Rank-Nullity Theorem - Linear Algebra

1. A linear transformation $T$ has a two-dimensional kernel and its image is a three-dimensional subspace. If $T$ is represented by a matrix $A$, which of the following could be the size of $A$?
   Answer: $4 \times 5$ The dimension of the kernel is the nullity, so $\text{nullity}(A) = 2$. The dimension of the image is the rank, so $\text{rank}(A) = 3$. The Rank-Nullity Theorem states that the number of columns, $n$, is the sum of the rank and nullity. Thus, $n = \text{rank}(A) + \text{nullity}(A) = 3 + 2 = 5$. The rank is also the dimension of the column space, which is a subspace of the codomain. If the matrix is $m \times n$, the codomain has dimension $m$. Therefore, the number of rows $m$ must be at least as large as the rank, so $m \ge 3$. The size of the matrix must be $m \times 5$ where $m \ge 3$. Of the choices, only $4 \times 5$ satisfies these conditions.
2. Consider the linear transformation $T: \mathbb{R}^8 \to \mathbb{R}^5$ defined by $T(\mathbf{x}) = A\mathbf{x}$ where $A$ is a $5 \times 8$ matrix. If the dimension of the range of $T$ is 3 and $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent vectors in the null space of $T$, what is the minimum number of additional vectors needed to form a basis for the null space?
   Answer: Exactly 2 additional vectors are needed Since $\dim(\text{range}(T)) = 3$, we have $\text{rank}(A) = 3$. By the rank-nullity theorem, $\text{nullity}(A) = 8 - 3 = 5$. Since the null space has dimension 5 and we already have 3 linearly independent vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, we need exactly $5 - 3 = 2$ additional vectors to complete a basis for the null space.
3. Let $P$ and $Q$ be $n \times n$ matrices such that $PQ = 0$ but neither $P$ nor $Q$ is the zero matrix. If $\text{rank}(P) = r$ and $\text{rank}(Q) = s$, which constraint must be satisfied?
   Answer: $r + s \leq n$ and both $r < n$ and $s < n$ Since $PQ = 0$, the range of $Q$ is contained in the null space of $P$. By the rank-nullity theorem, $\text{nullity}(P) = n - r$. Therefore, $s = \text{rank}(Q) \leq \text{nullity}(P) = n - r$, which gives $r + s \leq n$. Since neither matrix is zero, both $r \geq 1$ and $s \geq 1$. Also, since $PQ = 0$ with both matrices non-zero, neither can be invertible, so $r < n$ and $s < n$.
4. Let $A$ be a $6 \times 9$ matrix and $B$ be a $9 \times 4$ matrix. If $\text{rank}(A) = 4$, $\text{rank}(B) = 3$, and $\text{rank}(AB) = 2$, what can be concluded about the intersection of the null space of $A$ and the range of $B$?
   Answer: $\dim(\text{null}(A) \cap \text{range}(B)) = 1$ 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: $\dim(\text{range}(B)) = \dim(\text{range}(AB)) + \dim(\text{null}(A) \cap \text{range}(B))$. This formula tells us how the range of $B$ gets "lost" when multiplied by $A$ on the left. Since $\text{rank}(B) = 3$, we have $\dim(\text{range}(B)) = 3$. Since $\text{rank}(AB) = 2$, we have $\dim(\text{range}(AB)) = 2$. Substituting into our formula: $3 = 2 + \dim(\text{null}(A) \cap \text{range}(B))$, which gives us $\dim(\text{null}(A) \cap \text{range}(B)) = 1$. Looking at the wrong answers: Choice A suggests the intersection is trivial, but this would mean $A$ doesn't "kill" any vectors from the range of $B$, making $\text{rank}(AB) = \text{rank}(B) = 3$. Choice B claims the entire range of $B$ lies in the null space of $A$, which would make $AB = 0$, so $\text{rank}(AB) = 0$. Choice C would require $\text{rank}(AB) = 3 - 2 = 1$, 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.
5. Let $A$ be a $7 \times 9$ matrix. Which statement about the linear transformation $T(\vec{x}) = A\vec{x}$ must be true?
   Answer: The transformation cannot be one-to-one. The transformation maps from $\mathbb{R}^9$ to $\mathbb{R}^7$, so the dimension of the domain is $n=9$. The rank of the matrix $A$ (the dimension of the column space) is at most the number of rows, so $\text{rank}(A) \le 7$. By the Rank-Nullity Theorem, $\text{rank}(A) + \text{nullity}(A) = 9$. Since the maximum rank is 7, the minimum nullity is $9 - 7 = 2$. 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.
6. Let $A$ be a $6 \times 4$ matrix. If the null space of $A$ consists only of the zero vector, what can be concluded about the columns of $A$?
   Answer: The columns of $A$ are linearly independent. The matrix $A$ is $6 \times 4$, so it has $n=4$ columns. The null space consisting only of the zero vector, $\{\vec{0}\}$, means that the nullity of $A$ is 0. By the Rank-Nullity Theorem, $\text{rank}(A) + \text{nullity}(A) = n$. Substituting the values, we get $\text{rank}(A) + 0 = 4$, so $\text{rank}(A) = 4$. 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.
7. Let $A$ be a matrix such that the equation $A\vec{x} = \vec{0}$ has a solution set that can be described as a line passing through the origin in $\mathbb{R}^5$. What is the rank of matrix $A$?
   Answer: 4 The solution set of $A\vec{x} = \vec{0}$ is the null space of $A$. A line passing through the origin is a 1-dimensional subspace. This means the nullity of $A$ is 1. The vectors $\vec{x}$ are in $\mathbb{R}^5$, so the number of columns in $A$ is $n=5$. According to the Rank-Nullity Theorem, $\text{rank}(A) + \text{nullity}(A) = n$. Substituting the known values, we have $\text{rank}(A) + 1 = 5$. Therefore, the rank of $A$ is 4.
8. Which of the following describes an impossible scenario for a $5 \times 8$ matrix $A$?
   Answer: The rank of $A$ is 6 and the nullity of $A$ is 2. For a $5 \times 8$ matrix, the number of columns is $n=8$ and the number of rows is $m=5$. The Rank-Nullity Theorem requires that $\text{rank}(A) + \text{nullity}(A) = 8$. Additionally, the rank of a matrix cannot exceed the number of its rows or columns. In this case, $\text{rank}(A) \le \min(5, 8) = 5$. 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) $5+3=8$ and $5 \le 5$; B) $4+4=8$ and $4 \le 5$; C) $2+6=8$ and $2 \le 5$.
9. A linear transformation $S: \mathbb{R}^4 \to \mathbb{R}^6$ is defined by a matrix $B$. Which statement regarding the transformation $S$ must be true?
   Answer: The transformation cannot be onto. The transformation maps from $\mathbb{R}^4$ to $\mathbb{R}^6$. The dimension of the domain is $n=4$. The rank of the matrix $B$ is the dimension of its column space, which is a subspace of the codomain $\mathbb{R}^6$. The rank is also limited by the number of columns, so $\text{rank}(B) \le 4$. For the transformation to be onto, its image (column space) must be all of $\mathbb{R}^6$, which would require the rank to be 6. Since the rank is at most 4, the transformation cannot be onto.
10. A linear transformation $T: \mathbb{R}^5 \to \mathbb{R}^4$ is represented by a matrix $A$. If the column space of $A$ is a 3-dimensional subspace of $\mathbb{R}^4$, what is the dimension of the kernel of $T$?
    Answer: 2 The Rank-Nullity Theorem states that for a linear transformation $T: V \to W$, 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 $\mathbb{R}^5$, so its dimension is 5. The dimension of the column space is the rank, which is given as 3. Therefore, $\text{rank}(T) + \text{nullity}(T) = \dim(\text{domain})$. Substituting the values, we get $3 + \text{nullity}(T) = 5$. Solving for the nullity (the dimension of the kernel) gives $\text{nullity}(T) = 2$.
11. Let $V$ be a vector space of dimension 7 and $W$ be a vector space of dimension 5. For a linear transformation $T: V \to W$, the dimension of the image of $T$ is 4. What is the dimension of the kernel of $T$?
    Answer: 3 The Rank-Nullity Theorem states that $\dim(\text{Image}(T)) + \dim(\text{Kernel}(T)) = \dim(\text{Domain})$. In this problem, the domain is the vector space $V$, so $\dim(\text{Domain}) = 7$. The dimension of the image is the rank, which is given as 4. Plugging these values into the theorem gives $4 + \dim(\text{Kernel}(T)) = 7$. Solving for the dimension of the kernel yields $\dim(\text{Kernel}(T)) = 3$.
12. Consider the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ that orthogonally projects every vector onto the xy-plane. What is the dimension of the null space of the matrix representing $T$?
    Answer: 1 The transformation projects vectors from $\mathbb{R}^3$ onto the xy-plane. The image of this transformation is the entire xy-plane, which is a 2-dimensional subspace of $\mathbb{R}^3$. Therefore, the rank of the transformation is 2. The domain of the transformation is $\mathbb{R}^3$, so its dimension is $n=3$. By the Rank-Nullity Theorem, $\text{rank}(T) + \text{nullity}(T) = n$. So, $2 + \text{nullity}(T) = 3$. 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.
13. Consider two $4 \times 6$ matrices $E$ and $F$ such that $\text{rank}(E + F) = 3$, $\text{rank}(E) = 2$, and $\text{rank}(F) = 2$. Using the rank-nullity theorem, what can be determined about the dimension of the intersection of the row spaces of $E$ and $F$?
    Answer: $\dim(\text{row}(E) \cap \text{row}(F)) = 1$ 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: $\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V)$ for vector spaces. Here, the row spaces of $E$ and $F$ are subspaces of $\mathbb{R}^6$. Since $\text{rank}(E) = 2$, we have $\dim(\text{row}(E)) = 2$. Similarly, $\dim(\text{row}(F)) = 2$. The crucial observation is that $\text{rank}(E + F) = \dim(\text{row}(E + F)) = \dim(\text{row}(E) + \text{row}(F)) = 3$, 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: $3 = 2 + 2 - \dim(\text{row}(E) \cap \text{row}(F))$, so $\dim(\text{row}(E) \cap \text{row}(F)) = 1$. Option A suggests no intersection, which would give $\text{rank}(E + F) = 4$, contradicting the given information. Option B claims the intersection has dimension 2, which would mean the row spaces are identical, giving $\text{rank}(E + F) = 2$, 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 $\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V)$ is your primary tool. Practice identifying when the row space of a sum equals the sum of row spaces.
14. A $4 \times 7$ matrix $A$ has a null space of dimension 3. What is the dimension of the column space of its transpose, $A^T$?
    Answer: 4 The matrix $A$ is $4 \times 7$, so the dimension of its domain is the number of columns, $n=7$. We are given that its nullity is 3. By the Rank-Nullity Theorem for $A$, $\text{rank}(A) + \text{nullity}(A) = n$, so $\text{rank}(A) + 3 = 7$. This gives $\text{rank}(A) = 4$. A fundamental theorem of linear algebra states that the rank of a matrix is equal to the rank of its transpose, i.e., $\text{rank}(A) = \text{rank}(A^T)$. The dimension of the column space of $A^T$ is, by definition, the rank of $A^T$. Therefore, the dimension of the column space of $A^T$ is 4.
15. The homogeneous system $A\vec{x} = \vec{0}$ has a general solution that depends on four free parameters. If $A$ is a matrix with 10 columns, what is the rank of $A$?
    Answer: 6 The number of free parameters in the solution to a homogeneous system corresponds to the dimension of the null space (the nullity). Therefore, $\text{nullity}(A) = 4$. The matrix $A$ has 10 columns, which means the dimension of the domain is $n=10$. Applying the Rank-Nullity Theorem, $\text{rank}(A) + \text{nullity}(A) = n$, we get $\text{rank}(A) + 4 = 10$. Solving for the rank gives $\text{rank}(A) = 6$.
16. Consider a linear transformation $T: \mathbb{R}^{10} \to \mathbb{R}^{10}$ such that $T^2 = T$ ($T$ is idempotent). If the null space of $T$ has dimension 6, what is the relationship between the range and null space of $T$?
    Answer: $\text{range}(T) \cap \text{null}(T) = \{\mathbf{0}\}$ and $\dim(\text{range}(T)) = 4$ Since $\dim(\text{null}(T)) = 6$, by the rank-nullity theorem, $\text{rank}(T) = 10 - 6 = 4$, so $\dim(\text{range}(T)) = 4$. For an idempotent transformation $T^2 = T$, if $\mathbf{v} \in \text{range}(T)$, then $\mathbf{v} = T(\mathbf{u})$ for some $\mathbf{u}$, and $T(\mathbf{v}) = T(T(\mathbf{u})) = T^2(\mathbf{u}) = T(\mathbf{u}) = \mathbf{v}$. So every vector in the range is an eigenvector with eigenvalue 1. If $\mathbf{v} \in \text{range}(T) \cap \text{null}(T)$, then $T(\mathbf{v}) = \mathbf{v}$ and $T(\mathbf{v}) = \mathbf{0}$, so $\mathbf{v} = \mathbf{0}$. Also, $\mathbb{R}^{10} = \text{range}(T) \oplus \text{null}(T)$.

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