity requires a one-dimensional kernel because 3 = 2 + 1.

Chapters

0:00 Testing Three Vectors in R3 for Linear Dependence

The vectors (5, -2, 1), (1, -2, 5), and (1, 1, -4) are tested by setting a linear combination equal to the zero vector.
The resulting homogeneous system is encoded by a 3×3 coefficient matrix and reduced using elementary row operations.
The calculation finds the nonzero coefficients (-3, 7, 8); direct substitution verifies their linear combination is (0, 0, 0), proving dependence.

11:36 Choosing a Basis from Three Vectors in R2

The vectors (1, 1), (1, 2), and (2, 1) lie in R2, so the set of three must be dependent.
The relation (1, 2) + (2, 1) = 3(1, 1) makes the dependence explicit.
The first two vectors are independent and span the same subspace as all three, so they form a basis for that span.

16:35 A Zero Vector Reduces a Spanning Set to One Basis Vector

The example set in R2 includes (1, 1), (2, 2), and the zero vector, making it dependent.
The vector (2, 2) is twice (1, 1), while any set containing the zero vector is dependent.
The span is the line generated by the nonzero vector (1, 1), so a basis consists of that single vector.

20:15 Matrix–Vector Multiplication as a Combination of Columns

For an m×n matrix A and vector x = (x1, …, xn), Ax equals x1 times column 1 plus … plus xn times column n.
For A = [[2, −3, 0], [5, 11, −9]], multiplying by (x, y, z) produces (2x − 3y, 5x + 11y − 9z).
As x ranges over Rn, Ax ranges over all linear combinations of A’s columns: the column space of A.

27:29 Matrices Define Linear Transformations Between Rn and Rm

An m×n matrix A defines a function TA: Rn → Rm by TA(x) = Ax.
With A = [[2, −3, 0], [5, 11, −9]], the input (1, 2, 3) maps to (−4, 0).
Linearity requires both T(v + w) = T(v) + T(w) and T(cv) = cT(v).

31:44 Constructing a Matrix for Any Linear Map Rn → Rm

The standard basis vectors e1 through en express every x ∈ Rn uniquely as x1e1 + … + xnen.
Build an m×n matrix A by placing T(ej) in its jth column for each standard basis vector.
Linearity then gives T(x) = Σ xjT(ej) = Ax, showing every linear transformation Rn → Rm is represented by a matrix.

41:35 Why a Linear Transformation Sends Zero to Zero

For a linear map T, scalar compatibility gives T(0v) = 0T(v).
Because zero times any vector is the zero vector, T(0) = 0.
This fact underlies the definitions and subspace proofs for both the kernel and image.

45:41 Defining the Kernel and Proving It Is a Subspace

The kernel ker(T) consists of all vectors v in the domain V for which T(v) = 0 in the codomain W.
If v1 and v2 are in ker(T), then T(v1 + v2) = T(v1) + T(v2) = 0.
For any scalar c, T(cv1) = cT(v1) = 0; together with 0 ∈ ker(T), these properties establish that the kernel is a subspace of V.

49:16 Finding a Kernel by Solving a Homogeneous System

For the example matrix A = [[2, −3, 0], [5, 11, −9]], ker(TA) is found by solving Ax = 0 for x ∈ R3.
The kernel calculation turns matrix multiplication into a homogeneous system with two equations and three unknowns.
Row reduction is used to describe the solution set as a one-dimensional subspace, illustrating how a kernel can be computed from a matrix.

57:34 Defining the Image of a Linear Transformation

The image im(T) consists of all w ∈ W for which there is some v ∈ V with T(v) = w.
Unlike the kernel, which is a subset of the domain, the image is a subset of the codomain.
The image records exactly which outputs the transformation can produce.

59:20 Proving the Image Is a Subspace of the Codomain

The zero vector belongs to im(T) because T(0) = 0.
If w1 = T(v1) and w2 = T(v2), then w1 + w2 = T(v1 + v2), so the image is closed under addition.
For any scalar c, cw = cT(v) = T(cv), proving closure under scalar multiplication.

1:03:11 Computing the Image from Matrix Columns

For TA: R3 → R2 defined by A = [[2, −3, 0], [5, 11, −9]], the image is the column space of A.
The first column (2, 5) and third column (0, −9) are independent, as their linear combination can equal zero only when both coefficients are zero.
Two independent vectors in R2 span R2, so this transformation is onto and im(TA) = R2.

1:07:14 Rank-Nullity: Domain Dimension Splits Between Kernel and Image

In the R3-to-R2 example, the image has dimension 2 and the kernel has dimension 1.
Their dimensions add to 3, the dimension of the domain R3.
The general theorem states that for finite-dimensional V, dim(ker T) + dim(im T) = dim(V).

1:09:46 Building a Candidate Basis for the Domain

Choose a basis f1, …, fk for im(T), and choose preimages e1, …, ek in V such that T(ei) = fi.
Let e(k+1), …, e(k+l) be a basis for ker(T), where l = dim(ker T).
The proof aims to show that these k + l vectors together form a basis for V.

1:14:49 Proving the Preimages and Kernel Basis Are Independent

Apply T to a zero linear combination of e1, …, ek and the kernel basis vectors e(k+1), …, e(k+l).
The kernel terms map to zero, leaving a linear combination of f1, …, fk, which forces the first k coefficients to be zero because the fi form a basis.
The remaining coefficients must also vanish because the kernel basis vectors are linearly independent.

1:20:16 Spanning the Domain and Concluding Rank-Nullity

For any v ∈ V, T(v) lies in im(T) and can be written as a combination of f1, …, fk.
Subtracting the matching combination of preimages e1, …, ek from v leaves a vector in ker(T), expressible using the kernel basis.
The combined set is therefore a basis for V, giving dim(V) = k + l = dim(im T) + dim(ker T); the theorem is identified as important new material, not on the midterm.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Nathanson math lectures.