Linear Algebra Quiz: Rank Nullity Theorem
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Rank Nullity TheoremQuestion 1 of 16

Consider the linear transformation T:R8→R5T: \mathbb{R}^8 \to \mathbb{R}^5 defined by T(x)=AxT(\mathbf{x}) = A\mathbf{x} where AA is a 5×85 \times 8 matrix. If the dimension of the range of TT is 3 and v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 are linearly independent vectors in the null space of TT, what is the minimum number of additional vectors needed to form a basis for the null space?

Exactly 2 additional vectors are needed
Either 1 or 2 additional vectors are needed
Either 2 or 3 additional vectors are needed
Exactly 3 additional vectors are needed
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Linear Algebra Quiz

Linear Algebra Quiz: Rank Nullity Theorem

Practice Rank Nullity Theorem in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rank Nullity Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Consider the linear transformation T:R8→R5T: \mathbb{R}^8 \to \mathbb{R}^5 defined by T(x)=AxT(\mathbf{x}) = A\mathbf{x} where AA is a 5×85 \times 8 matrix. If the dimension of the range of TT is 3 and v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 are linearly independent vectors in the null space of TT, what is the minimum number of additional vectors needed to form a basis for the null space?

  1. Exactly 2 additional vectors are needed (correct answer)
  2. Either 1 or 2 additional vectors are needed
  3. Either 2 or 3 additional vectors are needed
  4. Exactly 3 additional vectors are needed
Explanation: Since dim⁡(range(T))=3\dim(\text{range}(T)) = 3, we have rank(A)=3\text{rank}(A) = 3. By the rank-nullity theorem, nullity(A)=8−3=5\text{nullity}(A) = 8 - 3 = 5. Since the null space has dimension 5 and we already have 3 linearly independent vectors v1,v2,v3\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, we need exactly 5−3=25 - 3 = 2 additional vectors to complete a basis for the null space.

Question 2

A linear transformation TT has a two-dimensional kernel and its image is a three-dimensional subspace. If TT is represented by a matrix AA, which of the following could be the size of AA?

  1. 3×23 \times 2
  2. 5×35 \times 3
  3. 4×54 \times 5 (correct answer)
  4. 5×65 \times 6
Explanation: The dimension of the kernel is the nullity, so nullity(A)=2\text{nullity}(A) = 2. The dimension of the image is the rank, so rank(A)=3\text{rank}(A) = 3. The Rank-Nullity Theorem states that the number of columns, nn, is the sum of the rank and nullity. Thus, n=rank(A)+nullity(A)=3+2=5n = \text{rank}(A) + \text{nullity}(A) = 3 + 2 = 5. The rank is also the dimension of the column space, which is a subspace of the codomain. If the matrix is m×nm \times n, the codomain has dimension mm. Therefore, the number of rows mm must be at least as large as the rank, so m≥3m \ge 3. The size of the matrix must be m×5m \times 5 where m≥3m \ge 3. Of the choices, only 4×54 \times 5 satisfies these conditions.

Question 3

Let VV be a vector space of dimension 7 and WW be a vector space of dimension 5. For a linear transformation T:V→WT: V \to W, the dimension of the image of TT is 4. What is the dimension of the kernel of TT?

  1. 1
  2. 3 (correct answer)
  3. 4
  4. 5
Explanation: The Rank-Nullity Theorem states that dim⁡(Image(T))+dim⁡(Kernel(T))=dim⁡(Domain)\dim(\text{Image}(T)) + \dim(\text{Kernel}(T)) = \dim(\text{Domain}). In this problem, the domain is the vector space VV, so dim⁡(Domain)=7\dim(\text{Domain}) = 7. The dimension of the image is the rank, which is given as 4. Plugging these values into the theorem gives 4+dim⁡(Kernel(T))=74 + \dim(\text{Kernel}(T)) = 7. Solving for the dimension of the kernel yields dim⁡(Kernel(T))=3\dim(\text{Kernel}(T)) = 3.

Question 4

Let AA be a 7×97 \times 9 matrix. Which statement about the linear transformation T(x⃗)=Ax⃗T(\vec{x}) = A\vec{x} must be true?

  1. The transformation must be one-to-one.
  2. The transformation must be onto.
  3. The transformation cannot be one-to-one. (correct answer)
  4. The columns of AA must be linearly independent.
Explanation: The transformation maps from R9\mathbb{R}^9 to R7\mathbb{R}^7, so the dimension of the domain is n=9n=9. The rank of the matrix AA (the dimension of the column space) is at most the number of rows, so rank(A)≤7\text{rank}(A) \le 7. By the Rank-Nullity Theorem, rank(A)+nullity(A)=9\text{rank}(A) + \text{nullity}(A) = 9. Since the maximum rank is 7, the minimum nullity is 9−7=29 - 7 = 2. A transformation is one-to-one if and only if its nullity is 0. Since the nullity must be at least 2, the transformation cannot be one-to-one.

Question 5

A linear transformation T:R5→R4T: \mathbb{R}^5 \to \mathbb{R}^4 is represented by a matrix AA. If the column space of AA is a 3-dimensional subspace of R4\mathbb{R}^4, what is the dimension of the kernel of TT?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: The Rank-Nullity Theorem states that for a linear transformation T:V→WT: V \to W, the sum of the dimension of its image (rank) and the dimension of its kernel (nullity) equals the dimension of its domain. Here, the domain is R5\mathbb{R}^5, so its dimension is 5. The dimension of the column space is the rank, which is given as 3. Therefore, rank(T)+nullity(T)=dim⁡(domain)\text{rank}(T) + \text{nullity}(T) = \dim(\text{domain}). Substituting the values, we get 3+nullity(T)=53 + \text{nullity}(T) = 5. Solving for the nullity (the dimension of the kernel) gives nullity(T)=2\text{nullity}(T) = 2.

Question 6

Let AA be a matrix such that the equation Ax⃗=0⃗A\vec{x} = \vec{0} has a solution set that can be described as a line passing through the origin in R5\mathbb{R}^5. What is the rank of matrix AA?

  1. 1
  2. 3
  3. 4 (correct answer)
  4. 5
Explanation: The solution set of Ax⃗=0⃗A\vec{x} = \vec{0} is the null space of AA. A line passing through the origin is a 1-dimensional subspace. This means the nullity of AA is 1. The vectors x⃗\vec{x} are in R5\mathbb{R}^5, so the number of columns in AA is n=5n=5. According to the Rank-Nullity Theorem, rank(A)+nullity(A)=n\text{rank}(A) + \text{nullity}(A) = n. Substituting the known values, we have rank(A)+1=5\text{rank}(A) + 1 = 5. Therefore, the rank of AA is 4.

Question 7

Let AA be a 6×46 \times 4 matrix. If the null space of AA consists only of the zero vector, what can be concluded about the columns of AA?

  1. The columns of AA are linearly dependent.
  2. The columns of AA are linearly independent. (correct answer)
  3. The columns of AA span R6\mathbb{R}^6.
  4. There are 6 linearly independent columns in AA.
Explanation: The matrix AA is 6×46 \times 4, so it has n=4n=4 columns. The null space consisting only of the zero vector, {0⃗}\{\vec{0}\}, means that the nullity of AA is 0. By the Rank-Nullity Theorem, rank(A)+nullity(A)=n\text{rank}(A) + \text{nullity}(A) = n. Substituting the values, we get rank(A)+0=4\text{rank}(A) + 0 = 4, so rank(A)=4\text{rank}(A) = 4. The rank of a matrix is the dimension of its column space, which also equals the maximum number of linearly independent columns. Since the matrix has 4 columns and its rank is 4, all 4 columns must be linearly independent.

Question 8

A 4×74 \times 7 matrix AA has a null space of dimension 3. What is the dimension of the column space of its transpose, ATA^T?

  1. 0
  2. 1
  3. 3
  4. 4 (correct answer)
Explanation: The matrix AA is 4×74 \times 7, so the dimension of its domain is the number of columns, n=7n=7. We are given that its nullity is 3. By the Rank-Nullity Theorem for AA, rank(A)+nullity(A)=n\text{rank}(A) + \text{nullity}(A) = n, so rank(A)+3=7\text{rank}(A) + 3 = 7. This gives rank(A)=4\text{rank}(A) = 4. A fundamental theorem of linear algebra states that the rank of a matrix is equal to the rank of its transpose, i.e., rank(A)=rank(AT)\text{rank}(A) = \text{rank}(A^T). The dimension of the column space of ATA^T is, by definition, the rank of ATA^T. Therefore, the dimension of the column space of ATA^T is 4.

Question 9

Consider a linear transformation T:R10→R10T: \mathbb{R}^{10} \to \mathbb{R}^{10} such that T2=TT^2 = T (TT is idempotent). If the null space of TT has dimension 6, what is the relationship between the range and null space of TT?

  1. range(T)∩null(T)={0}\text{range}(T) \cap \text{null}(T) = \{\mathbf{0}\} and dim⁡(range(T))=4\dim(\text{range}(T)) = 4 (correct answer)
  2. range(T)=null(T)\text{range}(T) = \text{null}(T) and both have dimension 5
  3. range(T)⊆null(T)\text{range}(T) \subseteq \text{null}(T) and dim⁡(range(T))=6\dim(\text{range}(T)) = 6
  4. null(T)⊆range(T)\text{null}(T) \subseteq \text{range}(T) and dim⁡(null(T))=4\dim(\text{null}(T)) = 4
Explanation: Since dim⁡(null(T))=6\dim(\text{null}(T)) = 6, by the rank-nullity theorem, rank(T)=10−6=4\text{rank}(T) = 10 - 6 = 4, so dim⁡(range(T))=4\dim(\text{range}(T)) = 4. For an idempotent transformation T2=TT^2 = T, if v∈range(T)\mathbf{v} \in \text{range}(T), then v=T(u)\mathbf{v} = T(\mathbf{u}) for some u\mathbf{u}, and T(v)=T(T(u))=T2(u)=T(u)=vT(\mathbf{v}) = T(T(\mathbf{u})) = T^2(\mathbf{u}) = T(\mathbf{u}) = \mathbf{v}. So every vector in the range is an eigenvector with eigenvalue 1. If v∈range(T)∩null(T)\mathbf{v} \in \text{range}(T) \cap \text{null}(T), then T(v)=vT(\mathbf{v}) = \mathbf{v} and T(v)=0T(\mathbf{v}) = \mathbf{0}, so v=0\mathbf{v} = \mathbf{0}. Also, R10=range(T)⊕null(T)\mathbb{R}^{10} = \text{range}(T) \oplus \text{null}(T).

Question 10

Let AA be a 6×96 \times 9 matrix and BB be a 9×49 \times 4 matrix. If rank(A)=4\text{rank}(A) = 4, rank(B)=3\text{rank}(B) = 3, and rank(AB)=2\text{rank}(AB) = 2, what can be concluded about the intersection of the null space of AA and the range of BB?

  1. dim⁡(null(A)∩range(B))=0\dim(\text{null}(A) \cap \text{range}(B)) = 0
  2. dim⁡(null(A)∩range(B))=3\dim(\text{null}(A) \cap \text{range}(B)) = 3
  3. dim⁡(null(A)∩range(B))=2\dim(\text{null}(A) \cap \text{range}(B)) = 2
  4. dim⁡(null(A)∩range(B))=1\dim(\text{null}(A) \cap \text{range}(B)) = 1 (correct answer)
Explanation: This problem tests your understanding of how matrix operations affect subspaces and their dimensions. When dealing with products of matrices and their ranks, you need to connect the rank-nullity theorem with properties of matrix multiplication. The key insight is using the fundamental relationship: dim⁡(range(B))=dim⁡(range(AB))+dim⁡(null(A)∩range(B))\dim(\text{range}(B)) = \dim(\text{range}(AB)) + \dim(\text{null}(A) \cap \text{range}(B)). This formula tells us how the range of BB gets "lost" when multiplied by AA on the left. Since rank(B)=3\text{rank}(B) = 3, we have dim⁡(range(B))=3\dim(\text{range}(B)) = 3. Since rank(AB)=2\text{rank}(AB) = 2, we have dim⁡(range(AB))=2\dim(\text{range}(AB)) = 2. Substituting into our formula: 3=2+dim⁡(null(A)∩range(B))3 = 2 + \dim(\text{null}(A) \cap \text{range}(B)), which gives us dim⁡(null(A)∩range(B))=1\dim(\text{null}(A) \cap \text{range}(B)) = 1. Looking at the wrong answers: Choice A suggests the intersection is trivial, but this would mean AA doesn't "kill" any vectors from the range of BB, making rank(AB)=rank(B)=3\text{rank}(AB) = \text{rank}(B) = 3. Choice B claims the entire range of BB lies in the null space of AA, which would make AB=0AB = 0, so rank(AB)=0\text{rank}(AB) = 0. Choice C would require rank(AB)=3−2=1\text{rank}(AB) = 3 - 2 = 1, not 2. Remember: when you see questions about ranks of matrix products, think about how vectors in the range of the second matrix might get mapped to zero by the first matrix. The dimension formula for intersections of subspaces is your key tool.

Question 11

Let PP and QQ be n×nn \times n matrices such that PQ=0PQ = 0 but neither PP nor QQ is the zero matrix. If rank(P)=r\text{rank}(P) = r and rank(Q)=s\text{rank}(Q) = s, which constraint must be satisfied?

  1. r+s≤nr + s \leq n and at least one of rr or ss equals nn
  2. r+s≤nr + s \leq n and both r<nr < n and s<ns < n (correct answer)
  3. r+s=nr + s = n and exactly one of rr or ss equals nn
  4. rs≤nrs \leq n and both r≥1r \geq 1 and s≥1s \geq 1
Explanation: Since PQ=0PQ = 0, the range of QQ is contained in the null space of PP. By the rank-nullity theorem, nullity(P)=n−r\text{nullity}(P) = n - r. Therefore, s=rank(Q)≤nullity(P)=n−rs = \text{rank}(Q) \leq \text{nullity}(P) = n - r, which gives r+s≤nr + s \leq n. Since neither matrix is zero, both r≥1r \geq 1 and s≥1s \geq 1. Also, since PQ=0PQ = 0 with both matrices non-zero, neither can be invertible, so r<nr < n and s<ns < n.

Question 12

Consider two 4×64 \times 6 matrices EE and FF such that rank(E+F)=3\text{rank}(E + F) = 3, rank(E)=2\text{rank}(E) = 2, and rank(F)=2\text{rank}(F) = 2. Using the rank-nullity theorem, what can be determined about the dimension of the intersection of the row spaces of EE and FF?

  1. dim⁡(row(E)∩row(F))=0\dim(\text{row}(E) \cap \text{row}(F)) = 0
  2. dim⁡(row(E)∩row(F))=2\dim(\text{row}(E) \cap \text{row}(F)) = 2
  3. dim⁡(row(E)∩row(F))=1\dim(\text{row}(E) \cap \text{row}(F)) = 1 (correct answer)
  4. dim⁡(row(E)∩row(F))\dim(\text{row}(E) \cap \text{row}(F)) cannot be determined
Explanation: When you encounter questions about matrix ranks and row spaces, think about how the rank-nullity theorem and dimension formulas for vector space intersections work together. The key insight is using the formula: dim⁡(U+V)=dim⁡(U)+dim⁡(V)−dim⁡(U∩V)\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V) for vector spaces. Here, the row spaces of EE and FF are subspaces of R6\mathbb{R}^6. Since rank(E)=2\text{rank}(E) = 2, we have dim⁡(row(E))=2\dim(\text{row}(E)) = 2. Similarly, dim⁡(row(F))=2\dim(\text{row}(F)) = 2. The crucial observation is that rank(E+F)=dim⁡(row(E+F))=dim⁡(row(E)+row(F))=3\text{rank}(E + F) = \dim(\text{row}(E + F)) = \dim(\text{row}(E) + \text{row}(F)) = 3, since the row space of a sum is contained in the sum of the row spaces, and this containment is often an equality. Applying the dimension formula: 3=2+2−dim⁡(row(E)∩row(F))3 = 2 + 2 - \dim(\text{row}(E) \cap \text{row}(F)), so dim⁡(row(E)∩row(F))=1\dim(\text{row}(E) \cap \text{row}(F)) = 1. Option A suggests no intersection, which would give rank(E+F)=4\text{rank}(E + F) = 4, contradicting the given information. Option B claims the intersection has dimension 2, which would mean the row spaces are identical, giving rank(E+F)=2\text{rank}(E + F) = 2, again contradicting our data. Option D incorrectly suggests indeterminacy when we have sufficient information to calculate exactly. Study tip: Remember that for matrix problems involving ranks and intersections, the dimension formula dim⁡(U+V)=dim⁡(U)+dim⁡(V)−dim⁡(U∩V)\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V) is your primary tool. Practice identifying when the row space of a sum equals the sum of row spaces.

Question 13

Which of the following describes an impossible scenario for a 5×85 \times 8 matrix AA?

  1. The rank of AA is 5 and the nullity of AA is 3.
  2. The rank of AA is 4 and the nullity of AA is 4.
  3. The rank of AA is 2 and the nullity of AA is 6.
  4. The rank of AA is 6 and the nullity of AA is 2. (correct answer)
Explanation: For a 5×85 \times 8 matrix, the number of columns is n=8n=8 and the number of rows is m=5m=5. The Rank-Nullity Theorem requires that rank(A)+nullity(A)=8\text{rank}(A) + \text{nullity}(A) = 8. Additionally, the rank of a matrix cannot exceed the number of its rows or columns. In this case, rank(A)≤min⁡(5,8)=5\text{rank}(A) \le \min(5, 8) = 5. Choice D states that the rank is 6, which violates the condition that the rank cannot be greater than the number of rows (5). All other options satisfy both conditions: A) 5+3=85+3=8 and 5≤55 \le 5; B) 4+4=84+4=8 and 4≤54 \le 5; C) 2+6=82+6=8 and 2≤52 \le 5.

Question 14

A linear transformation S:R4→R6S: \mathbb{R}^4 \to \mathbb{R}^6 is defined by a matrix BB. Which statement regarding the transformation SS must be true?

  1. The transformation cannot be onto. (correct answer)
  2. The transformation must be one-to-one.
  3. The nullity of BB must be exactly 2.
  4. The rank of BB must be 6.
Explanation: The transformation maps from R4\mathbb{R}^4 to R6\mathbb{R}^6. The dimension of the domain is n=4n=4. The rank of the matrix BB is the dimension of its column space, which is a subspace of the codomain R6\mathbb{R}^6. The rank is also limited by the number of columns, so rank(B)≤4\text{rank}(B) \le 4. For the transformation to be onto, its image (column space) must be all of R6\mathbb{R}^6, which would require the rank to be 6. Since the rank is at most 4, the transformation cannot be onto.

Question 15

Consider the linear transformation T:R3→R3T: \mathbb{R}^3 \to \mathbb{R}^3 that orthogonally projects every vector onto the xy-plane. What is the dimension of the null space of the matrix representing TT?

  1. 0
  2. 1 (correct answer)
  3. 2
  4. 3
Explanation: The transformation projects vectors from R3\mathbb{R}^3 onto the xy-plane. The image of this transformation is the entire xy-plane, which is a 2-dimensional subspace of R3\mathbb{R}^3. Therefore, the rank of the transformation is 2. The domain of the transformation is R3\mathbb{R}^3, so its dimension is n=3n=3. By the Rank-Nullity Theorem, rank(T)+nullity(T)=n\text{rank}(T) + \text{nullity}(T) = n. So, 2+nullity(T)=32 + \text{nullity}(T) = 3. This implies the nullity, which is the dimension of the null space, is 1. Geometrically, the set of vectors that project to the zero vector is the z-axis, which is a 1-dimensional line.

Question 16

The homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0} has a general solution that depends on four free parameters. If AA is a matrix with 10 columns, what is the rank of AA?

  1. 4
  2. 6 (correct answer)
  3. 10
  4. 14
Explanation: The number of free parameters in the solution to a homogeneous system corresponds to the dimension of the null space (the nullity). Therefore, nullity(A)=4\text{nullity}(A) = 4. The matrix AA has 10 columns, which means the dimension of the domain is n=10n=10. Applying the Rank-Nullity Theorem, rank(A)+nullity(A)=n\text{rank}(A) + \text{nullity}(A) = n, we get rank(A)+4=10\text{rank}(A) + 4 = 10. Solving for the rank gives rank(A)=6\text{rank}(A) = 6.