# Kernel, Image, and Rank–Nullity

Mathematics I · Linear Algebra · https://tryals.app/learn/mathematics-i/kernel-image-and-rank-nullity

## Kernel and Image

Two subspaces describe any linear map $T : V \to W$. The **kernel** ($\ker T$) is everything $T$ sends to zero: $\ker T = \{v \in V : T(v) = \mathbf{0}\}$. It measures what the map collapses. **$T$ is injective** exactly when $\ker T = \{\mathbf{0}\}$.

The **image** ($\operatorname{im} T$) is everything $T$ can produce: $\operatorname{im} T = \{T(v) : v \in V\}$. It measures what the map reaches. The dimension of the image is the **rank**, and the dimension of the kernel is the **nullity**.

| Subspace | Where it lives | What it measures |
| --- | --- | --- |
| **Kernel** | Domain $V$ | What collapses to $\mathbf{0}$ |
| **Image** | Codomain $W$ | What the map reaches |

## Rank–Nullity Theorem

The **Rank–Nullity Theorem** links these dimensions to the domain:
$$\dim(\ker T) + \dim(\operatorname{im} T) = \dim V$$

**Nullity + rank = dimension of the domain.** Every dimension of the input is accounted for: some collapse (kernel) and the rest survive (image). If a map from $\mathbb{R}^5$ has a 2D kernel, its image must be 3D.

> **Pitfall:** Confusing the kernel (in the domain) with the image (in the codomain), or treating Rank–Nullity as optional. It is a hard accounting constraint: one forces the other.

## Practice questions

4 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. T is a linear map from five-dimensional space into three-dimensional space. Select every statement that MUST be true.

A. T can be injective
B. The rank of T is at most 3
C. T can be surjective
D. Rank plus nullity equals 5
E. The nullity of T is at least 2

**Answer:** B. The rank of T is at most 3; C. T can be surjective; D. Rank plus nullity equals 5; E. The nullity of T is at least 2

**Why:** The image lives in a 3-dimensional codomain, so rank is at most 3 and nullity is therefore at least 5 - 3 = 2. A nonzero kernel forbids injectivity, but a map of rank 3 does exist, so surjectivity is possible. Squeezing a 5-dimensional space into 3 dimensions must collapse something.

Page: https://tryals.app/practice/mathematics-i/kernel-image-and-rank-nullity/t-is-a-linear-map-from-five-dimensional-space-into-three-dimensional

### 2. A linear map from ℝ⁵ to ℝ³ can never be injective. Which theorem settles that immediately, and how?

A. Rank-Nullity: rank is at most 3, so the kernel is non-trivial
B. The invertible matrix theorem, which requires a square matrix
C. The rank theorem, since rank must equal the domain's dimension for injectivity
D. None, injectivity depends on the specific map, not the dimensions

**Answer:** A. Rank-Nullity: rank is at most 3, so the kernel is non-trivial

**Why:** The image lives in ℝ³ so rank ≤ 3, and rank + nullity = 5 forces nullity ≥ 2, a kernel that big means whole directions collapse to zero. Squeezing five dimensions into three must lose something, and Rank-Nullity is that intuition made exact.

Page: https://tryals.app/practice/mathematics-i/kernel-image-and-rank-nullity/a-linear-map-from-to-can-never-be-injective-which-theorem

### 3. Rank-Nullity says rank + nullity = dim(domain), with no mention of the codomain. Why does the theorem account for the DOMAIN's dimension rather than the space the map lands in?

A. Because the rank is defined as the dimension of the codomain, which is why it does not appear separately.
B. Because the theorem only holds when the codomain is at least as large as the domain, which is assumed throughout.
C. Because it is bookkeeping on the input side only.
D. Because the codomain always has the same dimension as the domain for linear maps.

**Answer:** C. Because it is bookkeeping on the input side only.

**Why:** The accounting happens on the input side: every dimension of the domain is either crushed to zero or survives into the image, and the two must sum to what you started with. The codomain is room rather than constraint, map three dimensions into a hundred and the image is still at most three-dimensional. A map from five to three has nullity at least 2, so it cannot be injective.

Page: https://tryals.app/practice/mathematics-i/kernel-image-and-rank-nullity/rank-nullity-says-rank-nullity-dim-domain-with-no-mention-of

### 4. Order the steps of computing a basis for the kernel of a matrix A.

**Answer:**

1. Write down the homogeneous system Ax = 0
2. Row reduce A to echelon form
3. Identify the pivot columns and hence the free variables
4. Express every solution in terms of the free variables
5. Read off one basis vector of the kernel for each free variable

**Why:** Each free variable contributes one basis vector of the kernel, so nullity equals the number of free variables and rank equals the number of pivots. Since every column is either a pivot column or a free column, the two counts add to the number of columns, which is Rank-Nullity, proved by bookkeeping.

Page: https://tryals.app/practice/mathematics-i/kernel-image-and-rank-nullity/order-the-steps-of-computing-a-basis-for-the-kernel-of-a-matrix-a