Linear Algebra Quiz: Linear Transformation Definition
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Linear Transformation DefinitionQuestion 1 of 20

Let P2P_2 be the vector space of polynomials of degree at most 2. Which of the following transformations T:P2RT: P_2 \to \mathbb{R} is linear?

T(p(x))=p(1)+1T(p(x)) = p(1) + 1
T(p(x))=01(p(x))2dxT(p(x)) = \int_0^1 (p(x))^2 dx
T(p(x))=p(0)p(0)T(p(x)) = p(0) \cdot p'(0)
T(p(x))=p(2)p(1)T(p(x)) = p(2) - p'(1)
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Linear Algebra Quiz

Linear Algebra Quiz: Linear Transformation Definition

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

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

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Question 1

Let P2P_2 be the vector space of polynomials of degree at most 2. Which of the following transformations T:P2RT: P_2 \to \mathbb{R} is linear?

  1. T(p(x))=p(1)+1T(p(x)) = p(1) + 1
  2. T(p(x))=01(p(x))2dxT(p(x)) = \int_0^1 (p(x))^2 dx
  3. T(p(x))=p(0)p(0)T(p(x)) = p(0) \cdot p'(0)
  4. T(p(x))=p(2)p(1)T(p(x)) = p(2) - p'(1) (correct answer)
Explanation: A transformation is linear if it preserves vector addition and scalar multiplication. (A) is not linear because it is a translation; for the zero polynomial p(x)=0p(x)=0, T(0)=0+1=10T(0) = 0+1=1 \neq 0. (B) is not linear due to the square; T(cp(x))=01(cp(x))2dx=c201(p(x))2dx=c2T(p(x))T(c \cdot p(x)) = \int_0^1 (c \cdot p(x))^2 dx = c^2 \int_0^1 (p(x))^2 dx = c^2 T(p(x)), which violates homogeneity. (C) is not linear due to the product; let p(x)=cxp(x)=cx. Then p(0)=0p(0)=0 and p(0)=cp'(0)=c. T(p(x))=0T(p(x))=0. But consider q(x)=x+1q(x)=x+1. T(q(x))=q(0)q(0)=11=1T(q(x)) = q(0)q'(0) = 1 \cdot 1 = 1. T(2q(x))=T(2x+2)=(2)(2)=4T(2q(x)) = T(2x+2) = (2)(2)=4, whereas 2T(q(x))=2(1)=22T(q(x)) = 2(1) = 2. Homogeneity fails. (D) is linear because differentiation and evaluation are linear operations, and so is their linear combination. T(p(x)+q(x))=(p(2)+q(2))(p(1)+q(1))=(p(2)p(1))+(q(2)q(1))=T(p(x))+T(q(x))T(p(x)+q(x)) = (p(2)+q(2))-(p'(1)+q'(1)) = (p(2)-p'(1)) + (q(2)-q'(1)) = T(p(x))+T(q(x)), and T(cp(x))=cp(2)cp(1)=c(p(2)p(1))=cT(p(x))T(c \cdot p(x)) = c \cdot p(2) - c \cdot p'(1) = c(p(2)-p'(1))=cT(p(x)).

Question 2

Consider a transformation T:VWT: V \to W between two vector spaces. The condition T(0V)=0WT(\vec{0}_V) = \vec{0}_W is a well-known property related to linear transformations. Which statement accurately describes the significance of this condition?

  1. The condition is sufficient to prove linearity, as it follows from the homogeneity property T(cu)=cT(u)T(c\vec{u}) = cT(\vec{u}) by setting c=0c=0.
  2. The condition is both necessary and sufficient, meaning any transformation mapping the zero vector to the zero vector is linear.
  3. The condition is necessary for linearity, but it is not sufficient, as non-linear transformations exist that also map the zero vector to the zero vector. (correct answer)
  4. The condition is not necessary for linearity, as some linear transformations, such as translations, do not map the zero vector to the zero vector.
Explanation: The condition T(0V)=0WT(\vec{0}_V) = \vec{0}_W is a necessary consequence of the homogeneity property of linear transformations (T(0u)=0T(u)=0WT(0 \cdot \vec{u}) = 0 \cdot T(\vec{u}) = \vec{0}_W). Therefore, if a transformation does not map the zero vector to the zero vector, it cannot be linear. However, this condition is not sufficient to prove linearity. For example, T(x,y)=(x2,y2)T(x,y) = (x^2, y^2) satisfies T(0,0)=(0,0)T(0,0)=(0,0), but it is not a linear transformation. Thus, the condition is necessary but not sufficient.

Question 3

Consider the transformation T:R2R2T: \mathbb{R}^2 \to \mathbb{R}^2 defined by T(x,y)=(x,y)T(x, y) = (x, |y|). This transformation is not linear. Which of the following calculations correctly demonstrates its non-linearity?

  1. For c=2c = -2 and u=(1,3)\vec{u} = (1, 3), we have T(cu)=T(2,6)=(2,6)T(c\vec{u}) = T(-2, -6) = (-2, 6), while cT(u)=2T(1,3)=2(1,3)=(2,6)cT(\vec{u}) = -2T(1, 3) = -2(1, 3) = (-2, -6). (correct answer)
  2. For u=(1,2)\vec{u} = (1, 2) and v=(3,4)\vec{v} = (3, 4), T(u+v)=T(4,6)=(4,6)T(\vec{u}+\vec{v}) = T(4, 6) = (4, 6) and T(u)+T(v)=(1,2)+(3,4)=(4,6)T(\vec{u}) + T(\vec{v}) = (1, 2) + (3, 4) = (4, 6), verifying additivity.
  3. T(1,2)=(1,2)T(1, 2) = (1, 2) and T(1,2)=(1,2)T(1, -2) = (1, 2), which shows that TT is not an injective (one-to-one) function.
  4. T(0,0)=(0,0)=(0,0)T(0, 0) = (0, |0|) = (0, 0), which satisfies a necessary condition for linearity, but does not prove it.
Explanation: To determine if a transformation is linear, you must verify two properties: additivity (T(u+v)=T(u)+T(v)T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})) and scalar multiplication (T(cu)=cT(u)T(c\vec{u}) = cT(\vec{u})). If either property fails, the transformation is not linear. Choice A correctly demonstrates non-linearity by testing scalar multiplication. With c=2c = -2 and u=(1,3)\vec{u} = (1, 3), we get T(cu)=T(2,6)=(2,6)T(c\vec{u}) = T(-2, -6) = (-2, 6) since the absolute value makes 6=6|-6| = 6. However, cT(u)=2T(1,3)=2(1,3)=(2,6)cT(\vec{u}) = -2T(1, 3) = -2(1, 3) = (-2, -6). Since (2,6)(2,6)(-2, 6) \neq (-2, -6), the scalar multiplication property fails, proving non-linearity. Choice B actually verifies additivity works for this specific example, which doesn't prove non-linearity. The calculation T(4,6)=(4,6)T(4, 6) = (4, 6) equals T(1,2)+T(3,4)=(1,2)+(3,4)=(4,6)T(1, 2) + T(3, 4) = (1, 2) + (3, 4) = (4, 6), but this single case doesn't establish linearity overall. Choice C discusses injectivity, which is irrelevant to linearity. A transformation can be linear without being one-to-one, and vice versa. While the observation about T(1,2)=T(1,2)=(1,2)T(1, 2) = T(1, -2) = (1, 2) is correct, it doesn't address the linearity properties. Choice D correctly notes that T(0,0)=(0,0)T(0, 0) = (0, 0), which is necessary for linearity, but acknowledges this alone doesn't prove linearity. Study tip: When testing linearity, focus on the two defining properties. Negative inputs are particularly useful test cases for transformations involving absolute values, squares, or other operations that change signs.

Question 4

Let M2×2M_{2 \times 2} be the vector space of 2×22 \times 2 matrices with real entries. Which of the following transformations is linear?

  1. T:M2×2RT: M_{2 \times 2} \to \mathbb{R} defined by T(A)=det(A)T(A) = \det(A)
  2. T:M2×2RT: M_{2 \times 2} \to \mathbb{R} defined by T(A)=(A2)11T(A) = (A^2)_{11}, the entry in the first row and first column of A2A^2
  3. T:M2×2M2×2T: M_{2 \times 2} \to M_{2 \times 2} defined by T(A)=A+IT(A) = A + I, where II is the identity matrix
  4. T:M2×2RT: M_{2 \times 2} \to \mathbb{R} defined by T(A)=trace(A)T(A) = \mathrm{trace}(A) (correct answer)
Explanation: The trace of a matrix is the sum of its diagonal elements. The trace is a linear transformation because trace(A+B)=trace(A)+trace(B)\mathrm{trace}(A+B) = \mathrm{trace}(A) + \mathrm{trace}(B) and trace(cA)=ctrace(A)\mathrm{trace}(cA) = c \cdot \mathrm{trace}(A). (A) The determinant is not linear; for a 2×22 \times 2 matrix AA, det(cA)=c2det(A)cdet(A)\det(cA) = c^2 \det(A) \neq c \det(A) in general. (B) Squaring a matrix is not a linear operation. For A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, (A2)11=a2+bc(A^2)_{11} = a^2+bc. T(kA)=(ka)2+(kb)(kc)=k2(a2+bc)=k2T(A)T(kA) = (ka)^2 + (kb)(kc) = k^2(a^2+bc) = k^2 T(A), which violates homogeneity. (C) Adding a fixed non-zero matrix (or vector) is a translation, which is not linear because T(0)=0+I=I0T(0) = 0+I = I \neq 0.

Question 5

Let T:R2RT: \mathbb{R}^2 \to \mathbb{R} be a linear transformation. Which of the following functions F:R2RF: \mathbb{R}^2 \to \mathbb{R} is not guaranteed to be linear?

  1. F(v)=(T(v))2F(\vec{v}) = (T(\vec{v}))^2 (correct answer)
  2. F(v)=T(v)+T(v)F(\vec{v}) = T(\vec{v}) + T(-\vec{v})
  3. F(v)=2T(v)F(\vec{v}) = 2T(\vec{v})
  4. F(v)=T(vv0)+T(v0)F(\vec{v}) = T(\vec{v} - \vec{v}_0) + T(\vec{v}_0) for a fixed vector v0R2\vec{v}_0 \in \mathbb{R}^2
Explanation: When testing whether a function is linear, you need to verify two properties: additivity (F(u+v)=F(u)+F(v)F(\vec{u} + \vec{v}) = F(\vec{u}) + F(\vec{v})) and scalar multiplication (F(cv)=cF(v)F(c\vec{v}) = cF(\vec{v})). The key insight is that combining linear transformations through linear operations preserves linearity, but nonlinear operations destroy it. Option A, F(v)=(T(v))2F(\vec{v}) = (T(\vec{v}))^2, fails both linearity tests. For additivity: (T(u+v))2=(T(u)+T(v))2=T(u)2+2T(u)T(v)+T(v)2(T(\vec{u} + \vec{v}))^2 = (T(\vec{u}) + T(\vec{v}))^2 = T(\vec{u})^2 + 2T(\vec{u})T(\vec{v}) + T(\vec{v})^2, which doesn't equal T(u)2+T(v)2T(\vec{u})^2 + T(\vec{v})^2. The cross-term 2T(u)T(v)2T(\vec{u})T(\vec{v}) breaks linearity. This is the correct answer. Option B is linear because T(v)+T(v)=T(v)+T(1v)=T(v)T(v)=0T(\vec{v}) + T(-\vec{v}) = T(\vec{v}) + T(-1 \cdot \vec{v}) = T(\vec{v}) - T(\vec{v}) = 0 for any linear transformation, making FF the zero function. Option C is clearly linear since scalar multiplication of a linear transformation remains linear: 2T(u+v)=2(T(u)+T(v))=2T(u)+2T(v)2T(\vec{u} + \vec{v}) = 2(T(\vec{u}) + T(\vec{v})) = 2T(\vec{u}) + 2T(\vec{v}). Option D is linear because F(v)=T(vv0)+T(v0)=T(v)T(v0)+T(v0)=T(v)F(\vec{v}) = T(\vec{v} - \vec{v}_0) + T(\vec{v}_0) = T(\vec{v}) - T(\vec{v}_0) + T(\vec{v}_0) = T(\vec{v}), which is just TT itself. Study tip: When checking if composite functions are linear, watch for nonlinear operations like squaring, multiplication of outputs, or absolute values—these always break linearity, even when applied to linear transformations.

Question 6

Let T:R2R3T: \mathbb{R}^2 \to \mathbb{R}^3 be a linear transformation such that T([10])=[213]T\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} and T([01])=[041]T\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) = \begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix}. What is the value of T([32])T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right)?

  1. [6511]\begin{bmatrix} 6 \\ 5 \\ 11 \end{bmatrix}
  2. [234]\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}
  3. [6117]\begin{bmatrix} 6 \\ -11 \\ 7 \end{bmatrix} (correct answer)
  4. [682]\begin{bmatrix} 6 \\ -8 \\ 2 \end{bmatrix}
Explanation: By the properties of linear transformations, we can write the input vector as a linear combination of the basis vectors and apply the transformation. Specifically, T(cu+dv)=cT(u)+dT(v)T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v}). We have [32]=3[10]2[01]\begin{bmatrix} 3 \\ -2 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 1 \end{bmatrix}. Therefore, T([32])=T(3[10]2[01])=3T([10])2T([01])T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right) = T\left(3\begin{bmatrix} 1 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) = 3T\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) - 2T\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right). Substituting the given values: 3[213]2[041]=[639][082]=[6117]3\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ -3 \\ 9 \end{bmatrix} - \begin{bmatrix} 0 \\ 8 \\ 2 \end{bmatrix} = \begin{bmatrix} 6 \\ -11 \\ 7 \end{bmatrix}.

Question 7

Let VV and WW be vector spaces. A function T:VWT: V \to W is a linear transformation if it satisfies additivity and homogeneity. Which of the following conditions is a valid, alternative, single-property definition for linearity?

  1. T(cu+dv)=cT(u)+dT(v)T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v}) for all vectors u,vV\vec{u}, \vec{v} \in V and all scalars c,dc, d. (correct answer)
  2. T(uv)=T(u)T(v)T(\vec{u} - \vec{v}) = T(\vec{u}) - T(\vec{v}) for all vectors u,vV\vec{u}, \vec{v} \in V.
  3. T(cu)=cT(u)T(c\vec{u}) = cT(\vec{u}) for all vectors uV\vec{u} \in V and all scalars cc.
  4. T(u+v)=T(u)+T(v)T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}) for all vectors u,vV\vec{u}, \vec{v} \in V.
Explanation: The standard definition of a linear transformation requires two properties to be checked: additivity (T(u+v)=T(u)+T(v)T(\vec{u}+\vec{v})=T(\vec{u})+T(\vec{v})) and homogeneity (T(cu)=cT(u)T(c\vec{u})=cT(\vec{u})). The property listed in choice A, sometimes called the superposition principle, combines these two into a single statement. If A holds, one can prove additivity by setting c=d=1c=d=1, and one can prove homogeneity by setting d=0d=0. Conversely, if additivity and homogeneity hold, one can prove A: T(cu+dv)=T(cu)+T(dv)=cT(u)+dT(v)T(c\vec{u} + d\vec{v}) = T(c\vec{u}) + T(d\vec{v}) = cT(\vec{u}) + dT(\vec{v}). Choices B, C, and D list properties that are necessary for linearity but are not sufficient on their own.

Question 8

An affine transformation T:RnRmT: \mathbb{R}^n \to \mathbb{R}^m is defined by the rule T(x)=Ax+bT(\vec{x}) = A\vec{x} + \vec{b}, where AA is an m×nm \times n matrix and b\vec{b} is a fixed vector in Rm\mathbb{R}^m. For TT to be a linear transformation, which condition must be met?

  1. The matrix AA must be the zero matrix.
  2. The vector b\vec{b} must be the zero vector. (correct answer)
  3. The matrix AA must be invertible.
  4. The vector b\vec{b} must be in the column space of AA.
Explanation: For a transformation TT to be linear, it must satisfy T(0)=0T(\vec{0}) = \vec{0}. Applying this to the affine transformation, we get T(0)=A0+b=0+b=bT(\vec{0}) = A\vec{0} + \vec{b} = \vec{0} + \vec{b} = \vec{b}. For T(0)T(\vec{0}) to be equal to 0\vec{0}, we must have b=0\vec{b} = \vec{0}. If b=0\vec{b}=\vec{0}, the transformation becomes T(x)=AxT(\vec{x}) = A\vec{x}, which is the definition of a matrix transformation, a primary example of a linear transformation. The properties of matrix AA (being zero, invertible, etc.) do not determine whether the affine map is linear; only the translation vector b\vec{b} does.

Question 9

Let T:VWT: V \to W be a linear transformation. Which of the following statements is always true?

  1. If {v1,,vk}\{\vec{v}_1, \dots, \vec{v}_k\} is a linearly independent set in VV, then {T(v1),,T(vk)}\{T(\vec{v}_1), \dots, T(\vec{v}_k)\} is a linearly independent set in WW.
  2. The dimension of the domain VV must be less than or equal to the dimension of the codomain WW.
  3. For any vectors u,vV\vec{u}, \vec{v} \in V, the image of the line segment connecting them is the line segment connecting T(u)T(\vec{u}) and T(v)T(\vec{v}). (correct answer)
  4. The transformation TT must be surjective (onto).
Explanation: A point on the line segment between u\vec{u} and v\vec{v} can be parameterized as p(t)=(1t)u+tv\vec{p}(t) = (1-t)\vec{u} + t\vec{v} for t[0,1]t \in [0,1]. Applying the linear transformation TT gives T(p(t))=T((1t)u+tv)=(1t)T(u)+tT(v)T(\vec{p}(t)) = T((1-t)\vec{u} + t\vec{v}) = (1-t)T(\vec{u}) + tT(\vec{v}) due to linearity. This expression parameterizes the line segment between T(u)T(\vec{u}) and T(v)T(\vec{v}), so statement C is correct. Statement A is false; if TT has a non-trivial kernel, it can map a non-zero vector (a linearly independent set of one) to the zero vector. Statement B is false, e.g., a projection from R3\mathbb{R}^3 to R2\mathbb{R}^2. Statement D is false, e.g., the zero transformation T(v)=0T(\vec{v}) = \vec{0} unless WW is the zero vector space.

Question 10

Consider the function G:R2RG: \mathbb{R}^2 \to \mathbb{R} defined by G(x,y)=3x4y+5G(x, y) = 3x - 4y + 5. A student argues that GG is 'almost linear' because it satisfies G(cv)=cG(v)+5(1c)G(c\mathbf{v}) = cG(\mathbf{v}) + 5(1-c) for any scalar cc and vector v\mathbf{v}. What is the fundamental issue with this reasoning?

  1. The student failed to verify the additivity property G(u+v)=G(u)+G(v)G(\mathbf{u} + \mathbf{v}) = G(\mathbf{u}) + G(\mathbf{v}), which is the more important condition
  2. The formula G(cv)=cG(v)+5(1c)G(c\mathbf{v}) = cG(\mathbf{v}) + 5(1-c) is mathematically incorrect and doesn't hold for this function
  3. Linear transformations must map from Rn\mathbb{R}^n to Rn\mathbb{R}^n, so functions with codomain R\mathbb{R} cannot be linear
  4. The presence of the constant term 5 means GG is affine, not linear, and the modified homogeneity condition is irrelevant to linearity (correct answer)
Explanation: When you encounter questions about whether a function is linear, you need to check two fundamental properties: additivity and homogeneity. A linear transformation must satisfy both T(u+v)=T(u)+T(v)T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) and T(cv)=cT(v)T(c\mathbf{v}) = cT(\mathbf{v}) for all vectors and scalars. The function G(x,y)=3x4y+5G(x, y) = 3x - 4y + 5 is an affine transformation, not a linear one, because of the constant term 5. This constant term prevents GG from satisfying true homogeneity: G(cv)=3cx4cy+5G(c\mathbf{v}) = 3cx - 4cy + 5, while cG(v)=c(3x4y+5)=3cx4cy+5ccG(\mathbf{v}) = c(3x - 4y + 5) = 3cx - 4cy + 5c. These are only equal when c=1c = 1. The student's modified formula G(cv)=cG(v)+5(1c)G(c\mathbf{v}) = cG(\mathbf{v}) + 5(1-c) is actually correct mathematically, but it describes an affine property, not linearity. Answer D correctly identifies this fundamental distinction. Answer A incorrectly suggests additivity is "more important" – both properties are equally essential for linearity. Answer B is wrong because the student's formula does hold; you can verify it algebraically. Answer C misunderstands the definition of linear transformations – they can absolutely map from Rn\mathbb{R}^n to Rm\mathbb{R}^m where nmn \neq m. Study tip: Remember that any function with a constant term is affine, not linear. True linear transformations must pass through the origin, meaning T(0)=0T(\mathbf{0}) = \mathbf{0}. When you see a constant term, immediately think "affine transformation."

Question 11

A function H:VWH: V \to W between vector spaces satisfies H(u+v)=H(u)+H(v)H(\mathbf{u} + \mathbf{v}) = H(\mathbf{u}) + H(\mathbf{v}) for all vectors u,vV\mathbf{u}, \mathbf{v} \in V. Under what additional condition can we conclude that HH is linear?

  1. HH must map linearly independent sets to linearly independent sets and preserve basis relationships
  2. HH must be continuous and differentiable with respect to the standard topology on both vector spaces
  3. HH must preserve the inner product structure: H(u),H(v)=u,v\langle H(\mathbf{u}), H(\mathbf{v}) \rangle = \langle \mathbf{u}, \mathbf{v} \rangle for all u,v\mathbf{u}, \mathbf{v}
  4. HH must also satisfy H(cv)=cH(v)H(c\mathbf{v}) = cH(\mathbf{v}) for all scalars cc and vectors vV\mathbf{v} \in V (correct answer)
Explanation: When you encounter questions about linear transformations, remember that linearity requires two specific properties working together. The question gives you one half of the definition and asks what's missing. A linear transformation must satisfy both additivity and homogeneity. You're told that H(u+v)=H(u)+H(v)H(\mathbf{u} + \mathbf{v}) = H(\mathbf{u}) + H(\mathbf{v}) (additivity), but this alone isn't sufficient for linearity. The missing piece is scalar multiplication: H(cv)=cH(v)H(c\mathbf{v}) = cH(\mathbf{v}) for all scalars cc and vectors v\mathbf{v}. Together, these two properties ensure that HH preserves all linear combinations, which is the essence of linearity. Answer D correctly identifies this missing condition. Option A confuses linearity with other properties. While linear transformations do preserve linear independence in many cases, this isn't part of the definition of linearity itself, and requiring this as an additional condition would be circular reasoning. Option B introduces concepts from analysis that aren't relevant to the algebraic definition of linearity. In finite-dimensional spaces, linearity actually implies continuity, not the reverse, and differentiability is unnecessary. Option C describes an inner product-preserving map (an isometry), which is a much stronger condition than linearity. Many linear transformations don't preserve inner products—for example, any scaling transformation with factor other than ±1. Remember this key insight: linearity always requires both additivity AND homogeneity. If you see one property in a problem, immediately think about whether the other is present or needs to be established.

Question 12

Let T:R3R2T: \mathbb{R}^3 \to \mathbb{R}^2 be defined by T(x,y,z)=(x+y,2z)T(x, y, z) = (x + y, 2z). A student wants to verify linearity by checking that T(au+bv)=aT(u)+bT(v)T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v}) for specific vectors u=(1,0,1)\mathbf{u} = (1, 0, 1), v=(0,1,1)\mathbf{v} = (0, 1, -1), and scalars a=2,b=3a = 2, b = -3. What does this verification actually establish?

  1. It demonstrates that TT satisfies the additivity property but says nothing about scalar multiplication
  2. It completely proves that TT is linear since the chosen vectors span a 2-dimensional subspace of R3\mathbb{R}^3
  3. It confirms linearity for one specific case, but linearity requires this property to hold for all possible vectors and scalars (correct answer)
  4. It shows that TT is linear on the subspace spanned by u\mathbf{u} and v\mathbf{v}, which is sufficient for global linearity
Explanation: When testing whether a transformation is linear, you need to understand what verification of the linearity condition T(au+bv)=aT(u)+bT(v)T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v}) actually proves. This fundamental property must hold for every possible choice of vectors and scalars, not just specific examples. Let's see what this particular verification establishes. Computing T(2(1,0,1)+(3)(0,1,1))=T(2,3,5)=(1,10)T(2(1,0,1) + (-3)(0,1,-1)) = T(2,-3,5) = (-1,10) and 2T(1,0,1)+(3)T(0,1,1)=2(1,2)+(3)(1,2)=(1,10)2T(1,0,1) + (-3)T(0,1,-1) = 2(1,2) + (-3)(1,-2) = (-1,10) confirms the property holds for these specific values. This is consistent with linearity, but it's only one data point. Choice A is wrong because this verification checks both additivity and scalar multiplication simultaneously—the linearity condition combines both properties. Choice B incorrectly assumes that checking linearity on vectors spanning a 2D subspace proves global linearity. Even if these vectors formed a basis for all of R3\mathbb{R}^3, you'd still need to verify the property for all scalar combinations. Choice D makes a similar error, suggesting that linearity on a subspace guarantees global linearity, which is false. Choice C correctly identifies that this verification confirms linearity for one specific case only. To prove linearity rigorously, you need either to check the property for arbitrary vectors and scalars (impossible in practice) or use the definition to show it algebraically for the general case. Study tip: When proving linearity, specific numerical examples can only disprove linearity (by finding counterexamples) or provide evidence—never complete proof. Always work with arbitrary variables for rigorous proofs.

Question 13

A transformation R:R2R2R: \mathbb{R}^2 \to \mathbb{R}^2 rotates every vector by 90°90° counterclockwise about the origin. Which approach most directly demonstrates that RR satisfies the definition of a linear transformation?

  1. Demonstrate that RR has a matrix representation and that matrix multiplication always produces linear transformations
  2. Verify that RR preserves lengths and angles, which are the fundamental requirements for linear transformations
  3. Show that R(cv)=cR(v)R(c\mathbf{v}) = cR(\mathbf{v}) and R(u+v)=R(u)+R(v)R(\mathbf{u} + \mathbf{v}) = R(\mathbf{u}) + R(\mathbf{v}) using geometric properties of rotation (correct answer)
  4. Show that RR maps the standard basis vectors to orthogonal unit vectors, confirming orthogonal transformation properties
Explanation: When you encounter questions about proving a transformation is linear, always return to the fundamental definition: a transformation is linear if and only if it satisfies two properties for all vectors and scalars. The correct approach is C because it directly verifies the two defining properties of linear transformations: additivity R(u+v)=R(u)+R(v)R(\mathbf{u} + \mathbf{v}) = R(\mathbf{u}) + R(\mathbf{v}) and homogeneity R(cv)=cR(v)R(c\mathbf{v}) = cR(\mathbf{v}). For a 90° counterclockwise rotation, you can verify these geometrically: rotating the sum of two vectors gives the same result as summing their individual rotations, and rotating a scaled vector equals scaling the rotated vector. A puts the cart before the horse—while linear transformations do have matrix representations, you need to prove linearity first before concluding the matrix exists. The existence of a matrix is a consequence, not a proof of linearity. B confuses linear transformations with isometries. Preserving lengths and angles makes a transformation orthogonal, but linear transformations can stretch, shrink, or shear while remaining linear. Length and angle preservation is neither necessary nor sufficient for linearity. D focuses on orthogonal transformation properties rather than linearity. While mapping standard basis vectors to orthogonal unit vectors tells you about the transformation's geometric nature, it doesn't directly establish the additivity and homogeneity required for linearity. Study tip: Always prove linearity by verifying the two fundamental properties directly. Don't be distracted by other geometric properties—they might be interesting consequences, but they're not what defines linearity.

Question 14

A student defines L:R2R3L: \mathbb{R}^2 \to \mathbb{R}^3 by L(x,y)=(xy,0,x+2y)L(x, y) = (x - y, 0, x + 2y) and claims to have verified linearity by showing that L(2e1)=2L(e1)L(2\mathbf{e}_1) = 2L(\mathbf{e}_1) and L(3e2)=3L(e2)L(3\mathbf{e}_2) = 3L(\mathbf{e}_2), where e1=(1,0)\mathbf{e}_1 = (1, 0) and e2=(0,1)\mathbf{e}_2 = (0, 1). What critical verification is missing from this approach?

  1. The student must confirm that LL maps the zero vector to the zero vector, which is a necessary condition for linearity
  2. The student must verify that L(au+bv)=aL(u)+bL(v)L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v}) for arbitrary vectors and scalars, not just basis vectors with specific scalars (correct answer)
  3. The student must verify that LL preserves linear independence by showing that linearly independent inputs produce linearly independent outputs
  4. The student must demonstrate that the transformation is invertible by constructing the inverse function L1L^{-1}
Explanation: When you encounter questions about verifying linearity, remember that a linear transformation must satisfy two fundamental properties for ALL vectors and scalars: L(cu)=cL(u)L(c\mathbf{u}) = cL(\mathbf{u}) (scalar multiplication) and L(u+v)=L(u)+L(v)L(\mathbf{u} + \mathbf{v}) = L(\mathbf{u}) + L(\mathbf{v}) (additivity). These can be combined into the single condition: L(au+bv)=aL(u)+bL(v)L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v}) for arbitrary vectors and scalars. The student's approach is dangerously incomplete. Testing only L(2e1)=2L(e1)L(2\mathbf{e}_1) = 2L(\mathbf{e}_1) and L(3e2)=3L(e2)L(3\mathbf{e}_2) = 3L(\mathbf{e}_2) verifies linearity for just two specific cases with basis vectors. To prove linearity rigorously, you must demonstrate that L(au+bv)=aL(u)+bL(v)L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v}) holds for any vectors u,v\mathbf{u}, \mathbf{v} and any scalars a,ba, b. This makes B correct. Option A is wrong because while L(0)=0L(\mathbf{0}) = \mathbf{0} is necessary for linearity, it's automatically satisfied if the general linearity condition holds. Option C confuses linearity with other properties—linear transformations don't need to preserve linear independence (they can have non-trivial null spaces). Option D is incorrect because invertibility isn't required for linearity; many linear transformations aren't invertible. Study tip: When verifying linearity, never rely on testing specific examples. Always prove the general property L(au+bv)=aL(u)+bL(v)L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v}) using arbitrary vectors and scalars—this is the gold standard that encompasses all linearity requirements.

Question 15

A transformation T:RnRmT: \mathbb{R}^n \to \mathbb{R}^m can be represented by the matrix equation T(x)=AxT(\mathbf{x}) = A\mathbf{x} for some matrix AA. Which statement about the relationship between this matrix representation and linearity is most accurate?

  1. Every linear transformation can be represented this way, and every such matrix representation defines a linear transformation (correct answer)
  2. Only bijective linear transformations can be represented as matrix multiplication, while others require additional constant terms
  3. Matrix representation is possible only when the domain and codomain have the same finite dimension
  4. The matrix representation exists if and only if the transformation preserves orthogonality and has determinant equal to ±1
Explanation: This is a fundamental theorem: every linear transformation between finite-dimensional vector spaces can be represented as multiplication by a matrix, and conversely, every matrix multiplication defines a linear transformation. Choice A correctly states this bidirectional relationship. Choice B incorrectly restricts to bijective transformations and mentions constant terms (which would make the transformation affine, not linear). Choice C incorrectly requires equal dimensions - linear transformations can map between spaces of different dimensions. Choice D confuses linear transformations with orthogonal transformations and unnecessarily restricts to invertible cases.

Question 16

Which of the following transformations T:R2R2T: \mathbb{R}^2 \to \mathbb{R}^2 is a linear transformation?

  1. T(x,y)=(x+y,x1)T(x, y) = (x+y, x-1)
  2. T(x,y)=(2x3y,x+5y)T(x, y) = (2x - 3y, x + 5y) (correct answer)
  3. T(x,y)=(xy,y)T(x, y) = (xy, y)
  4. T(x,y)=(x2,y2)T(x, y) = (x^2, y^2)
Explanation: A transformation TT is linear if it satisfies T(u+v)=T(u)+T(v)T(\vec{u}+\vec{v}) = T(\vec{u}) + T(\vec{v}) and T(cu)=cT(u)T(c\vec{u}) = cT(\vec{u}). Choice B represents a matrix transformation T(x)=AxT(\vec{x}) = A\vec{x} with A=(2315)A = \begin{pmatrix} 2 & -3 \\ 1 & 5 \end{pmatrix}, which is always linear. Choice A is not linear because T(0,0)=(0,1)0T(0, 0) = (0, -1) \neq \vec{0}. Choice C is not linear because of the product term xyxy; for example, T(2(1,1))=T(2,2)=(4,2)T(2(1,1)) = T(2,2) = (4,2), but 2T(1,1)=2(1,1)=(2,2)2T(1,1) = 2(1,1) = (2,2). Choice D is not linear because of the squared terms; for example, T((1,0)+(1,0))=T(2,0)=(4,0)T((1,0)+(1,0)) = T(2,0) = (4,0), but T(1,0)+T(1,0)=(1,0)+(1,0)=(2,0)T(1,0)+T(1,0) = (1,0)+(1,0) = (2,0).

Question 17

A function F:R3R2F: \mathbb{R}^3 \to \mathbb{R}^2 satisfies F(2u+3v)=2F(u)+3F(v)F(2\mathbf{u} + 3\mathbf{v}) = 2F(\mathbf{u}) + 3F(\mathbf{v}) for specific vectors u\mathbf{u} and v\mathbf{v}. What additional condition is necessary and sufficient to conclude that FF is linear?

  1. The given property must hold for all vectors u,vR3\mathbf{u}, \mathbf{v} \in \mathbb{R}^3 and all scalars, not just the specific case shown (correct answer)
  2. FF must also satisfy F(u×v)=F(u)×F(v)F(\mathbf{u} \times \mathbf{v}) = F(\mathbf{u}) \times F(\mathbf{v}) for the cross product operation between vectors
  3. The function must be continuous at the origin and differentiable throughout its domain with constant derivative
  4. FF must preserve orthogonality, meaning uv\mathbf{u} \perp \mathbf{v} implies F(u)F(v)F(\mathbf{u}) \perp F(\mathbf{v}) for all input vectors
Explanation: The given condition shows linearity for one specific case, but linearity requires F(au+bv)=aF(u)+bF(v)F(a\mathbf{u} + b\mathbf{v}) = aF(\mathbf{u}) + bF(\mathbf{v}) for ALL vectors u,v\mathbf{u}, \mathbf{v} in the domain and ALL scalars a,ba, b. Choice A correctly identifies this universal requirement. Choice B is incorrect because cross products are not preserved by general linear transformations (and the codomain R2\mathbb{R}^2 doesn't have a cross product). Choice C confuses linear transformations with differentiable functions. Choice D describes orthogonal transformations, which are a special subset of linear transformations.

Question 18

Which of the following geometric transformations of the Cartesian plane R2\mathbb{R}^2 is not a linear transformation?

  1. A rotation of 3030^\circ counter-clockwise about the origin.
  2. A reflection across the line y=xy=x.
  3. A projection onto the x-axis.
  4. A translation by the vector [11]\begin{bmatrix} 1 \\ 1 \end{bmatrix}. (correct answer)
Explanation: A key property of any linear transformation TT is that it must map the zero vector to the zero vector, i.e., T(0)=0T(\vec{0}) = \vec{0}. A translation by a non-zero vector v0\vec{v}_0 is given by T(x)=x+v0T(\vec{x}) = \vec{x} + \vec{v}_0. For this transformation, T(0)=0+v0=v0T(\vec{0}) = \vec{0} + \vec{v}_0 = \vec{v}_0. Since v0=[11]0\vec{v}_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \neq \vec{0}, this translation is not a linear transformation. Rotations about the origin, reflections across lines through the origin, and projections onto subspaces are all standard examples of linear transformations.

Question 19

Consider a function P:R2R2P: \mathbb{R}^2 \to \mathbb{R}^2 that projects vectors onto the line y=xy = x. If PP is linear, which of the following must be true about P(3,6)P(3, -6)?

  1. P(3,6)=(0,0)P(3, -6) = (0, 0) because the vector (3,6)(3, -6) is orthogonal to the line y=xy = x and linear projections map orthogonal vectors to zero
  2. P(3,6)=(1.5,1.5)P(3, -6) = (-1.5, -1.5) because linearity requires that the projection formula P(x,y)=(x+y2,x+y2)P(x,y) = \left(\frac{x+y}{2}, \frac{x+y}{2}\right) applies consistently (correct answer)
  3. P(3,6)=(3,3)P(3, -6) = (3, 3) because linear transformations preserve the xx-component when projecting onto lines through the origin
  4. P(3,6)P(3, -6) cannot be determined without additional information about how PP affects the standard basis vectors
Explanation: When you encounter linear projection problems, the key is understanding that projections onto lines through the origin follow specific geometric and algebraic rules that must be consistent with linearity. To find the projection formula onto the line y=xy = x, consider that any point projected onto this line must have equal xx and yy coordinates. The projection of vector (x,y)(x,y) onto line y=xy = x is the point on the line closest to (x,y)(x,y). Using the projection formula for the direction vector (1,1)(1,1), we get: P(x,y)=(x,y)(1,1)(1,1)(1,1)(1,1)=x+y2(1,1)=(x+y2,x+y2)P(x,y) = \frac{(x,y) \cdot (1,1)}{(1,1) \cdot (1,1)}(1,1) = \frac{x+y}{2}(1,1) = \left(\frac{x+y}{2}, \frac{x+y}{2}\right) Applying this to (3,6)(3,-6): P(3,6)=(3+(6)2,3+(6)2)=(1.5,1.5)P(3,-6) = \left(\frac{3+(-6)}{2}, \frac{3+(-6)}{2}\right) = (-1.5, -1.5) Choice A incorrectly claims that orthogonal vectors project to zero. While (3,6)(3,-6) is orthogonal to (1,1)(1,1), projections onto lines (not just vectors) don't map orthogonal vectors to the origin—they map them to points on the line. Choice C incorrectly assumes that linear transformations preserve the xx-component during projection, which isn't true for projections onto diagonal lines. Choice D suggests we need more information, but the projection formula is completely determined once we know we're projecting onto line y=xy = x. Remember: projections onto lines through the origin have unique, deterministic formulas that you can derive using the dot product projection formula. Don't confuse projections onto lines with projections onto single vectors.

Question 20

Consider a function T:R2R2T: \mathbb{R}^2 \to \mathbb{R}^2 defined by T(x,y)=(x+2y,3xy)T(x, y) = (x + 2y, 3x - y). Which property must be verified to confirm that TT is a linear transformation?

  1. TT preserves addition and scalar multiplication independently for all vectors in the domain (correct answer)
  2. TT maps the zero vector to itself and preserves the magnitude of all input vectors
  3. TT is bijective and preserves the dot product between any two vectors in the domain
  4. TT commutes with matrix multiplication and preserves linear independence of basis vectors
Explanation: A linear transformation must satisfy two fundamental properties: additivity (T(u + v) = T(u) + T(v)) and homogeneity (T(cu) = cT(u)). These can be combined as T(au + bv) = aT(u) + bT(v). Choice A correctly identifies these essential properties. Choice B incorrectly includes magnitude preservation, which is not required for linearity. Choice C confuses linear transformations with orthogonal transformations. Choice D mentions properties that are not part of the definition of linear transformations.