Linear Transformation Definition - Linear Algebra

Consider a function $P: \mathbb{R}^2 \to \mathbb{R}^2$ that projects vectors onto the line $y = x$ . If $P$ is linear, which of the following must be true about $P(3, -6)$ ?
Answer: $P(3, -6) = (-1.5, -1.5)$ because linearity requires that the projection formula $P(x,y) = \left(\frac{x+y}{2}, \frac{x+y}{2}\right)$ applies consistently 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 $y = x$ , consider that any point projected onto this line must have equal $x$ and $y$ coordinates. The projection of vector $(x,y)$ onto line $y = x$ is the point on the line closest to $(x,y)$ . Using the projection formula for the direction vector $(1,1)$ , we get: $P(x,y) = \frac{(x,y) \cdot (1,1)}{(1,1) \cdot (1,1)}(1,1) = \frac{x+y}{2}(1,1) = \left(\frac{x+y}{2}, \frac{x+y}{2}\right)$ Applying this to $(3,-6)$ : $P(3,-6) = \left(\frac{3+(-6)}{2}, \frac{3+(-6)}{2}\right) = (-1.5, -1.5)$ Choice A incorrectly claims that orthogonal vectors project to zero. While $(3,-6)$ is orthogonal to $(1,1)$ , 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 $x$ -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 $y = x$ . 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.
Let $T: \mathbb{R}^2 \to \mathbb{R}$ be a linear transformation. Which of the following functions $F: \mathbb{R}^2 \to \mathbb{R}$ is not guaranteed to be linear?
Answer: $F(\vec{v}) = (T(\vec{v}))^2$ When testing whether a function is linear, you need to verify two properties: additivity ( $F(\vec{u} + \vec{v}) = F(\vec{u}) + F(\vec{v})$ ) and scalar multiplication ( $F(c\vec{v}) = cF(\vec{v})$ ). The key insight is that combining linear transformations through linear operations preserves linearity, but nonlinear operations destroy it. Option A, $F(\vec{v}) = (T(\vec{v}))^2$ , fails both linearity tests. For additivity: $(T(\vec{u} + \vec{v}))^2 = (T(\vec{u}) + T(\vec{v}))^2 = T(\vec{u})^2 + 2T(\vec{u})T(\vec{v}) + T(\vec{v})^2$ , which doesn't equal $T(\vec{u})^2 + T(\vec{v})^2$ . The cross-term $2T(\vec{u})T(\vec{v})$ breaks linearity. This is the correct answer. Option B is linear because $T(\vec{v}) + T(-\vec{v}) = T(\vec{v}) + T(-1 \cdot \vec{v}) = T(\vec{v}) - T(\vec{v}) = 0$ for any linear transformation, making $F$ the zero function. Option C is clearly linear since scalar multiplication of a linear transformation remains linear: $2T(\vec{u} + \vec{v}) = 2(T(\vec{u}) + T(\vec{v})) = 2T(\vec{u}) + 2T(\vec{v})$ . Option D is linear because $F(\vec{v}) = T(\vec{v} - \vec{v}_0) + T(\vec{v}_0) = T(\vec{v}) - T(\vec{v}_0) + T(\vec{v}_0) = T(\vec{v})$ , which is just $T$ 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.
Consider the function $G: \mathbb{R}^2 \to \mathbb{R}$ defined by $G(x, y) = 3x - 4y + 5$ . A student argues that $G$ is 'almost linear' because it satisfies $G(c\mathbf{v}) = cG(\mathbf{v}) + 5(1-c)$ for any scalar $c$ and vector $\mathbf{v}$ . What is the fundamental issue with this reasoning?
Answer: The presence of the constant term 5 means $G$ is affine, not linear, and the modified homogeneity condition is irrelevant to linearity 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 $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ and $T(c\mathbf{v}) = cT(\mathbf{v})$ for all vectors and scalars. The function $G(x, y) = 3x - 4y + 5$ is an affine transformation, not a linear one, because of the constant term 5. This constant term prevents $G$ from satisfying true homogeneity: $G(c\mathbf{v}) = 3cx - 4cy + 5$ , while $cG(\mathbf{v}) = c(3x - 4y + 5) = 3cx - 4cy + 5c$ . These are only equal when $c = 1$ . The student's modified formula $G(c\mathbf{v}) = cG(\mathbf{v}) + 5(1-c)$ 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 $\mathbb{R}^n$ to $\mathbb{R}^m$ where $n \neq m$ . Study tip: Remember that any function with a constant term is affine, not linear. True linear transformations must pass through the origin, meaning $T(\mathbf{0}) = \mathbf{0}$ . When you see a constant term, immediately think "affine transformation."
Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $T(x, y, z) = (x + y, 2z)$ . A student wants to verify linearity by checking that $T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v})$ for specific vectors $\mathbf{u} = (1, 0, 1)$ , $\mathbf{v} = (0, 1, -1)$ , and scalars $a = 2, b = -3$ . What does this verification actually establish?
Answer: It confirms linearity for one specific case, but linearity requires this property to hold for all possible vectors and scalars When testing whether a transformation is linear, you need to understand what verification of the linearity condition $T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v})$ 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 $T(2(1,0,1) + (-3)(0,1,-1)) = T(2,-3,5) = (-1,10)$ and $2T(1,0,1) + (-3)T(0,1,-1) = 2(1,2) + (-3)(1,-2) = (-1,10)$ 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 $\mathbb{R}^3$ , 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.
Which of the following transformations $T: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation?
Answer: $T(x, y) = (2x - 3y, x + 5y)$ A transformation $T$ is linear if it satisfies $T(\vec{u}+\vec{v}) = T(\vec{u}) + T(\vec{v})$ and $T(c\vec{u}) = cT(\vec{u})$ . Choice B represents a matrix transformation $T(\vec{x}) = A\vec{x}$ with $A = \begin{pmatrix} 2 & -3 \\ 1 & 5 \end{pmatrix}$ , which is always linear. Choice A is not linear because $T(0, 0) = (0, -1) \neq \vec{0}$ . Choice C is not linear because of the product term $xy$ ; for example, $T(2(1,1)) = T(2,2) = (4,2)$ , but $2T(1,1) = 2(1,1) = (2,2)$ . Choice D is not linear because of the squared terms; for example, $T((1,0)+(1,0)) = T(2,0) = (4,0)$ , but $T(1,0)+T(1,0) = (1,0)+(1,0) = (2,0)$ .
Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be a linear transformation such that $T\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$ and $T\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) = \begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix}$ . What is the value of $T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right)$ ?
Answer: $\begin{bmatrix} 6 \\ -11 \\ 7 \end{bmatrix}$ 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, $T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v})$ . We have $\begin{bmatrix} 3 \\ -2 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ . Therefore, $T\left(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\right) = T\left(3\begin{bmatrix} 1 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) = 3T\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right) - 2T\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right)$ . Substituting the given values: $3\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ -3 \\ 9 \end{bmatrix} - \begin{bmatrix} 0 \\ 8 \\ 2 \end{bmatrix} = \begin{bmatrix} 6 \\ -11 \\ 7 \end{bmatrix}$ .
Let $V$ and $W$ be vector spaces. A function $T: V \to W$ is a linear transformation if it satisfies additivity and homogeneity. Which of the following conditions is a valid, alternative, single-property definition for linearity?
Answer: $T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v})$ for all vectors $\vec{u}, \vec{v} \in V$ and all scalars $c, d$ . The standard definition of a linear transformation requires two properties to be checked: additivity ( $T(\vec{u}+\vec{v})=T(\vec{u})+T(\vec{v})$ ) and homogeneity ( $T(c\vec{u})=cT(\vec{u})$ ). 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 $c=d=1$ , and one can prove homogeneity by setting $d=0$ . Conversely, if additivity and homogeneity hold, one can prove A: $T(c\vec{u} + d\vec{v}) = T(c\vec{u}) + T(d\vec{v}) = cT(\vec{u}) + dT(\vec{v})$ . Choices B, C, and D list properties that are necessary for linearity but are not sufficient on their own.
A student defines $L: \mathbb{R}^2 \to \mathbb{R}^3$ by $L(x, y) = (x - y, 0, x + 2y)$ and claims to have verified linearity by showing that $L(2\mathbf{e}_1) = 2L(\mathbf{e}_1)$ and $L(3\mathbf{e}_2) = 3L(\mathbf{e}_2)$ , where $\mathbf{e}_1 = (1, 0)$ and $\mathbf{e}_2 = (0, 1)$ . What critical verification is missing from this approach?
Answer: The student must verify that $L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v})$ for arbitrary vectors and scalars, not just basis vectors with specific scalars When you encounter questions about verifying linearity, remember that a linear transformation must satisfy two fundamental properties for ALL vectors and scalars: $L(c\mathbf{u}) = cL(\mathbf{u})$ (scalar multiplication) and $L(\mathbf{u} + \mathbf{v}) = L(\mathbf{u}) + L(\mathbf{v})$ (additivity). These can be combined into the single condition: $L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v})$ for arbitrary vectors and scalars. The student's approach is dangerously incomplete. Testing only $L(2\mathbf{e}_1) = 2L(\mathbf{e}_1)$ and $L(3\mathbf{e}_2) = 3L(\mathbf{e}_2)$ verifies linearity for just two specific cases with basis vectors. To prove linearity rigorously, you must demonstrate that $L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v})$ holds for any vectors $\mathbf{u}, \mathbf{v}$ and any scalars $a, b$ . This makes B correct. Option A is wrong because while $L(\mathbf{0}) = \mathbf{0}$ 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 $L(a\mathbf{u} + b\mathbf{v}) = aL(\mathbf{u}) + bL(\mathbf{v})$ using arbitrary vectors and scalars—this is the gold standard that encompasses all linearity requirements.
Consider a transformation $T: V \to W$ between two vector spaces. The condition $T(\vec{0}_V) = \vec{0}_W$ is a well-known property related to linear transformations. Which statement accurately describes the significance of this condition?
Answer: 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. The condition $T(\vec{0}_V) = \vec{0}_W$ is a necessary consequence of the homogeneity property of linear transformations ( $T(0 \cdot \vec{u}) = 0 \cdot T(\vec{u}) = \vec{0}_W$ ). 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, $T(x,y) = (x^2, y^2)$ satisfies $T(0,0)=(0,0)$ , but it is not a linear transformation. Thus, the condition is necessary but not sufficient.
Let $T: V \to W$ be a linear transformation. Which of the following statements is always true?
Answer: For any vectors $\vec{u}, \vec{v} \in V$ , the image of the line segment connecting them is the line segment connecting $T(\vec{u})$ and $T(\vec{v})$ . A point on the line segment between $\vec{u}$ and $\vec{v}$ can be parameterized as $\vec{p}(t) = (1-t)\vec{u} + t\vec{v}$ for $t \in [0,1]$ . Applying the linear transformation $T$ gives $T(\vec{p}(t)) = T((1-t)\vec{u} + t\vec{v}) = (1-t)T(\vec{u}) + tT(\vec{v})$ due to linearity. This expression parameterizes the line segment between $T(\vec{u})$ and $T(\vec{v})$ , so statement C is correct. Statement A is false; if $T$ 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 $\mathbb{R}^3$ to $\mathbb{R}^2$ . Statement D is false, e.g., the zero transformation $T(\vec{v}) = \vec{0}$ unless $W$ is the zero vector space.
A function $H: V \to W$ between vector spaces satisfies $H(\mathbf{u} + \mathbf{v}) = H(\mathbf{u}) + H(\mathbf{v})$ for all vectors $\mathbf{u}, \mathbf{v} \in V$ . Under what additional condition can we conclude that $H$ is linear?
Answer: $H$ must also satisfy $H(c\mathbf{v}) = cH(\mathbf{v})$ for all scalars $c$ and vectors $\mathbf{v} \in V$ 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 $H(\mathbf{u} + \mathbf{v}) = H(\mathbf{u}) + H(\mathbf{v})$ (additivity), but this alone isn't sufficient for linearity. The missing piece is scalar multiplication: $H(c\mathbf{v}) = cH(\mathbf{v})$ for all scalars $c$ and vectors $\mathbf{v}$ . Together, these two properties ensure that $H$ 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.
A transformation $R: \mathbb{R}^2 \to \mathbb{R}^2$ rotates every vector by $90°$ counterclockwise about the origin. Which approach most directly demonstrates that $R$ satisfies the definition of a linear transformation?
Answer: Show that $R(c\mathbf{v}) = cR(\mathbf{v})$ and $R(\mathbf{u} + \mathbf{v}) = R(\mathbf{u}) + R(\mathbf{v})$ using geometric properties of rotation 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 $R(\mathbf{u} + \mathbf{v}) = R(\mathbf{u}) + R(\mathbf{v})$ and homogeneity $R(c\mathbf{v}) = cR(\mathbf{v})$ . 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.
Let $P_2$ be the vector space of polynomials of degree at most 2. Which of the following transformations $T: P_2 \to \mathbb{R}$ is linear?
Answer: $T(p(x)) = p(2) - p'(1)$ 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 $p(x)=0$ , $T(0) = 0+1=1 \neq 0$ . (B) is not linear due to the square; $T(c \cdot p(x)) = \int_0^1 (c \cdot p(x))^2 dx = c^2 \int_0^1 (p(x))^2 dx = c^2 T(p(x))$ , which violates homogeneity. (C) is not linear due to the product; let $p(x)=cx$ . Then $p(0)=0$ and $p'(0)=c$ . $T(p(x))=0$ . But consider $q(x)=x+1$ . $T(q(x)) = q(0)q'(0) = 1 \cdot 1 = 1$ . $T(2q(x)) = T(2x+2) = (2)(2)=4$ , whereas $2T(q(x)) = 2(1) = 2$ . Homogeneity fails. (D) is linear because differentiation and evaluation are linear operations, and so is their linear combination. $T(p(x)+q(x)) = (p(2)+q(2))-(p'(1)+q'(1)) = (p(2)-p'(1)) + (q(2)-q'(1)) = T(p(x))+T(q(x))$ , and $T(c \cdot p(x)) = c \cdot p(2) - c \cdot p'(1) = c(p(2)-p'(1))=cT(p(x))$ .
A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ can be represented by the matrix equation $T(\mathbf{x}) = A\mathbf{x}$ for some matrix $A$ . Which statement about the relationship between this matrix representation and linearity is most accurate?
Answer: Every linear transformation can be represented this way, and every such matrix representation defines a linear transformation 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.
A function $F: \mathbb{R}^3 \to \mathbb{R}^2$ satisfies $F(2\mathbf{u} + 3\mathbf{v}) = 2F(\mathbf{u}) + 3F(\mathbf{v})$ for specific vectors $\mathbf{u}$ and $\mathbf{v}$ . What additional condition is necessary and sufficient to conclude that $F$ is linear?
Answer: The given property must hold for all vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$ and all scalars, not just the specific case shown The given condition shows linearity for one specific case, but linearity requires $F(a\mathbf{u} + b\mathbf{v}) = aF(\mathbf{u}) + bF(\mathbf{v})$ for ALL vectors $\mathbf{u}, \mathbf{v}$ in the domain and ALL scalars $a, b$ . Choice A correctly identifies this universal requirement. Choice B is incorrect because cross products are not preserved by general linear transformations (and the codomain $\mathbb{R}^2$ 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.
Which of the following geometric transformations of the Cartesian plane $\mathbb{R}^2$ is not a linear transformation?
Answer: A translation by the vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ . A key property of any linear transformation $T$ is that it must map the zero vector to the zero vector, i.e., $T(\vec{0}) = \vec{0}$ . A translation by a non-zero vector $\vec{v}_0$ is given by $T(\vec{x}) = \vec{x} + \vec{v}_0$ . For this transformation, $T(\vec{0}) = \vec{0} + \vec{v}_0 = \vec{v}_0$ . Since $\vec{v}_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \neq \vec{0}$ , 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.
Consider the transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x, y) = (x, |y|)$ . This transformation is not linear. Which of the following calculations correctly demonstrates its non-linearity?
Answer: For $c = -2$ and $\vec{u} = (1, 3)$ , we have $T(c\vec{u}) = T(-2, -6) = (-2, 6)$ , while $cT(\vec{u}) = -2T(1, 3) = -2(1, 3) = (-2, -6)$ . To determine if a transformation is linear, you must verify two properties: additivity ( $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$ ) and scalar multiplication ( $T(c\vec{u}) = cT(\vec{u})$ ). If either property fails, the transformation is not linear. Choice A correctly demonstrates non-linearity by testing scalar multiplication. With $c = -2$ and $\vec{u} = (1, 3)$ , we get $T(c\vec{u}) = T(-2, -6) = (-2, 6)$ since the absolute value makes $|-6| = 6$ . However, $cT(\vec{u}) = -2T(1, 3) = -2(1, 3) = (-2, -6)$ . Since $(-2, 6) \neq (-2, -6)$ , 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 $T(4, 6) = (4, 6)$ equals $T(1, 2) + T(3, 4) = (1, 2) + (3, 4) = (4, 6)$ , 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 $T(1, 2) = T(1, -2) = (1, 2)$ is correct, it doesn't address the linearity properties. Choice D correctly notes that $T(0, 0) = (0, 0)$ , 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.
Consider a function $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x, y) = (x + 2y, 3x - y)$ . Which property must be verified to confirm that $T$ is a linear transformation?
Answer: $T$ preserves addition and scalar multiplication independently for all vectors in the domain 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.
Let $M_{2 \times 2}$ be the vector space of $2 \times 2$ matrices with real entries. Which of the following transformations is linear?
Answer: $T: M_{2 \times 2} \to \mathbb{R}$ defined by $T(A) = \mathrm{trace}(A)$ The trace of a matrix is the sum of its diagonal elements. The trace is a linear transformation because $\mathrm{trace}(A+B) = \mathrm{trace}(A) + \mathrm{trace}(B)$ and $\mathrm{trace}(cA) = c \cdot \mathrm{trace}(A)$ . (A) The determinant is not linear; for a $2 \times 2$ matrix $A$ , $\det(cA) = c^2 \det(A) \neq c \det(A)$ in general. (B) Squaring a matrix is not a linear operation. For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , $(A^2)_{11} = a^2+bc$ . $T(kA) = (ka)^2 + (kb)(kc) = k^2(a^2+bc) = k^2 T(A)$ , which violates homogeneity. (C) Adding a fixed non-zero matrix (or vector) is a translation, which is not linear because $T(0) = 0+I = I \neq 0$ .
An affine transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is defined by the rule $T(\vec{x}) = A\vec{x} + \vec{b}$ , where $A$ is an $m \times n$ matrix and $\vec{b}$ is a fixed vector in $\mathbb{R}^m$ . For $T$ to be a linear transformation, which condition must be met?
Answer: The vector $\vec{b}$ must be the zero vector. For a transformation $T$ to be linear, it must satisfy $T(\vec{0}) = \vec{0}$ . Applying this to the affine transformation, we get $T(\vec{0}) = A\vec{0} + \vec{b} = \vec{0} + \vec{b} = \vec{b}$ . For $T(\vec{0})$ to be equal to $\vec{0}$ , we must have $\vec{b} = \vec{0}$ . If $\vec{b}=\vec{0}$ , the transformation becomes $T(\vec{x}) = A\vec{x}$ , which is the definition of a matrix transformation, a primary example of a linear transformation. The properties of matrix $A$ (being zero, invertible, etc.) do not determine whether the affine map is linear; only the translation vector $\vec{b}$ does.

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