Kernel & Range - Linear Algebra

1. Let $V = M_{2 \times 2}$ be the vector space of $2 \times 2$ real matrices. Consider the linear transformation $L: V \to \mathbb{R}$ defined by $L(A) = \text{tr}(A)$, where the trace is the sum of the main diagonal elements. What is the range of $L$?
   Answer: $\mathbb{R}$ When you encounter questions about the range of a linear transformation, you need to determine all possible output values the transformation can produce. The transformation $L(A) = \text{tr}(A)$ takes any $2 \times 2$ matrix and outputs the sum of its diagonal entries. For a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we have $L(A) = a + d$. To find the range, ask: can $L$ produce any real number? Yes! For any real number $r$, you can construct a matrix whose trace equals $r$. For example, the matrix $\begin{pmatrix} r & 0 \\ 0 & 0 \end{pmatrix}$ has trace $r + 0 = r$. Since you can achieve any real number as a trace, the range is all of $\mathbb{R}$. Choice B suggests the range is only $\{0\}$. This would mean every matrix has trace zero, which is clearly false—the identity matrix has trace 2. Choice C claims the range is only non-negative numbers. This ignores that diagonal entries can be negative. The matrix $\begin{pmatrix} -5 & 0 \\ 0 & 0 \end{pmatrix}$ has trace $-5$, showing negative outputs are possible. Choice D suggests the range is $\mathbb{R}^2$, but $L$ maps to real numbers (scalars), not ordered pairs. The codomain is $\mathbb{R}$, so the range must be a subset of $\mathbb{R}$. Study tip: For range problems, construct specific examples to test whether particular outputs are achievable. If you can systematically produce any element in a set, that set is likely your range.
2. The range of a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ is the plane spanned by the vectors $\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$. Which of the following could be the standard matrix for $T$?
   Answer: $\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ -1 & 2 & 1 \end{pmatrix}$ The range of $T$ is its column space. Therefore, each column of the standard matrix must be a linear combination of the spanning vectors $\vec{v}_1$ and $\vec{v}_2$. Let's check the columns of matrix A. The first column is $\vec{v}_1$. The second column is $\vec{v}_2$. The third column is $\vec{v}_1 + \vec{v}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$. Since all three columns are in the span of $\vec{v}_1$ and $\vec{v}_2$, and the first two columns are linearly independent, the column space is precisely the plane spanned by $\vec{v}_1$ and $\vec{v}_2$. This matrix is a valid candidate. Distractor B has a third column not in the span. Distractor C has a range of $\mathbb{R}^3$. Distractor D has a range of dimension 1.
3. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation that orthogonally projects vectors onto the plane defined by $x - 2y + 3z = 0$. Which statement accurately describes the kernel and range of $T$?
   Answer: The kernel of $T$ is the line spanned by the vector $\begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$, and the range of $T$ is the plane $x - 2y + 3z = 0$. The kernel of an orthogonal projection is the set of vectors that are mapped to the zero vector. For a projection onto a plane, these are precisely the vectors orthogonal to the plane. The normal vector to the plane $x - 2y + 3z = 0$ is $\vec{n} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$, so the kernel is the line spanned by this vector. The range of the projection is the set of all possible outputs, which is the plane itself, $x - 2y + 3z = 0$.
4. Let $A$ be an $m \times n$ matrix representing a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$. Which of the following statements is false?
   Answer: If the columns of $A$ are linearly independent, then $\text{ker}(T) = \mathbb{R}^n$. If the columns of matrix $A$ are linearly independent, the equation $A\vec{x} = \vec{0}$ has only the trivial solution, $\vec{x} = \vec{0}$. By definition, the kernel of $T$ is the set of all solutions to this equation. Therefore, if the columns are linearly independent, $\text{ker}(T) = \{\vec{0}\}$. The statement that $\text{ker}(T) = \mathbb{R}^n$ means that every vector in the domain maps to zero, which only happens if $A$ is the zero matrix (whose columns are linearly dependent). Therefore, this statement is false. The other three statements are fundamental truths: A and B are from the definition of kernel and range, and D is the Rank-Nullity Theorem.
5. Let $T: V \to W$ be a linear transformation between vector spaces $V$ and $W$. Which of the following conditions is sufficient to guarantee that $T$ is injective (one-to-one)?
   Answer: The only solution to the equation $T(\vec{x}) = \vec{0}_W$ is the vector $\vec{x} = \vec{0}_V$. A linear transformation $T$ is injective if and only if its kernel (or null space) contains only the zero vector. The statement 'The only solution to the equation $T(\vec{x}) = \vec{0}_W$ is the vector $\vec{x} = \vec{0}_V$' is the definition of $\text{ker}(T) = \{\vec{0}_V\}$. This condition is both necessary and sufficient for injectivity. Choice A is necessary for finite-dimensional spaces but not sufficient. Choices B and D are definitions of a surjective (onto) transformation.
6. Let $T: V \to W$ be a linear transformation and let $\mathcal{B} = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n\}$ be a basis for the vector space $V$. Which of the following sets is guaranteed to be a spanning set for the range of $T$?
   Answer: $\{T(\vec{v}_1), T(\vec{v}_2), \dots, T(\vec{v}_n)\}$ When you encounter questions about the range of a linear transformation, focus on understanding what vectors can be "reached" by applying the transformation. The range of $T$ consists of all possible outputs $T(\vec{v})$ where $\vec{v} \in V$. Since $\mathcal{B} = \{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n\}$ is a basis for $V$, every vector $\vec{v} \in V$ can be written as $\vec{v} = c_1\vec{v}_1 + c_2\vec{v}_2 + \cdots + c_n\vec{v}_n$ for some scalars $c_i$. When we apply $T$ to this vector, linearity gives us: $T(\vec{v}) = T(c_1\vec{v}_1 + \cdots + c_n\vec{v}_n) = c_1T(\vec{v}_1) + \cdots + c_nT(\vec{v}_n)$ This shows that every vector in the range of $T$ is a linear combination of $\{T(\vec{v}_1), T(\vec{v}_2), \dots, T(\vec{v}_n)\}$, making this set a spanning set for the range. Answer D is correct. Answer A is wrong because any basis for $W$ spans all of $W$, but the range of $T$ might be a proper subspace of $W$, so this could contain unnecessary vectors. Answer B fails because removing vectors from $\mathcal{B}$ means we lose the ability to represent all vectors in $V$, so we can't guarantee we'll span the entire range. Answer C is incorrect because any proper subset of $\{T(\vec{v}_1), \dots, T(\vec{v}_n)\}$ might miss some vectors needed to span the range. Study tip: Remember that the image of a basis under a linear transformation always spans the range, even though it might not be linearly independent itself.
7. Let $S: U \to V$ and $T: V \to W$ be linear transformations between finite-dimensional vector spaces. Which of the following statements about the composite transformation $T \circ S: U \to W$ is always true?
   Answer: $\text{ker}(S) \subseteq \text{ker}(T \circ S)$ Let $\vec{u}$ be a vector in the kernel of $S$, so $S(\vec{u}) = \vec{0}_V$. To check if $\vec{u}$ is in the kernel of $T \circ S$, we compute $(T \circ S)(\vec{u}) = T(S(\vec{u}))$. Since $S(\vec{u}) = \vec{0}_V$, this becomes $T(\vec{0}_V)$. Because $T$ is a linear transformation, $T(\vec{0}_V) = \vec{0}_W$. Thus, any vector in $\text{ker}(S)$ is also in $\text{ker}(T \circ S)$, which means $\text{ker}(S) \subseteq \text{ker}(T \circ S)$. The other inclusions are not always true; for example, the range of the composition is a subset of the range of $T$, not the other way around.
8. Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be a linear transformation such that $T((1,0,1,0)) = (1,2,1)$, $T((0,1,0,1)) = (2,1,3)$, and $\ker(T) = \text{span}\{(1,1,-1,-1), (2,0,1,-1)\}$. Which vector is in $\text{range}(T)$?
   Answer: $(5,4,7)$ Since $\text{dim}(\ker(T)) = 2$, by rank-nullity theorem, $\text{dim}(\text{range}(T)) = 4 - 2 = 2$. The range is spanned by $T((1,0,1,0)) = (1,2,1)$ and $T((0,1,0,1)) = (2,1,3)$. We need to check which answer choice can be written as $a(1,2,1) + b(2,1,3)$ for some scalars $a,b$. For choice C: $(5,4,7) = a(1,2,1) + b(2,1,3)$ gives $a + 2b = 5$, $2a + b = 4$, $a + 3b = 7$. From the first two equations: $a = 5 - 2b$ and $2(5-2b) + b = 4$, so $10 - 4b + b = 4$, thus $b = 2$ and $a = 1$. Checking: $1 + 3(2) = 7$ ✓. The other choices don't satisfy this system.
9. Consider linear transformations $S: \mathbb{R}^3 \to \mathbb{R}^2$ and $T: \mathbb{R}^2 \to \mathbb{R}^3$ such that $\text{range}(S) = \mathbb{R}^2$ and $\ker(T) = \{(0,0)\}$. What is $\text{dim}(\text{range}(T \circ S))$?
   Answer: 2 When you see composition of linear transformations, think about how the dimensions of kernels and ranges interact. The key insight is understanding how information flows through the composite transformation $T \circ S$. Since $S: \mathbb{R}^3 \to \mathbb{R}^2$ has $\text{range}(S) = \mathbb{R}^2$, we know $S$ is surjective (onto). This means $S$ maps $\mathbb{R}^3$ onto all of $\mathbb{R}^2$. By the rank-nullity theorem, $\dim(\text{range}(S)) = 2$, so $\dim(\ker(S)) = 3 - 2 = 1$. Now, $T: \mathbb{R}^2 \to \mathbb{R}^3$ with $\ker(T) = \{(0,0)\}$ means $T$ is injective (one-to-one). Since $\dim(\ker(T)) = 0$, we have $\dim(\text{range}(T)) = 2 - 0 = 2$. For the composition $T \circ S$, since $S$ outputs all possible vectors in $\mathbb{R}^2$ and $T$ is injective on $\mathbb{R}^2$, the transformation $T$ will map the entire range of $S$ (which is $\mathbb{R}^2$) injectively into $\mathbb{R}^3$. Therefore, $\dim(\text{range}(T \circ S)) = \dim(\mathbb{R}^2) = 2$. Answer (A) 2 is correct. Answer (B) 1 incorrectly assumes some dimension is lost in composition. Answer (C) 3 wrongly suggests the composition could be surjective onto $\mathbb{R}^3$, but this is impossible since we start from $\mathbb{R}^3$ and pass through the "bottleneck" of $\mathbb{R}^2$. Answer (D) is incorrect because the given conditions completely determine the answer. Remember: in compositions, the maximum possible output dimension is limited by the smallest intermediate space, but injectivity preserves whatever dimension makes it through.
10. Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation where $n > m$. If $T$ is surjective, which statement about the relationship between $\ker(T)$ and the standard basis vectors of $\mathbb{R}^n$ must be true?
    Answer: The kernel $\ker(T)$ is non-trivial but may not contain any standard basis vectors Since $T$ is surjective and $n > m$, by rank-nullity theorem: $\text{dim}(\ker(T)) = n - \text{dim}(\text{range}(T)) = n - m > 0$. So $\ker(T)$ is non-trivial. However, $\ker(T)$ need not contain any standard basis vectors. For example, if $T: \mathbb{R}^3 \to \mathbb{R}^2$ with $T(x,y,z) = (x+y, y+z)$, then $\ker(T) = \text{span}\{(1,-1,1)\}$, which contains no standard basis vectors. Choice A assumes necessity, choice B assumes a specific structure that's not required, choice C contradicts the possibility of standard basis vectors being in the kernel.
11. Let $T: V \to V$ be a linear operator on a vector space $V$. Which statement provides a correct interpretation of the kernel of $T$ in the context of eigenvalues and eigenvectors?
    Answer: The kernel of $T$ is the eigenspace corresponding to the eigenvalue $\lambda = 0$. When you encounter questions connecting kernels and eigenvalues, you're dealing with a fundamental relationship in linear algebra. The key insight is recognizing what happens when you apply a linear operator to vectors in its kernel. The kernel of $T$, denoted $\ker(T)$, consists of all vectors $v$ such that $T(v) = 0$. Now, recall that an eigenvector $v$ with eigenvalue $\lambda$ satisfies $T(v) = \lambda v$. If $v \in \ker(T)$, then $T(v) = 0$, which means $\lambda v = 0$. Since eigenvectors are non-zero by definition, we must have $\lambda = 0$. This shows that every vector in the kernel is an eigenvector with eigenvalue 0, and conversely, every eigenvector with eigenvalue 0 is in the kernel. Therefore, $\ker(T)$ is precisely the eigenspace for $\lambda = 0$, making D correct. A is wrong because the kernel only contains eigenvectors corresponding to $\lambda = 0$, not all eigenvectors. B incorrectly identifies the eigenvalue as 1 instead of 0. C creates a logical contradiction—if the kernel is non-trivial (contains non-zero vectors), then 0 is an eigenvalue of $T$, not the absence of zero eigenvalues. Study tip: Remember that "kernel" and "eigenvalue zero" are intimately connected. Whenever you see kernel questions, immediately think about the eigenvalue 0. This connection also explains why invertible matrices have trivial kernels—they have no zero eigenvalues.
12. Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation. If $n > m$ and $T$ is known to be surjective, what must be the dimension of the kernel of $T$?
    Answer: $n - m$ By the Rank-Nullity Theorem, $\dim(\text{domain}) = \text{rank}(T) + \text{nullity}(T)$. The domain is $\mathbb{R}^n$, so its dimension is $n$. The transformation $T$ is surjective (onto), which means its range is the entire codomain $\mathbb{R}^m$. Therefore, the dimension of the range, $\text{rank}(T)$, is equal to $m$. Substituting these into the theorem gives $n = m + \text{nullity}(T)$. Solving for the nullity (the dimension of the kernel) yields $\text{nullity}(T) = n - m$.
13. Let $A$ be a $3 \times 4$ matrix and let $T_A: \mathbb{R}^4 \to \mathbb{R}^3$ be the associated linear transformation $T_A(x) = Ax$. If the columns of $A$ are linearly dependent and $\text{range}(T_A) = \mathbb{R}^3$, what can be concluded about $\ker(T_A)$?
    Answer: $\text{dim}(\ker(T_A)) = 1$ and $\ker(T_A)$ is one-dimensional When you encounter questions about linear transformations and their properties, focus on the fundamental relationship between the kernel, range, and dimensions given by the rank-nullity theorem. Here's the key insight: The rank-nullity theorem states that for any linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$, we have $\dim(\text{domain}) = \dim(\ker(T)) + \dim(\text{range}(T))$. Since $T_A: \mathbb{R}^4 \to \mathbb{R}^3$ and $\text{range}(T_A) = \mathbb{R}^3$, we know $\dim(\text{range}(T_A)) = 3$. Therefore: $4 = \dim(\ker(T_A)) + 3$, which gives us $\dim(\ker(T_A)) = 1$. The condition that columns are linearly dependent confirms this makes sense—if all four columns were linearly independent, the range couldn't be 3-dimensional (it would be at most the number of independent columns). With exactly 3 linearly independent columns spanning $\mathbb{R}^3$, we get a 1-dimensional kernel. Option B incorrectly assumes surjective transformations always have trivial kernels—this is only true when the domain and codomain have equal dimensions. Option C makes the same error with different wording. Option D misinterprets linear dependence; while the columns being dependent guarantees a non-trivial kernel, it doesn't determine the exact dimension without considering the range. Study tip: Always use rank-nullity theorem for kernel/range dimension questions. The dimensions must add up to the domain's dimension, regardless of whether the transformation is injective or surjective.
14. Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be defined by $T(x_1, x_2, x_3, x_4) = (x_1 + 2x_2 - x_3, 2x_1 + x_2 + x_4, x_1 - x_2 + x_3 + 2x_4)$. If $\text{dim}(\ker(T)) = 2$, what is $\text{dim}(\text{range}(T))$?
    Answer: 2 By the rank-nullity theorem, $\text{dim}(\text{domain}) = \text{dim}(\ker(T)) + \text{dim}(\text{range}(T))$. Since $T: \mathbb{R}^4 \to \mathbb{R}^3$, we have $\text{dim}(\text{domain}) = 4$. Given $\text{dim}(\ker(T)) = 2$, we get $4 = 2 + \text{dim}(\text{range}(T))$, so $\text{dim}(\text{range}(T)) = 2$. Choice A assumes nullity 3, choice C confuses range dimension with codomain dimension, choice D incorrectly uses domain dimension.
15. What is a basis for the kernel of the linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^3$ represented by the matrix $A$? $A = \begin{pmatrix} 1 & 2 & 0 & 3 \\ 2 & 4 & 1 & 8 \\ 1 & 2 & 1 & 5 \end{pmatrix}$
    Answer: $\{\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -3 \\ 0 \\ -2 \\ 1 \end{pmatrix}\}$ When you encounter questions about finding a basis for the kernel (null space) of a linear transformation, you need to find all vectors x such that Ax = 0. The kernel represents the "information lost" by the transformation. To find the kernel of matrix $A$, set up the homogeneous system $A\mathbf{x} = \mathbf{0}$ and row reduce: $\begin{pmatrix} 1 & 2 & 0 & 3 \\ 2 & 4 & 1 & 8 \\ 1 & 2 & 1 & 5 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ The system becomes: $x_1 + 2x_2 + 3x_4 = 0$ and $x_3 + 2x_4 = 0$. With free variables $x_2$ and $x_4$, we get $x_1 = -2x_2 - 3x_4$ and $x_3 = -2x_4$. Setting $x_2 = 1, x_4 = 0$ gives $\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix}$. Setting $x_2 = 0, x_4 = 1$ gives $\begin{pmatrix} -3 \\ 0 \\ -2 \\ 1 \end{pmatrix}$. These vectors span the kernel, so A is correct. B is wrong because these vectors have the wrong dimension (they're in $\mathbb{R}^3$, not $\mathbb{R}^4$) and actually come from the row space of $A$. C is wrong for the same reason - these are 3-dimensional vectors that cannot be in the kernel of a transformation from $\mathbb{R}^4$. D contains vectors of the correct dimension, but they don't satisfy $A\mathbf{x} = \mathbf{0}$. You can verify this by direct multiplication. Study tip: Always check your kernel vectors by substituting them back into $A\mathbf{x} = \mathbf{0}$. Also remember that kernel vectors must have the same dimension as the domain of the transformation.
16. A linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^3$ is represented by a $3 \times 4$ matrix $A$. If the transformation $T$ is surjective (onto), what must be true about the reduced row echelon form (RREF) of $A$?
    Answer: The RREF of $A$ has a pivot in every row. For $T$ to be surjective, its range must be all of the codomain, $\mathbb{R}^3$. This means the dimension of the range, which is the rank of matrix $A$, must be 3. The rank of a matrix is equal to the number of pivots in its RREF. Since $A$ is a $3 \times 4$ matrix, having a rank of 3 means there must be 3 pivots. As there are only 3 rows, this implies there must be a pivot position in every row.
17. Let $S: \mathbb{R}^3 \to \mathbb{R}^3$ and $T: \mathbb{R}^3 \to \mathbb{R}^3$ be linear transformations where $\text{dim}(\ker(S)) = 1$ and $\text{dim}(\ker(T)) = 2$. What can be concluded about $\text{dim}(\ker(S \circ T))$?
    Answer: $\text{dim}(\ker(S \circ T)) \geq 2$ and $\text{dim}(\ker(S \circ T)) \leq 3$ We know $\ker(T) \subseteq \ker(S \circ T)$ since if $T(x) = 0$, then $(S \circ T)(x) = S(T(x)) = S(0) = 0$. Thus $\text{dim}(\ker(S \circ T)) \geq \text{dim}(\ker(T)) = 2$. Also, $\text{dim}(\ker(S \circ T)) \leq 3$ since the domain is 3-dimensional. The exact value depends on how $\text{range}(T)$ intersects $\ker(S)$. Choice A assumes equality, choice B claims maximum always occurs, choice D ignores the contribution from $\ker(T)$.
18. Define $T: M_{2 \times 2} \to \mathbb{R}^2$ by $T\begin{pmatrix} a & b \\ c & d \end{pmatrix} = (a + d, b - c)$, where $M_{2 \times 2}$ is the space of $2 \times 2$ matrices. What is $\text{dim}(\ker(T)) + \text{dim}(\text{range}(T))$?
    Answer: 4 The domain $M_{2 \times 2}$ has dimension 4. By the rank-nullity theorem, $\text{dim}(\ker(T)) + \text{dim}(\text{range}(T)) = \text{dim}(M_{2 \times 2}) = 4$. To find $\ker(T)$: we need $a + d = 0$ and $b - c = 0$, so $d = -a$ and $c = b$. Thus $\ker(T) = \left\{\begin{pmatrix} a & b \\ b & -a \end{pmatrix} : a,b \in \mathbb{R}\right\}$, which has dimension 2. Therefore $\text{dim}(\text{range}(T)) = 4 - 2 = 2$. The sum is $2 + 2 = 4$. The other choices miscalculate either the domain dimension or misapply rank-nullity.
19. Consider the linear transformation $D: P_3 \to P_2$ defined by $D(p(x)) = p'(x)$, where $P_n$ is the vector space of polynomials of degree at most $n$. What is a basis for the kernel of $D$?
    Answer: $\{1\}$ The kernel of $D$ consists of all polynomials $p(x)$ in $P_3$ such that $D(p(x)) = p'(x) = 0$. The only polynomials whose derivative is zero are the constant polynomials. A general polynomial in $P_3$ is $p(x) = ax^3 + bx^2 + cx + d$. Its derivative is $p'(x) = 3ax^2 + 2bx + c$. For this to be the zero polynomial, we must have $a=0$, $b=0$, and $c=0$. This leaves $p(x) = d$, where $d$ is any constant. The space of constant polynomials is a one-dimensional subspace of $P_3$ spanned by the polynomial $p(x)=1$. Thus, a basis for the kernel is $\{1\}$.

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