LA-3.3

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Rank-Nullity Theorem

The Rank-Nullity Theorem is the crown jewel of linear map theory. This fundamental dimension formula connects the kernel and image, revealing that dimension is conserved under linear maps.

3-4 hours Core Level 10 Objectives

Learning Objectives

State the Rank-Nullity Theorem precisely
Prove the Rank-Nullity Theorem using basis extension
Apply the theorem to compute dimensions of kernel and image
Understand consequences for injectivity and surjectivity
Use the theorem to solve existence/uniqueness problems
Apply the theorem to matrix equations Ax = b
Prove impossibility results (no injective/surjective maps)
Connect to the dimension formula for subspaces
Apply to quotient spaces and the First Isomorphism Theorem
Understand the theorem's role in linear systems

Prerequisites

Kernel and image of linear maps (LA-3.2)
Basis and dimension (LA-2.4)
Linear independence and span (LA-2.3)
Basis extension theorem
Injectivity and surjectivity characterizations

Historical Context

The Rank-Nullity Theorem is also known as the Dimension Theorem or the Fundamental Theorem of Linear Maps. It represents one of the deepest insights in linear algebra, connecting the "information lost" (kernel) with the "information preserved" (image) under a linear transformation.

The theorem's proof technique—extending a basis of a subspace to a basis of the whole space—is a fundamental method that appears throughout linear algebra. This "extend the small to the large" approach is central to many dimension-counting arguments.

1. The Rank-Nullity Theorem

This theorem provides the fundamental relationship between the dimensions of the kernel and image of a linear map. It is the cornerstone for understanding linear systems.

Definition 3.7: Rank and Nullity

For a linear map $T: V \to W$ from a finite-dimensional space $V$ :

The rank of $T$ is $\text{rank}(T) = \dim(\text{im } T)$
The nullity of $T$ is $\text{nullity}(T) = \dim(\ker T)$

Theorem 3.25: Rank-Nullity Theorem (Dimension Formula)

Let $T: V \to W$ be a linear map from a finite-dimensional vector space $V$ . Then:

\dim(V) = \dim(\ker T) + \dim(\text{im } T)

Equivalently:

\dim(V) = \text{nullity}(T) + \text{rank}(T)

Proof:

The proof uses the "extend a basis" technique. Let $k = \dim(\ker T)$ .

Step 1: Let $\{v_1, \ldots, v_k\}$ be a basis for $\ker T$ . Since $\ker T \subseteq V$ , we can extend this to a basis of $V$ :

\{v_1, \ldots, v_k, u_1, \ldots, u_r\}

where $k + r = \dim V$ .

Step 2: We claim $\{T(u_1), \ldots, T(u_r)\}$ is a basis for $\text{im } T$ .

Spanning: Any $w \in \text{im } T$ equals $T(v)$ for some $v \in V$ . Write $v = \sum_{i=1}^{k} a_i v_i + \sum_{j=1}^{r} b_j u_j$ . Then:

w = T(v) = \sum_{i=1}^{k} a_i T(v_i) + \sum_{j=1}^{r} b_j T(u_j) = \sum_{j=1}^{r} b_j T(u_j)

since $T(v_i) = 0$ for each $v_i \in \ker T$ .

Linear independence: Suppose $\sum_{j=1}^{r} c_j T(u_j) = 0$ . Then:

T\left(\sum_{j=1}^{r} c_j u_j\right) = 0

So $\sum_{j=1}^{r} c_j u_j \in \ker T$ . Since $\{v_1, \ldots, v_k\}$ is a basis for $\ker T$ :

\sum_{j=1}^{r} c_j u_j = \sum_{i=1}^{k} a_i v_i

Rearranging: $\sum_{i=1}^{k} a_i v_i - \sum_{j=1}^{r} c_j u_j = 0$ . Since $\{v_1, \ldots, v_k, u_1, \ldots, u_r\}$ is a basis of $V$ , all coefficients are 0. In particular, $c_1 = \cdots = c_r = 0$ .

Conclusion: $\dim(\text{im } T) = r = \dim V - k = \dim V - \dim(\ker T)$ .

∎

Remark 3.10: Intuition

Think of $V$ as having $n$ "degrees of freedom." The kernel measures how many degrees of freedom get "collapsed" to zero. The image measures how many "survive" to the output. These must add up to $n$ .

2. Fundamental Consequences

Corollary 3.3: Injectivity, Surjectivity, and Bijectivity

Let $T: V \to W$ be linear with $\dim V = \dim W = n$ . The following are equivalent:

$T$ is injective ( $\ker T = \{0\}$ )
$T$ is surjective ( $\text{im } T = W$ )
$T$ is bijective (an isomorphism)
$\text{rank}(T) = n$
$\text{nullity}(T) = 0$

Proof:

By Rank-Nullity: $n = \text{rank}(T) + \text{nullity}(T)$ .

$T$ is injective iff $\text{nullity}(T) = 0$ iff $\text{rank}(T) = n$
$T$ is surjective iff $\text{rank}(T) = \dim W = n$ iff $\text{nullity}(T) = 0$

So injective ⟺ surjective ⟺ bijective when dimensions are equal.

∎

Corollary 3.4: Impossibility Results

Let $T: V \to W$ be linear.

If $\dim V > \dim W$ : $T$ cannot be injective.
If $\dim V < \dim W$ : $T$ cannot be surjective.

Proof:

Case 1: If $\dim V > \dim W$ , then $\text{rank}(T) \leq \dim W < \dim V$ .

By Rank-Nullity: $\text{nullity}(T) = \dim V - \text{rank}(T) > 0$ .

So $\ker T \neq \{0\}$ , meaning $T$ is not injective.

Case 2: If $\dim V < \dim W$ , then:

$\text{rank}(T) \leq \dim V < \dim W$

So $\text{im } T \neq W$ , meaning $T$ is not surjective.

∎

Corollary 3.5: Bounds on Rank

For any linear map $T: V \to W$ :

\text{rank}(T) \leq \min\{\dim V, \dim W\}

3. Examples

Example 3.15: Differentiation

Problem: Let $D: P_3 \to P_2$ be differentiation. Find rank and nullity.

Solution:

$\ker(D) = \{c : c \in \mathbb{R}\}$ (constant polynomials)
$\text{nullity}(D) = 1$
By Rank-Nullity: $\text{rank}(D) = \dim P_3 - 1 = 4 - 1 = 3$
$\text{im}(D) = P_2$ (D is surjective since rank = dim P₂)

Example 3.16: Projection

Problem: Let $P: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $P(x, y, z) = (x, y, 0)$ .

Solution:

$\ker(P) = \{(0, 0, z) : z \in \mathbb{R}\}$ , so $\text{nullity}(P) = 1$
$\text{im}(P) = \{(x, y, 0) : x, y \in \mathbb{R}\}$ , so $\text{rank}(P) = 2$
Check: $1 + 2 = 3 = \dim \mathbb{R}^3$ ✓

Example 3.17: Matrix Map

Problem: Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be defined by $T(x) = Ax$ where

A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 1 & 2 & 1 & 3 \end{pmatrix}

Solution: Row reduce to find rank:

A \to \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}

$\text{rank}(T) = 2$ (2 pivot rows)

$\text{nullity}(T) = 4 - 2 = 2$ (2 free variables)

Example 3.18: Existence Question

Problem: Can there exist an injective linear map $T: \mathbb{R}^5 \to \mathbb{R}^3$ ?

Solution: No! If $T$ were injective, then $\text{nullity}(T) = 0$ .

By Rank-Nullity: $\text{rank}(T) = 5$ .

But $\text{rank}(T) \leq \dim \mathbb{R}^3 = 3 < 5$ . Contradiction!

4. Applications to Linear Systems

The Rank-Nullity Theorem directly applies to understanding solutions of $Ax = b$ .

Theorem 3.26: Solution Structure

For an $m \times n$ matrix $A$ with $\text{rank}(A) = r$ :

Homogeneous system $Ax = 0$ : Solution space has dimension $n - r$ (nullity)
Non-homogeneous $Ax = b$ : Solutions exist iff $b \in \text{col}(A)$
Uniqueness: If solutions exist, they form $x_p + \text{null}(A)$
Solution is unique iff $\text{nullity}(A) = 0$ , i.e., $r = n$

Example 3.19: System Analysis

Problem: A $3 \times 5$ matrix $A$ has $\text{rank}(A) = 2$ . Describe solutions of $Ax = 0$ .

Solution:

$\text{nullity}(A) = 5 - 2 = 3$

The solution space is 3-dimensional: $\text{null}(A) = \text{span}\{v_1, v_2, v_3\}$

General solution: $x = c_1 v_1 + c_2 v_2 + c_3 v_3$ for $c_1, c_2, c_3 \in \mathbb{R}$

5. Advanced Results

Theorem 3.27: Subspace Version

If $T: V \to V$ is linear and $W \subseteq V$ is a subspace, then:

\dim T(W) + \dim(\ker T \cap W) = \dim W

Theorem 3.28: Idempotent Decomposition

If $T: V \to V$ satisfies $T^2 = T$ (idempotent), then:

V = \ker T \oplus \text{im } T

Proof:

Step 1: Show $\ker T \cap \text{im } T = \{0\}$ .

Let $v \in \ker T \cap \text{im } T$ . Then $T(v) = 0$ and $v = T(u)$ for some $u$ .

Thus $v = T(u) = T^2(u) = T(T(u)) = T(v) = 0$ .

Step 2: Show $V = \ker T + \text{im } T$ .

For any $v \in V$ : $v = (v - T(v)) + T(v)$

Check: $T(v - T(v)) = T(v) - T^2(v) = T(v) - T(v) = 0$

So $v - T(v) \in \ker T$ and $T(v) \in \text{im } T$ .

Conclusion: By Rank-Nullity, dimensions add up, confirming $V = \ker T \oplus \text{im } T$ .

∎

Theorem 3.29: Kernel and Image Chains

For $T: V \to V$ linear, there exists $m$ such that:

\ker T^m = \ker T^{m+1} = \ker T^{m+2} = \cdots

\text{im } T^m = \text{im } T^{m+1} = \text{im } T^{m+2} = \cdots

Moreover, $m \leq \dim V$ .

Theorem 3.30: First Isomorphism Theorem

For any linear map $T: V \to W$ :

V / \ker T \cong \text{im } T

Taking dimensions: $\dim V - \dim(\ker T) = \dim(\text{im } T)$ , which is the Rank-Nullity Theorem!

6. Worked Examples

Example: Trace Map

Problem: Find rank and nullity of $\text{tr}: M_n \to \mathbb{R}$ .

Solution:

$\text{im}(\text{tr}) = \mathbb{R}$ since $\text{tr}(cI) = cn$ .

$\text{rank}(\text{tr}) = 1$

By Rank-Nullity: $\text{nullity}(\text{tr}) = n^2 - 1$

Example: Is T Injective?

Problem: Is $T: \mathbb{R}^3 \to \mathbb{R}^3$ with $T(x,y,z) = (x+y, y+z, x+z)$ injective?

Solution: Find $\ker T$ : solve $x+y = y+z = x+z = 0$ .

Adding equations: $2(x+y+z) = 0$ , so $x+y+z = 0$ .

Combined with $x+y = 0$ : $z = 0$ , then $y = -x$ .

Check: $x + (-x) = 0$ ✓, $(-x) + 0 = -x \neq 0$ unless $x = 0$ .

So $\ker T = \{0\}$ and $T$ is injective.

Example: Dimension Constraint

Problem: If $T: V \to V$ and $\ker T = \text{im } T$ , what can we say about $\dim V$ ?

Solution: Let $k = \dim(\ker T) = \dim(\text{im } T)$ .

By Rank-Nullity: $\dim V = k + k = 2k$

Conclusion: $\dim V$ must be even.

7. Common Mistakes

Mistake 1: Confusing Domain and Codomain

The theorem says $\dim V = \text{rank} + \text{nullity}$ , where $V$ is the domain. The codomain $W$ doesn't appear directly in the formula!

Mistake 2: Assuming Injectivity Implies Surjectivity

This is only true when $\dim V = \dim W$ ! A map $T: \mathbb{R}^2 \to \mathbb{R}^3$ can be injective but never surjective.

Mistake 3: Infinite Dimensions

The theorem requires $V$ to be finite-dimensional! In infinite dimensions, the statement doesn't make sense as stated.

Mistake 4: Rank Can Exceed dim(W)

Remember: $\text{rank}(T) = \dim(\text{im } T) \leq \dim W$ . The rank is bounded by both domain and codomain dimensions.

8. Key Takeaways

The Formula

$\dim V = \text{rank}(T) + \text{nullity}(T)$

Equal Dimensions

When $\dim V = \dim W$ : injective ⟺ surjective ⟺ bijective

Bounds

$\text{rank}(T) \leq \min\{\dim V, \dim W\}$

Matrix Application

For $m \times n$ matrix: nullity = n - rank = # free variables

9. Additional Practice Problems

Problem 1

Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be linear with $\ker T = \text{span}\{(1,1,0,0), (0,0,1,1)\}$ . What is $\text{rank}(T)$ ?

Problem 2

If $T: P_5 \to P_3$ is surjective, what is $\dim(\ker T)$ ?

Problem 3

Prove that if $T: V \to V$ satisfies $T^2 = 0$ , then $\text{rank}(T) \leq \frac{1}{2}\dim V$ .

Problem 4

Let $T: M_{2 \times 2} \to M_{2 \times 2}$ be defined by $T(A) = A - A^T$ . Find $\ker T$ and $\text{im } T$ , and verify Rank-Nullity.

Problem 5

If $\dim V = 10$ and $T: V \to V$ has $\text{rank}(T^2) = 3$ , what are the possible values of $\text{rank}(T)$ ?

Problem 6

Let $S, T: V \to V$ be linear operators with $ST = 0$ . Prove that $\text{rank}(S) + \text{rank}(T) \leq \dim V$ .

10. More Worked Examples

Example: Symmetric Part

Problem: Let $T: M_n \to M_n$ be defined by $T(A) = \frac{1}{2}(A + A^T)$ . Find rank and nullity.

Solution:

$\ker T = \{A : A + A^T = 0\} = \{A : A^T = -A\}$ = skew-symmetric matrices

$\dim(\ker T) = \frac{n(n-1)}{2}$

$\text{im } T = \{B : B = B^T\}$ = symmetric matrices

$\dim(\text{im } T) = \frac{n(n+1)}{2}$

Verification: $\frac{n(n-1)}{2} + \frac{n(n+1)}{2} = n^2 = \dim M_n$ ✓

Example: Integration Operator

Problem: Let $T: P_n \to P_{n+1}$ be defined by $T(p)(x) = \int_0^x p(t)\,dt$ . Find rank and nullity.

Solution:

If $T(p) = 0$ , then $\int_0^x p(t)\,dt = 0$ for all $x$ .

Differentiating: $p(x) = 0$ .

So $\ker T = \{0\}$ and $\text{nullity}(T) = 0$ .

By Rank-Nullity: $\text{rank}(T) = \dim P_n = n + 1$ .

$T$ is injective but not surjective (constant polynomials are not in $\text{im } T$ ).

Example: Composition Rank

Problem: If $S: \mathbb{R}^5 \to \mathbb{R}^4$ has rank 3 and $T: \mathbb{R}^4 \to \mathbb{R}^3$ has rank 2, what can we say about $\text{rank}(T \circ S)$ ?

Solution:

$\text{rank}(T \circ S) \leq \min\{\text{rank}(T), \text{rank}(S)\} = \min\{2, 3\} = 2$

Also, $\text{rank}(T \circ S) \geq \text{rank}(T) + \text{rank}(S) - \dim \mathbb{R}^4 = 2 + 3 - 4 = 1$

Conclusion: $1 \leq \text{rank}(T \circ S) \leq 2$

Example: Shift Operator

Problem: Let $T: P_n \to P_n$ be defined by $T(p)(x) = p(x+1) - p(x)$ . Find rank and nullity.

Solution:

For $p(x) = a_0 + a_1 x + \cdots + a_n x^n$ :

$T(p)$ has degree $n-1$ (leading term $na_n x^{n-1}}$ )

$\ker T = \{\text{constant polynomials}\}$ , so $\text{nullity}(T) = 1$

$\text{im } T = P_{n-1}$ , so $\text{rank}(T) = n$

Check: $1 + n = n + 1 = \dim P_n$ ✓

11. Theoretical Insights

Theorem 3.31: Rank of Composition

For linear maps $T: V \to W$ and $S: W \to U$ :

\text{rank}(S \circ T) \leq \min\{\text{rank}(S), \text{rank}(T)\}

Moreover (Sylvester's inequality):

\text{rank}(S \circ T) \geq \text{rank}(S) + \text{rank}(T) - \dim W

Theorem 3.32: Rank and Powers

For $T: V \to V$ linear:

\text{rank}(T) \geq \text{rank}(T^2) \geq \text{rank}(T^3) \geq \cdots

The sequence stabilizes: there exists $m$ such that $\text{rank}(T^m) = \text{rank}(T^{m+1}) = \cdots$

Theorem 3.33: Nilpotent Operators

If $T: V \to V$ is nilpotent ( $T^k = 0$ for some $k$ ), then:

\text{rank}(T) \leq \dim V - \dim V / k

In particular, if $T^2 = 0$ , then $\text{rank}(T) \leq \frac{\dim V}{2}$ .

Proof:

If $T^2 = 0$ , then $\text{im } T \subseteq \ker T$ .

So $\text{rank}(T) = \dim(\text{im } T) \leq \dim(\ker T) = \text{nullity}(T)$ .

By Rank-Nullity: $\dim V = \text{rank}(T) + \text{nullity}(T) \geq 2 \cdot \text{rank}(T)$ .

Thus $\text{rank}(T) \leq \frac{\dim V}{2}$ .

∎

Remark 3.11: Equivalence of Conditions

For $T: V \to V$ with $\dim V = n$ , the following are equivalent:

$V = \ker T \oplus \text{im } T$
$\ker T \cap \text{im } T = \{0\}$
$\ker T = \ker T^2$
$\text{im } T = \text{im } T^2$
$\text{rank}(T) = \text{rank}(T^2)$

12. Applications

Differential Equations

For the linear ODE $L(y) = f$ where $L$ is a differential operator:

General solution = particular solution + $\ker L$
$\dim(\ker L)$ = order of the ODE
Uniqueness requires $n$ initial conditions (where $n$ = order)

Data Analysis

In Principal Component Analysis (PCA):

Rank = number of significant principal components
Nullity = redundant dimensions in data
Low-rank approximation discards null space

Control Theory

Controllability and observability depend on rank conditions:

System is controllable iff controllability matrix has full rank
Rank deficiency indicates uncontrollable modes

Cryptography

In linear codes:

Code dimension = rank of generator matrix
Syndrome space dimension = rank of parity check matrix
Error correction capability depends on these ranks

13. Quick Reference Summary

Result	Statement
Rank-Nullity	$\dim V = \text{rank}(T) + \text{nullity}(T)$
Equal dim	injective ⟺ surjective ⟺ bijective
$\dim V > \dim W$	$T$ cannot be injective
$\dim V < \dim W$	$T$ cannot be surjective
Rank bound	$\text{rank}(T) \leq \min\{\dim V, \dim W\}$
Matrix	nullity = n - rank = # free variables
Idempotent	$T^2 = T \Rightarrow V = \ker T \oplus \text{im } T$

14. Challenge Problems

Challenge 1: Rank-Nullity for Composition

Problem: Let $T: V \to V$ be linear. Prove that for all $k \geq 1$ :

\text{rank}(T^k) - \text{rank}(T^{k+1}) \leq \text{rank}(T^{k-1}) - \text{rank}(T^k)

In other words, the rank decreases by smaller and smaller amounts.

Challenge 2: Fitting Dimensions

Problem: Let $V$ have dimension 6. Find all possible values of $(\text{rank}(T), \text{rank}(T^2), \text{rank}(T^3))$ for a linear operator $T: V \to V$ with $T^3 = 0$ .

Challenge 3: Sum of Operators

Problem: Let $S, T: V \to V$ be linear. Prove:

\text{rank}(S + T) \leq \text{rank}(S) + \text{rank}(T)

Give an example where equality holds and one where it fails.

15. Study Tips

Tip 1: Always Check with Rank-Nullity

After finding kernel and image separately, verify $\dim(\ker T) + \dim(\text{im } T) = \dim V$ . This catches computational errors.

Tip 2: Use Dimension Comparisons

Before computing anything, compare $\dim V$ and $\dim W$ to determine what's possible (injective? surjective? neither?).

Tip 3: Equal Dimensions Simplify

When $\dim V = \dim W$ , you only need to check ONE of: injective, surjective, $\ker T = \{0\}$ , or $\text{rank}(T) = n$ . They're all equivalent!

Tip 4: For Matrices, Count Free Variables

Nullity = number of free variables = n - rank. Row reduce to find pivot columns, then count non-pivot columns.

16. Geometric Interpretation

The Rank-Nullity Theorem has a beautiful geometric meaning that helps build intuition.

Collapsing Dimensions

Think of the linear map $T$ as "projecting" or "flattening" the domain:

Nullity: Dimensions that get "collapsed" to zero
Rank: Dimensions that "survive" in the output
Total dimensions = collapsed + surviving

Projection Example

Consider projecting $\mathbb{R}^3$ onto the xy-plane:

The z-axis (1 dimension) gets collapsed to zero → nullity = 1
The xy-plane (2 dimensions) survives → rank = 2
Check: 1 + 2 = 3 ✓

Rotation Example

Consider a rotation in $\mathbb{R}^3$ :

Nothing gets collapsed (rotation is distance-preserving) → nullity = 0
All 3 dimensions survive → rank = 3
Check: 0 + 3 = 3 ✓
This shows rotation is an isomorphism!

17. Connection to Other Topics

Quotient Spaces

The First Isomorphism Theorem gives $V/\ker T \cong \text{im } T$ .

Taking dimensions: $\dim V - \dim(\ker T) = \dim(\text{im } T)$

This is exactly Rank-Nullity! The theorem is the "dimension version" of the First Isomorphism Theorem.

Exact Sequences

In homological algebra, Rank-Nullity generalizes to the statement that:

0 \to \ker T \to V \to \text{im } T \to 0

is an exact sequence. The dimensions must "add up" correctly.

Eigenspaces

For an eigenvalue $\lambda$ of $T$ :

The eigenspace $E_\lambda = \ker(T - \lambda I)$
$\dim E_\lambda = \text{nullity}(T - \lambda I)$
This is the geometric multiplicity of $\lambda$

18. Proof Techniques Using Rank-Nullity

Many linear algebra proofs use Rank-Nullity as a key step. Here are common techniques:

Technique 1: Dimension Counting

To show $T$ is an isomorphism when $\dim V = \dim W$ :

Show $\ker T = \{0\}$ (easier to check), or
Show $T$ is surjective (if that's easier)
Either condition implies the other by Rank-Nullity!

Technique 2: Impossibility Arguments

To prove something cannot exist:

Suppose it exists and has certain properties
Use Rank-Nullity to derive a dimension contradiction
Example: No injective map from $\mathbb{R}^5$ to $\mathbb{R}^3$

Technique 3: Subspace Dimension Bounds

To bound the dimension of a subspace:

Express it as $\ker T$ or $\text{im } T$ for some appropriate $T$
Use Rank-Nullity to relate dimensions
This is especially useful for solution spaces of equations

What's Next?

With the Rank-Nullity Theorem mastered, you're ready for:

Isomorphisms: Bijective linear maps that show when spaces are "the same"
Matrix Representation: Converting linear maps to matrices
Change of Basis: How matrices transform under basis changes
Dual Spaces: The space of linear functionals

The Rank-Nullity Theorem will continue to appear throughout linear algebra—it's truly foundational. Mastering it now will pay dividends in every topic that follows.

Rank-Nullity Theorem Practice

Questions

Correct

Accuracy

T: \mathbb{R}^5 \to \mathbb{R}^3

and

\dim(\ker T) = 2

, what is

\dim(\text{im } T)

Easy

Not attempted

T: \mathbb{R}^4 \to \mathbb{R}^4

is surjective, is it injective?

Medium

Not attempted

Can a linear map

T: \mathbb{R}^3 \to \mathbb{R}^5

be surjective?

Medium

Not attempted

\text{rank}(T) = \dim(V)

, then

T

is:

Medium

Not attempted

The nullity of the zero map

0: V \to W

is:

Easy

Not attempted

T: V \to V

and

\ker(T) = \text{im}(T)

, what is

\dim(V)

Hard

Not attempted

T: P_3 \to P_2

is differentiation. What is

\text{nullity}(T)

Medium

Not attempted

A

m \times n

and

\text{rank}(A) = r

, the solution space of

Ax = 0

has dimension:

Medium

Not attempted

T: \mathbb{R}^n \to \mathbb{R}^n

has

\text{rank}(T) = n - 1

, then

\dim(\ker T) =

Easy

Not attempted

For

T: V \to W

, if

\dim V = 7

and

\dim W = 4

, what is the minimum possible

\dim(\ker T)

Medium

Not attempted

T^2 = T

(idempotent), then

V =

Hard

Not attempted

\ker T = \ker T^2

, then:

Hard

Not attempted

Frequently Asked Questions

Why is the Rank-Nullity theorem so important?

It's the fundamental constraint on linear maps. It tells us that 'dimension lost' (kernel) plus 'dimension gained' (image) equals the starting dimension. Everything about existence and uniqueness of solutions follows from this.

What's the relationship to solving Ax = b?

For the matrix A: nullity = # free variables, rank = # basic variables. A solution exists iff b ∈ im(A). The solution is unique iff nullity = 0. The general solution is x = x_p + null(A).

How does this relate to the dimension formula for subspaces?

The Rank-Nullity theorem can be seen as a special case of dimension formulas. For the map T, we're decomposing V based on what happens under T. The proof technique ('extend a basis') is the same.

What happens in infinite dimensions?

The theorem needs modification. It holds for finite-dimensional V, but in infinite dimensions, both ker and im can be infinite-dimensional. Functional analysis handles these cases with concepts like Fredholm operators.

Can I use this to prove a map is an isomorphism?

Yes! If T: V → W with dim(V) = dim(W), then T is an isomorphism iff ker(T) = {0} iff T is surjective. You only need to check one condition—the other follows automatically.

What's the geometric intuition?

Think of the domain V as having n 'degrees of freedom'. Some are 'used up' by being sent to 0 (the nullity). The rest 'survive' to form the image (the rank). These must add up to n.

How does Rank-Nullity connect to matrix rank?

For a matrix A representing T, rank(A) = rank(T) = dim(im T) = column rank = row rank. The nullity equals the number of free variables, which is n - rank(A).

What's the First Isomorphism Theorem connection?

The First Isomorphism Theorem states V/ker(T) ≅ im(T). Taking dimensions: dim(V) - dim(ker T) = dim(im T), which is exactly the Rank-Nullity Theorem!

Can rank(T) ever exceed dim(V)?

No! rank(T) = dim(im T), and im(T) is spanned by the images of a basis of V, so rank(T) ≤ dim(V). Similarly, rank(T) ≤ dim(W).

What if both V and W are infinite-dimensional?

The statement 'dim(V) = rank(T) + nullity(T)' may not make sense as written. In functional analysis, one studies the 'index' of an operator: ind(T) = dim(ker T) - dim(coker T), which can be finite even when ker and coker are infinite-dimensional.