Rank-Nullity & Dimension Arguments Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice rank-nullity and dimension arguments: computing \(\dim\ker T\) and \(\dim\operatorname{Im}T\), using \(\dim V=\operatorname{rank}T+\operatorname{nullity}T\), remembering that rank is at most \(\min(\dim V,\dim W)\), deciding when maps can be injective or surjective, recognizing that finite-dimensional isomorphic spaces have the same dimension, reading matrix maps by domain dimension and rank, and proving impossibility statements before doing algebra. If you want a refresher, open the lesson for mentally followable examples and checks.
Answer the question set and review your mistakes at the end.
How this rank-nullity practice works
1. Take the practice set: answer rank, nullity, injectivity, surjectivity, and matrix-dimension questions below.
2. Open the lesson: review the theorem, rank bounds, square-map equivalences, and common traps with worked examples.
3. Retry: return to the question set and count dimensions before solving systems.
What you will learn in the rank-nullity & dimension arguments lesson
Purpose: Use rank-nullity as a dimension-counting tool. The goal is to move quickly between a linear map \(T:V\to W\), its kernel, its image, and statements like injective, surjective, full rank, rank deficient, and invertible.
Success criteria
State rank-nullity in the form \(\dim V=\dim\ker T+\dim\operatorname{Im}T\).
Remember that the domain dimension, not the codomain dimension, is on the left.
Compute missing rank or nullity values from small data.
Use \(\operatorname{rank}T\le \min(\dim V,\dim W)\) to reject impossible claims.
Recognize injectivity as nullity \(0\), and surjectivity as rank \(\dim W\).
Use equal finite dimensions to turn injective into surjective and surjective into injective.
Use isomorphism to conclude that finite-dimensional spaces have the same dimension.
For an \(m\times n\) matrix, use \(n\) as the domain dimension.
Build short dimension arguments before doing coordinate calculations.
Key vocabulary
Kernel: \(\ker T=\{v\in V:T(v)=0\}\).
Image: \(\operatorname{Im}T=\{T(v):v\in V\}\).
Rank: \(\dim\operatorname{Im}T\).
Nullity: \(\dim\ker T\).
Full column rank: rank equals the number of columns, so the matrix map is injective.
Full row rank: rank equals the number of rows, so the matrix map is surjective onto the codomain.
Quick pre-check
Pre-check 1: If \(T:\mathbb{R}^4\to\mathbb{R}^2\) has rank \(2\), what is \(\dim\ker T\)?
Hint: Use \(4=2+\dim\ker T\).
Pre-check 2: Can a linear map \(T:\mathbb{R}^2\to\mathbb{R}^3\) be surjective?
Hint: The image dimension cannot exceed the domain dimension.
Rank-nullity counts domain dimensions
Learning goal: Use rank-nullity to find a missing dimension without solving the entire linear system.
Key idea
For a linear map \(T:V\to W\) with finite-dimensional domain, \[\dim V=\dim\ker T+\dim\operatorname{Im}T.\] The image dimension is the rank, and the kernel dimension is the nullity. The theorem partitions the domain into directions that collapse to \(0\) and directions that produce independent outputs.
Recognition checklist
Identify the domain \(V\).
Find the domain dimension \(\dim V\).
Translate image dimension into rank.
Translate kernel dimension into nullity.
Solve the one-step equation \(\dim V=\operatorname{rank}T+\operatorname{nullity}T\).
Worked example
Example: If \(T:\mathbb{R}^7\to\mathbb{R}^3\) is surjective, what is \(\dim\ker T\)?
Surjective means \(\operatorname{Im}T=\mathbb{R}^3\), so \(\operatorname{rank}T=3\). Rank-nullity gives \[7=\dim\ker T+3,\] hence \(\dim\ker T=4\).
Try it
Try it 1: If \(T:\mathbb{R}^6\to\mathbb{R}^4\) has rank \(4\), what is its nullity?
Hint: Rank and nullity add to the domain dimension \(6\).
Try it 2: A linear map from a \(9\)-dimensional space has rank \(4\). What is its nullity?
Hint: Use \(9=4+\operatorname{nullity}T\).
Rank cannot exceed either side
Learning goal: Use dimension bounds to decide whether a proposed rank, injection, or surjection is possible.
Key idea
For \(T:V\to W\), the image is created from vectors in \(V\) but lives inside \(W\). Therefore \[\operatorname{rank}T\le \dim V,\qquad \operatorname{rank}T\le \dim W,\] so \(\operatorname{rank}T\le \min(\dim V,\dim W)\).
Worked example
Example: Can a linear map \(T:\mathbb{R}^5\to\mathbb{R}^3\) have rank \(4\)?
No. The image is a subspace of \(\mathbb{R}^3\), so its dimension is at most \(3\). Rank \(4\) is impossible before any formula for \(T\) is known.
Try it
Try it 1: The rank of a linear map \(V\to W\) is at most what?
Hint: The image cannot have more independent directions than the domain provides or the codomain can contain.
Try it 2: If \(T:\mathbb{R}^3\to\mathbb{R}^7\) is injective, what is its nullity?
Hint: Injective means no nonzero vector is sent to \(0\).
Summary
Rank is never bigger than the domain dimension.
Rank is never bigger than the codomain dimension.
Small domain to large codomain can be injective but not surjective.
Large domain to small codomain can be surjective but not injective.
Use rank and nullity to test maps
Learning goal: Translate injective and surjective into rank-nullity language.
Key idea
A linear map \(T:V\to W\) is injective exactly when \(\ker T=\{0\}\), so nullity is \(0\) and rank is \(\dim V\). It is surjective exactly when \(\operatorname{Im}T=W\), so rank is \(\dim W\). These are different requirements unless \(\dim V=\dim W\).
Worked example
Example: Can a linear map \(T:\mathbb{R}^4\to\mathbb{R}^2\) be injective?
No. Injectivity would require nullity \(0\), so rank \(4\). But the rank is at most the codomain dimension \(2\). Therefore some nonzero vector must lie in the kernel.
Try it
Try it 1: Can a linear map \(T:\mathbb{R}^5\to\mathbb{R}^3\) be surjective?
Hint: Surjectivity only needs rank \(3\), which is within the possible rank bound.
Try it 2: Can a linear map \(T:\mathbb{R}^3\to\mathbb{R}^2\) be injective?
Hint: Injectivity would need rank \(3\), but the codomain only has dimension \(2\).
For \(m\times n\) matrices, the domain has dimension \(n\)
Learning goal: Apply rank-nullity to matrix maps without mixing up rows and columns.
Key idea
An \(m\times n\) matrix \(A\) defines \(x\mapsto Ax\) from \(\mathbb{R}^n\) to \(\mathbb{R}^m\). Rank-nullity uses the domain dimension \(n\): \[n=\operatorname{rank}A+\operatorname{nullity}A.\] Rows describe the codomain size; columns describe input coordinates.
Worked example
Example: If a \(2\times5\) matrix has rank \(2\), what is its nullity?
The map goes from \(\mathbb{R}^5\) to \(\mathbb{R}^2\), so the domain dimension is \(5\). Rank-nullity gives \(5=2+\operatorname{nullity}A\), so nullity is \(3\).
Try it
Try it 1: If a \(5\times2\) matrix has rank \(2\), what can be said about its column map?
Hint: Rank equals the number of columns, so the nullity is \(0\).
Try it 2: If a \(4\times4\) matrix has rank \(3\), what is the dimension of its nullspace?
Hint: The domain dimension is \(4\).
Square maps collapse many tests into one
Learning goal: Use finite-dimensional isomorphism and equal-dimension tests to connect injectivity, surjectivity, full rank, and invertibility.
Key idea
Any finite-dimensional isomorphism \(T:V\to W\) forces \(\dim V=\dim W\). Conversely, when \(\dim V=\dim W=n\), the following conditions are equivalent: \(T\) is injective, \(T\) is surjective, \(\operatorname{rank}T=n\), \(\operatorname{nullity}T=0\), and \(T\) is an isomorphism. For an \(n\times n\) matrix, these are also equivalent to invertibility.
Worked example
Example: If \(T:\mathbb{R}^3\to\mathbb{R}^3\) is surjective, what else follows?
Surjective means rank \(3\). Rank-nullity gives \(3=3+\operatorname{nullity}T\), so nullity is \(0\). Therefore \(T\) is injective as well.
Try it
Try it 1: A linear map \(T:\mathbb{R}^n\to\mathbb{R}^n\) has kernel \(\{0\}\). What follows?
Hint: Zero kernel gives injectivity; equal finite dimensions convert that into onto.
Try it 2: For a square \(n\times n\) matrix, rank \(n\) is equivalent to what?
Hint: Full rank square matrices have zero kernel and reach the whole codomain.
Try it 3: If \(\dim V=3\) and \(\dim W=5\), can \(V\) and \(W\) be isomorphic?
Hint: Finite-dimensional isomorphic spaces must have the same dimension.
Prove impossibility before computing
Learning goal: Turn rank-nullity into short arguments about what a map can or cannot do.
Key idea
A dimension argument often has this shape: assume the desired property, translate it into rank or nullity, then compare with the rank bound. This is faster than solving for every vector when the dimensions already decide the question.
Worked example
Example: Explain why no linear map from a \(2\)-dimensional space to a \(3\)-dimensional space can be surjective.
The image is generated by inputs from a \(2\)-dimensional domain, so its dimension is at most \(2\). Surjectivity onto a \(3\)-dimensional codomain would require rank \(3\). Since \(3>2\), surjectivity is impossible.
Try it
Try it 1: If \(\dim V=4\) and \(T:V\to W\) is injective, what is \(\operatorname{rank}T\)?
Hint: Injectivity means nullity \(0\), so all four domain dimensions appear in the image.
Try it 2: If \(\dim V=6\) and \(\dim\ker T=6\), what is \(\operatorname{Im}T\)?
Hint: The kernel already uses all domain dimensions, leaving rank \(0\).
Avoid the common dimension traps
Learning goal: Keep the domain, codomain, rank, and nullity roles separate.
Common traps
Wrong side: rank-nullity uses \(\dim V\), the domain dimension.
Too small to cover: if \(\dim V<\dim W\), no map \(V\to W\) is surjective.
Too large to inject: if \(\dim V>\dim W\), no map \(V\to W\) is injective.
Full rank means full possible rank: for rectangular matrices this can mean injective or surjective, not always both.
Rank \(0\): the image is \(\{0\}\), so a linear map is the zero map.
Worked example
Example: If \(\dim V=4\) and \(\operatorname{rank}T=0\), what is \(T\)?
Rank \(0\) means the image has dimension \(0\), so the image is \(\{0\}\). Every input maps to \(0\), which is exactly the zero map.
Try it
Try it 1: If \(T:\mathbb{R}^4\to\mathbb{R}^4\) has nullity \(2\), can it be surjective?
Hint: Nullity \(2\) gives rank \(2\), not rank \(4\).
Try it 2: If \(\dim V=5\) and \(\operatorname{rank}T=0\), what is \(\dim\ker T\)?
Rank is \(\dim\operatorname{Im}T\); nullity is \(\dim\ker T\).
Injective means nullity \(0\); surjective means rank \(\dim W\).
Rank is at most \(\min(\dim V,\dim W)\).
An \(m\times n\) matrix uses \(n\) as the domain dimension.
In equal finite dimensions, injective and surjective are equivalent.
Finite-dimensional isomorphic spaces have the same dimension.
Next step: Close this lesson and try the quiz again. Before calculating, ask: what is the domain dimension, what is the maximum possible rank, and which condition is the problem asking for?
Practice set
Rang-nulheid en dimensieargumenten practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
Als \(T:\mathbb{R}^4\to\mathbb{R}^2\) rang \(2\) heeft, wat is dan \(\dim(\ker T)\)?
Correct answer: B. \(2\)
Explanation: De rang-nulheidstelling geeft \(4=2+\dim(\ker T)\).
Question 2Not answered
Kan een lineaire afbeelding \(T:\mathbb{R}^2\to\mathbb{R}^3\) surjectief zijn?
Correct answer: B. Nee, omdat het domein te klein is
Explanation: Nee. De rang is hoogstens \(2\), maar surjectiviteit op \(\mathbb{R}^3\) vereist rang \(3\).
Question 3Not answered
Kan een lineaire afbeelding \(T:\mathbb{R}^3\to\mathbb{R}^2\) injectief zijn?
Correct answer: B. Nee, de kernel kan niet \(\{0\}\) zijn
Explanation: Nee. De rang is hoogstens \(2\), dus de nuliteit is minstens \(1\).
Question 4Not answered
Als \(T:\mathbb{R}^5\to W\) nuliteit \(3\) heeft, wat is dan de rang?
Correct answer: D. \(2\)
Explanation: De rang-nulheidstelling geeft \(5=\operatorname{rank}T+3\).
Question 5Not answered
Als \(T:V\to W\) injectief is en \(\dim V=4\), wat is dan \(\operatorname{rank}T\)?
Correct answer: D. \(4\)
Explanation: Injectief betekent nuliteit \(0\), dus de rang is gelijk aan \(\dim V\).
Question 6Not answered
Als de lineaire afbeelding \(T:\mathbb{R}^3\to\mathbb{R}^3\) surjectief is, wat kunnen we dan zeggen over \(T\)?
Correct answer: D. Zij is injectief
Explanation: Voor lineaire afbeeldingen tussen gelijke eindigdimensionale ruimten zijn surjectief en injectief equivalent.
Question 7Not answered
Als \(\dim V=6\) en \(\dim(\ker T)=6\), wat is dan \(\operatorname{Im}T\)?
Correct answer: D. \(\{0\}\)
Explanation: De nuliteit gebruikt al heel \(V\), dus de rang is \(0\).
Question 8Not answered
Als \(T:\mathbb{R}^4\to\mathbb{R}^4\) rang \(3\) heeft, is \(T\) dan injectief?
Correct answer: D. Nee, omdat de nuliteit \(1\) is
Explanation: Nee. De nuliteit is \(1\), dus de kernel is niet alleen \(\{0\}\).
Question 9Not answered
Als een lineaire afbeelding vanuit een \(3\)-dimensionale ruimte een \(1\)-dimensionaal beeld heeft, wat is dan de nuliteit?
Correct answer: A. \(2\)
Explanation: De rang-nulheidstelling geeft \(3=1+\text{nuliteit}\).
Question 10Not answered
Een lineaire afbeelding \(T:\mathbb{R}^n\to\mathbb{R}^n\) heeft kernel \(\{0\}\). Wat volgt daaruit?
Correct answer: B. \(T\) is surjectief
Explanation: Voor gelijke eindige dimensies impliceert injectief surjectief.
