Column Space & Null Space - Linear Algebra

Consider a $5 \times 4$ matrix $J$ where every vector in $\mathbb{R}^4$ can be written uniquely as $\mathbf{v} + \mathbf{w}$ where $\mathbf{v} \in$ null( $J$ ) and $J\mathbf{w} \in$ column( $J$ ). What must be true about the rank of $J$ ?
Answer: The rank must be 4, because this decomposition property requires $J$ to have full column rank The given condition describes the fundamental theorem of linear algebra: every vector in the domain can be uniquely decomposed into components from the null space and row space. For this decomposition to be unique for every vector in $\mathbb{R}^4$ , we need null( $J$ ) and row( $J$ ) to be complementary subspaces that together span $\mathbb{R}^4$ . This happens if and only if $J$ has full column rank, i.e., rank( $J$ ) = 4. If rank < 4, then null( $J$ ) has dimension > 0 and row( $J$ ) has dimension < 4, so they can't span all of $\mathbb{R}^4$ . Choice B incorrectly assumes equal dimensions are required, Choices C and D don't recognize the full rank requirement.
Let $A = \begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ -1 & 2 & 4 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}$ . Which statement accurately describes the vector $\vec{v}$ ?
Answer: The vector $\vec{v}$ is in neither the null space nor the column space of $A$ . To check if $\vec{v}$ is in the null space, we compute $A\vec{v}$ . $A\vec{v} = \begin{pmatrix} 1(2) + 0(-1) - 2(4) \\ 0(2) + 1(-1) + 1(4) \\ -1(2) + 2(-1) + 4(4) \end{pmatrix} = \begin{pmatrix} -6 \\ 3 \\ 12 \end{pmatrix}$ . Since $A\vec{v} \neq \vec{0}$ , $\vec{v}$ is not in the null space. To check if $\vec{v}$ is in the column space, we check if $A\vec{x} = \vec{v}$ is consistent by row reducing the augmented matrix $[A|\vec{v}]$ : $\begin{pmatrix} 1 & 0 & -2 & 2 \\ 0 & 1 & 1 & -1 \\ -1 & 2 & 4 & 4 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -2 & 2 \\ 0 & 1 & 1 & -1 \\ 0 & 2 & 2 & 6 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -2 & 2 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 8 \end{pmatrix}$ The last row corresponds to the equation $0 = 8$ , which is a contradiction. The system is inconsistent, so $\vec{v}$ is not in the column space.
A $3 \times 4$ matrix $D$ has the property that its null space contains the vector $\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix}$ . If the column space of $D$ is 2-dimensional, what can be concluded about the system $D\mathbf{x} = \mathbf{0}$ ?
Answer: It has exactly two free variables, and the solution space includes but is not limited to scalar multiples of the given vector Since $D$ is $3 \times 4$ with a 2-dimensional column space, rank( $D$ ) = 2. By the rank-nullity theorem, nullity( $D$ ) = $4 - 2 = 2$ , meaning there are 2 free variables. The null space is 2-dimensional, so while the given vector is in the null space, it doesn't span the entire null space (which would require 2 linearly independent vectors). Choice A is wrong about the number of free variables, Choice C miscalculates the rank, Choice D makes an unjustified claim about component relationships, Choice B correctly identifies 2 free variables and that the solution space is larger than just scalar multiples of one vector.
Let $A$ be a $3 \times 4$ matrix. Which of the following combinations for the dimensions of the column space and null space of $A$ is impossible?
Answer: $\text{dim}(\text{Col}(A)) = 4$ and $\text{dim}(\text{Nul}(A)) = 0$ . For an $m \times n$ matrix, the dimension of the column space (rank) cannot exceed the number of rows or columns. For a $3 \times 4$ matrix, the rank must be less than or equal to $\min(3, 4) = 3$ . Therefore, it is impossible for the dimension of the column space to be 4. Additionally, the Rank-Nullity Theorem states that $\text{rank}(A) + \text{nullity}(A) = n$ . For choices A, B, and C, the sum is $3+1=4$ , $2+2=4$ , and $1+3=4$ , which is consistent with $n=4$ . For choice D, the sum is $4+0=4$ , but the rank of 4 is impossible.
Let $F$ be a $3 \times 5$ matrix such that the system $F\mathbf{x} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ has infinitely many solutions. What is the maximum possible dimension of the column space of $F$ ?
Answer: 3, because $F$ maps from $\mathbb{R}^5$ to $\mathbb{R}^3$ and the given vector is in the column space If the system $F\mathbf{x} = \mathbf{b}$ has infinitely many solutions, then it has at least one solution (consistency) and the null space of $F$ is non-trivial (infinitely many solutions). However, this doesn't prevent $F$ from having full row rank. A $3 \times 5$ matrix can have rank at most 3, and having infinitely many solutions just means nullity ≥ 1, so rank ≤ 4. Since the maximum rank is 3 anyway, the column space can indeed be 3-dimensional. The given vector being in the column space doesn't restrict the dimension. Choice B incorrectly assumes infinitely many solutions severely limits rank, Choices C and D underestimate the maximum possible rank.
Let $K$ be a $4 \times 4$ matrix with the property that $K^2 = K$ . If the null space of $K$ has dimension 2, what is the dimension of the intersection of the column space and null space of $K$ ?
Answer: 0, because idempotent matrices with 2-dimensional null spaces have complementary column and null spaces When you encounter idempotent matrices (where $K^2 = K$ ), think about how this special property creates a fundamental relationship between the column space and null space that differs from general matrices. For any idempotent matrix, every vector $\mathbf{v}$ in $\mathbb{R}^4$ can be uniquely decomposed as $\mathbf{v} = K\mathbf{v} + (I-K)\mathbf{v}$ , where $K\mathbf{v}$ lies in the column space and $(I-K)\mathbf{v}$ lies in the null space. This decomposition proves that $\mathbb{R}^4 = \text{Col}(K) \oplus \text{Null}(K)$ — the column space and null space are complementary subspaces that intersect only at the zero vector. Since $\dim(\text{Null}(K)) = 2$ and these spaces are complementary in $\mathbb{R}^4$ , we have $\dim(\text{Col}(K)) = 4 - 2 = 2$ . Because they're complementary, $\dim(\text{Col}(K) \cap \text{Null}(K)) = 0$ . Option A is wrong because while it reaches the correct conclusion, the reasoning is false — general matrices can have overlapping column and null spaces. Option B incorrectly assumes the idempotent property forces exactly one shared vector beyond zero. Option D makes the major error of claiming the null space is contained in the column space, which would contradict their complementary relationship. Study tip: Remember that idempotent matrices act like "projections" — they split the space into two complementary pieces (column space and null space) that only meet at zero. This complementary relationship is the key insight for solving these problems.
The set of solutions to a homogeneous system $A\vec{x} = \vec{0}$ is given by the parametric vector form $\vec{x} = s \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 5 \\ 0 \\ -3 \\ 1 \end{pmatrix}$ for $s, t \in \mathbb{R}$ . Which of the following matrices could be the reduced row echelon form of $A$ ?
Answer: $\begin{pmatrix} 1 & 2 & 0 & -5 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ The solution shows that $x_2$ and $x_4$ are free variables, let $x_2=s$ and $x_4=t$ . The pivot variables are $x_1$ and $x_3$ . From the solution vector, we can write the equations for the pivot variables in terms of the free variables: $x_1 = -2s + 5t$ and $x_3 = 0s - 3t$ . Rearranging these to match the RREF system gives $x_1 + 2s - 5t = 0 \implies x_1 + 2x_2 - 5x_4 = 0$ , and $x_3 + 3t = 0 \implies x_3 + 3x_4 = 0$ . These two equations correspond to the first two rows of the RREF matrix. The first row must be $\begin{pmatrix} 1 & 2 & 0 & -5 \end{pmatrix}$ and the second row must be $\begin{pmatrix} 0 & 0 & 1 & 3 \end{pmatrix}$ .
The null space of a matrix $A$ is spanned by the single vector $\vec{v} = \begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \end{pmatrix}$ . Which of the following statements about matrix $A$ must be true?
Answer: The dimension of the column space of $A$ is 3. The null space is spanned by a single non-zero vector, so its dimension, $\text{dim}(\text{Nul}(A))$ , is 1. The vectors in the null space are in $\mathbb{R}^4$ , which means the matrix $A$ must have $n=4$ columns for the product $A\vec{x}$ to be defined. By the Rank-Nullity Theorem, $\text{rank}(A) + \text{dim}(\text{Nul}(A)) = n$ . Substituting the known values, we get $\text{rank}(A) + 1 = 4$ , which implies $\text{rank}(A) = 3$ . The rank of a matrix is the dimension of its column space. Therefore, the dimension of the column space of $A$ must be 3.
A $5 \times 9$ matrix $A$ has a column space with dimension 4. What is the dimension of the null space of $A$ ?
Answer: 5 The Rank-Nullity Theorem states that for an $m \times n$ matrix $A$ , the sum of the dimension of the column space (rank) and the dimension of the null space (nullity) is equal to the number of columns, $n$ . Here, $A$ is a $5 \times 9$ matrix, so $n=9$ . We are given that $\text{dim}(\text{Col}(A)) = \text{rank}(A) = 4$ . Therefore, $4 + \text{dim}(\text{Nul}(A)) = 9$ . Solving for the dimension of the null space gives $\text{dim}(\text{Nul}(A)) = 9 - 4 = 5$ .
Consider the matrix $A = \begin{pmatrix} 1 & 0 & 3 & 2 \\ -2 & 1 & -6 & -3 \\ 1 & 1 & 3 & 3 \end{pmatrix}$ . Which of the following sets of vectors forms a basis for the null space of $A$ ?
Answer: $\left\{ \begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -2 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right\}$ To find a basis for the null space, we solve $A\vec{x} = \vec{0}$ by row reducing the matrix $A$ . $\begin{pmatrix} 1 & 0 & 3 & 2 \\ -2 & 1 & -6 & -3 \\ 1 & 1 & 3 & 3 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 3 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 3 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ The system is $x_1 + 3x_3 + 2x_4 = 0$ and $x_2 + x_4 = 0$ . The variables $x_3$ and $x_4$ are free. Let $x_3 = s$ and $x_4 = t$ . Then $x_1 = -3s - 2t$ and $x_2 = -t$ . The solution in parametric vector form is $\vec{x} = s \begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} -2 \\ -1 \\ 0 \\ 1 \end{pmatrix}$ . The two vectors in this form make up the basis for the null space.
Let $C$ be a $4 \times 5$ matrix with rank 3. If vectors $\mathbf{u}$ and $\mathbf{v}$ are both in the null space of $C$ , and $\mathbf{w} = 2\mathbf{u} - 3\mathbf{v}$ , which of the following must be true about $\mathbf{w}$ ?
Answer: $\mathbf{w}$ is orthogonal to every vector in the column space of $C$ and lies in a 2-dimensional subspace Since $C$ has rank 3 and is $4 \times 5$ , the null space has dimension $5 - 3 = 2$ . The null space is a subspace, so any linear combination of vectors in it (like $\mathbf{w} = 2\mathbf{u} - 3\mathbf{v}$ ) is also in the null space. Vectors in the null space are orthogonal to all row vectors of $C$ , which span the row space. For matrices, the row space and column space have the same dimension, and the null space is orthogonal to the row space. Choice A is wrong (null space and column space are generally disjoint), Choice B is wrong about magnitude, Choice C is correct about null space membership but wrong about spanning (this depends on linear independence), Choice D correctly identifies both properties.
Let $A = \begin{pmatrix} 1 & -2 & 0 & 3 \\ 2 & -4 & 1 & 8 \\ -3 & 6 & 2 & 3 \end{pmatrix}$ . What is the dimension of the null space of $A$ , denoted $\text{dim}(\text{Nul}(A))$ ?
Answer: 1 To find the dimension of the null space, we determine the number of free variables in the solution to $A\vec{x} = \vec{0}$ . We row reduce the matrix $A$ : $\begin{pmatrix} 1 & -2 & 0 & 3 \\ 2 & -4 & 1 & 8 \\ -3 & 6 & 2 & 3 \end{pmatrix} \sim \begin{pmatrix} 1 & -2 & 0 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 2 & 12 \end{pmatrix} \sim \begin{pmatrix} 1 & -2 & 0 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 8 \end{pmatrix}$ This matrix has three pivot columns (columns 1, 3, and 4). The number of columns is 4. The dimension of the null space is the number of non-pivot columns, which is $4 - 3 = 1$ . The second column corresponds to the free variable.
Let $A = \begin{pmatrix} 1 & 3 & 5 & 0 \\ -2 & -5 & -8 & 1 \\ 0 & 1 & 2 & 1 \end{pmatrix}$ . The reduced row echelon form of $A$ is $\begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ . Which of the following sets is a basis for the column space of $A$ ?
Answer: $\left\{ \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ -5 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right\}$ A basis for the column space of a matrix $A$ is formed by the pivot columns of the original matrix $A$ . The reduced row echelon form (RREF) shows that the pivot positions are in columns 1, 2, and 4. Therefore, a basis for the column space of $A$ is the set of the first, second, and fourth columns of the original matrix $A$ .
Matrix $E$ has columns $\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3, \mathbf{c}_4$ where $\mathbf{c}_1 + 2\mathbf{c}_2 - \mathbf{c}_3 = \mathbf{0}$ and $\mathbf{c}_4$ is not in the span of $\{\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3\}$ . If $E$ is a $5 \times 4$ matrix, what is the dimension of the null space of $E$ ?
Answer: 1, because the rank is 3 and we apply the rank-nullity theorem The given relation $\mathbf{c}_1 + 2\mathbf{c}_2 - \mathbf{c}_3 = \mathbf{0}$ shows that $\mathbf{c}_3$ is in the span of $\{\mathbf{c}_1, \mathbf{c}_2\}$ , so these three vectors span at most a 2-dimensional space. Since $\mathbf{c}_4$ is not in span $\{\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3\} =$ span $\{\mathbf{c}_1, \mathbf{c}_2\}$ , we have that $\{\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_4\}$ are linearly independent (assuming $\mathbf{c}_1, \mathbf{c}_2$ are linearly independent). Thus rank( $E$ ) = 3. By rank-nullity theorem: nullity = $4 - 3 = 1$ . Choice A gives the right answer but wrong reasoning, Choice B miscalculates the rank, Choice C confuses the concept, Choice D gives correct reasoning and answer.
For what value of $h$ is the vector $\vec{b} = \begin{pmatrix} 1 \\ 5 \\ h \end{pmatrix}$ in the column space of the matrix $A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 0 & 2 \\ -1 & 2 & -3 \end{pmatrix}$ ?
Answer: $h = 0.5$ The vector $\vec{b}$ is in the column space of $A$ if the system $A\vec{x} = \vec{b}$ is consistent. We analyze the augmented matrix $[A|\vec{b}]$ via row reduction: $\begin{pmatrix} 1 & -1 & 2 & 1 \\ 2 & 0 & 2 & 5 \\ -1 & 2 & -3 & h \end{pmatrix} \xrightarrow{R_2-2R_1, R_3+R_1} \begin{pmatrix} 1 & -1 & 2 & 1 \\ 0 & 2 & -2 & 3 \\ 0 & 1 & -1 & h+1 \end{pmatrix}$ Now, we perform the operation $R_3 - \frac{1}{2}R_2$ : $\begin{pmatrix} 1 & -1 & 2 & 1 \\ 0 & 2 & -2 & 3 \\ 0 & 0 & 0 & (h+1) - \frac{1}{2}(3) \end{pmatrix}$ For the system to be consistent, the last entry in the last row must be zero. So, $(h+1) - 1.5 = 0$ , which simplifies to $h - 0.5 = 0$ . Therefore, $h=0.5$ .
Matrix $G$ is formed by taking the first three columns of a $4 \times 6$ matrix $H$ that has rank 4. If the null space of $G$ is 2-dimensional, what can be concluded about the relationship between the columns of $H$ ?
Answer: At least one of columns 4, 5, or 6 of $H$ must be linearly independent from the first three columns Since $G$ is $4 \times 3$ with a 2-dimensional null space, rank( $G$ ) = $3 - 2 = 1$ . This means the first three columns of $H$ span only a 1-dimensional space. Since $H$ has rank 4, its columns span a 4-dimensional space. Therefore, columns 4, 5, and 6 must contribute additional dimensions beyond what the first three columns provide. At least one of the last three columns must be linearly independent from the first three. Choice A is wrong (would make rank of $H$ at most 1), Choice C is wrong (first three columns have rank 1, not 3), Choice D is incorrect about equal spans.
A linear transformation $T: \mathbb{R}^5 \to \mathbb{R}^3$ is defined by $T(\vec{x}) = A\vec{x}$ . If $T$ is surjective (onto), what is the dimension of the kernel of $T$ ?
Answer: 2 The linear transformation $T$ maps from $\mathbb{R}^5$ to $\mathbb{R}^3$ , so its standard matrix $A$ is $3 \times 5$ . The kernel of $T$ is the null space of $A$ , and the range of $T$ is the column space of $A$ . If $T$ is surjective, its range is the entire codomain, $\mathbb{R}^3$ . This means the dimension of the range (and thus the dimension of the column space, or rank of $A$ ) is 3. By the Rank-Nullity Theorem, $\text{rank}(A) + \text{dim}(\text{Nul}(A)) = n$ . Here, $\text{rank}(A) = 3$ and $n=5$ . So, $3 + \text{dim}(\text{Nul}(A)) = 5$ . The dimension of the null space (kernel) is $5 - 3 = 2$ .
Let $A$ be a $4 \times 3$ matrix with columns $\vec{a}_1, \vec{a}_2, \vec{a}_3$ . Suppose the set $\{\vec{a}_1, \vec{a}_2\}$ is linearly independent and $\vec{a}_3 = 2\vec{a}_1 - \vec{a}_2$ . What is the dimension of the null space of $A$ ?
Answer: 1 The column space of $A$ is spanned by its columns, $\{\vec{a}_1, \vec{a}_2, \vec{a}_3\}$ . Since $\vec{a}_3$ is a linear combination of $\vec{a}_1$ and $\vec{a}_2$ , it does not add to the dimension of the space spanned by $\{\vec{a}_1, \vec{a}_2\}$ . We are given that $\{\vec{a}_1, \vec{a}_2\}$ is a linearly independent set, so it forms a basis for the column space of $A$ . Thus, the dimension of the column space, $\text{rank}(A)$ , is 2. The matrix $A$ is $4 \times 3$ , so it has $n=3$ columns. By the Rank-Nullity Theorem, $\text{rank}(A) + \text{dim}(\text{Nul}(A)) = n$ , which gives $2 + \text{dim}(\text{Nul}(A)) = 3$ . Therefore, the dimension of the null space is $3 - 2 = 1$ .
Let $A$ be a $4 \times 6$ matrix. Which of the following statements correctly describes the column space, $\text{Col}(A)$ , and the null space, $\text{Nul}(A)$ ?
Answer: $\text{Col}(A)$ is a subspace of $\mathbb{R}^4$ and $\text{Nul}(A)$ is a subspace of $\mathbb{R}^6$ . For an $m \times n$ matrix $A$ , the column space is the span of the column vectors. Since $A$ is $4 \times 6$ , its columns are vectors in $\mathbb{R}^4$ . Thus, $\text{Col}(A)$ is a subspace of $\mathbb{R}^4$ . The null space is the set of all vectors $\vec{x}$ such that $A\vec{x} = \vec{0}$ . For the matrix-vector product $A\vec{x}$ to be defined, $\vec{x}$ must have as many entries as $A$ has columns. Since $A$ has 6 columns, $\vec{x}$ must be in $\mathbb{R}^6$ . Thus, $\text{Nul}(A)$ is a subspace of $\mathbb{R}^6$ .

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