Scaling & Stretching - Linear Algebra

1. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is defined by the matrix $A = \begin{pmatrix} 5 & 0 \\ 0 & 0.2 \end{pmatrix}$. Which statement correctly describes the eigenvalues and eigenvectors of this transformation?
   Answer: The eigenvalues are 5 and 0.2, with corresponding eigenvectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$. For a diagonal matrix, the eigenvalues are the entries on the main diagonal. In this case, the eigenvalues are $\lambda_1 = 5$ and $\lambda_2 = 0.2$. The eigenvectors of a diagonal scaling matrix are the standard basis vectors. The vector $\vec{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is scaled by 5, so it is the eigenvector for $\lambda_1 = 5$. The vector $\vec{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ is scaled by 0.2, so it is the eigenvector for $\lambda_2 = 0.2$.
2. Consider the linear transformation $S: \mathbb{R}^3 \to \mathbb{R}^3$ that stretches by a factor of 4 along the z-axis while leaving the x and y coordinates unchanged. If the vector $\vec{u} = (2, -1, 3)$ is transformed to $S(\vec{u})$, and then $S(\vec{u})$ is transformed again by the same transformation $S$, what is the resulting vector?
   Answer: $(2, -1, 48)$ after applying $S$ twice sequentially The transformation matrix is $S = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix}$. First application: $S(\vec{u}) = (2, -1, 12)$. Second application: $S(S(\vec{u})) = S(2, -1, 12) = (2, -1, 48)$. The composition $S \circ S$ has the effect of stretching by factor $4^2 = 16$ along the z-axis. Choice A gives the correct result but incorrectly describes it as composition notation. Choice B incorrectly applies scaling to all components. Choice D incorrectly applies scaling to x and y components.
3. Let $S$ be a linear transformation that scales the y-coordinate by a factor of 4, and let $T$ be a linear transformation that scales the x-coordinate by a factor of $1/2$. What is the standard matrix for the composite transformation $T \circ S$ (which applies $S$ first)?
   Answer: $\begin{pmatrix} 1/2 & 0 \\ 0 & 4 \end{pmatrix}$ The standard matrix for $S$ (scaling y by 4) is $A_S = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}$. The standard matrix for $T$ (scaling x by $1/2$) is $A_T = \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix}$. The composite transformation $T \circ S$ is represented by the matrix product $A_T A_S$. $A_T A_S = \begin{pmatrix} 1/2 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} (1/2)(1) + (0)(0) & (1/2)(0) + (0)(4) \\ (0)(1) + (1)(0) & (0)(0) + (1)(4) \end{pmatrix} = \begin{pmatrix} 1/2 & 0 \\ 0 & 4 \end{pmatrix}$
4. Consider the transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $T(x,y) = (ax, by)$ where $a, b > 0$. If $T$ maps the triangle with vertices $(0,0)$, $(4,0)$, and $(0,6)$ to a triangle with area 36 square units, and the transformed triangle has a base (along the x-axis) that is twice as long as its height (along the y-axis), what are the values of $a$ and $b$?
   Answer: $a = 3$ and $b = 1$, satisfying both the area and proportion conditions The original triangle has vertices $(0,0)$, $(4,0)$, $(0,6)$ with area $\frac{1}{2} \cdot 4 \cdot 6 = 12$. After transformation: $(0,0)$, $(4a,0)$, $(0,6b)$. The new area is $\frac{1}{2} \cdot 4a \cdot 6b = 12ab = 36$, so $ab = 3$. The base is $4a$ and height is $6b$. The condition "base is twice the height" gives $4a = 2(6b)$, so $4a = 12b$, thus $a = 3b$. Substituting: $(3b) \cdot b = 3$, so $3b^2 = 3$, giving $b = 1$ and $a = 3$. Choices B and D have incorrect calculations. Choice C gives the same answer but with less precise reasoning in the description.
5. A scaling transformation $S$ stretches vectors by factor $m$ along a line making angle $\theta$ with the positive x-axis, while leaving vectors perpendicular to this line unchanged. If $\theta = 45°$ and $S$ maps the vector $\vec{v} = (1,1)$ to $\vec{w} = (3,3)$, what does $S$ do to the vector $\vec{u} = (1,-1)$?
   Answer: Maps $\vec{u}$ to $(1,-1)$ since it's perpendicular to the stretching direction At $\theta = 45°$, the stretching direction is along $(1,1)$ (or any scalar multiple). Since $\vec{v} = (1,1)$ maps to $(3,3)$, the scaling factor $m = 3$ along this direction. The vector $\vec{u} = (1,-1)$ is perpendicular to $(1,1)$ since their dot product is $1 \cdot 1 + (-1) \cdot 1 = 0$. Vectors perpendicular to the stretching direction remain unchanged, so $S(\vec{u}) = (1,-1)$. Choice B incorrectly applies uniform scaling. Choice C incorrectly suggests rotation. Choice D gives the correct answer but incorrectly describes the reasoning (invariant direction vs. perpendicular to stretching direction).
6. The rectangle with vertices at $(0,0), (-6,0), (-6,5),$ and $(0,5)$ is the image of the unit square (vertices at $(0,0), (1,0), (1,1), (0,1)$) under a linear transformation $T$ that scales along the coordinate axes. What is the standard matrix for $T$?
   Answer: $\begin{pmatrix} -6 & 0 \\ 0 & 5 \end{pmatrix}$ Let the scaling matrix be $A = \begin{pmatrix} k_x & 0 \\ 0 & k_y \end{pmatrix}$. The transformation maps the vertices of the unit square to the vertices of a rectangle. Specifically, $T(1,0) = (k_x, 0)$ and $T(0,1) = (0, k_y)$. The vertices of the image rectangle are $(0,0), (-6,0), (0,5),$ and $(-6,5)$. The two vectors defining the sides of the rectangle from the origin are $(-6,0)$ and $(0,5)$. These must be the images of the standard basis vectors. Thus, $T(1,0) = (-6,0)$ and $T(0,1) = (0,5)$. This implies $k_x = -6$ and $k_y = 5$. The standard matrix is therefore $\begin{pmatrix} -6 & 0 \\ 0 & 5 \end{pmatrix}$.
7. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation represented by the standard matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}$. Which statement best describes the geometric effect of this transformation?
   Answer: A vertical stretch by a factor of 4 and no change horizontally. The matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}$ transforms a vector $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ to $A\vec{v} = \begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ 4y \end{pmatrix}$. This means the x-coordinate is unchanged, while the y-coordinate is multiplied by 4. This corresponds to a vertical stretch by a factor of 4.
8. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ scales vectors along the coordinate axes. Given that $T\left(\begin{pmatrix} 3 \\ -2 \end{pmatrix}\right) = \begin{pmatrix} -9 \\ -1 \end{pmatrix}$, what is the scaling factor for the x-axis?
   Answer: -3 Let the standard matrix for the scaling be $A = \begin{pmatrix} k_x & 0 \\ 0 & k_y \end{pmatrix}$. Applying this to the given vector yields $\begin{pmatrix} k_x & 0 \\ 0 & k_y \end{pmatrix} \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3k_x \\ -2k_y \end{pmatrix}$. We are given that this result is equal to $\begin{pmatrix} -9 \\ -1 \end{pmatrix}$. By comparing the components, we get two equations: $3k_x = -9$ and $-2k_y = -1$. Solving the first equation gives $k_x = -9/3 = -3$. The scaling factor for the x-axis is -3.
9. A linear transformation $T$ in $\mathbb{R}^2$ compresses vectors horizontally to one-third of their original length and stretches them vertically to five times their original length. What is the standard matrix for $T$?
   Answer: $\begin{pmatrix} 1/3 & 0 \\ 0 & 5 \end{pmatrix}$ A horizontal compression by a factor of $1/3$ means the x-coordinate is multiplied by $1/3$. A vertical stretch by a factor of 5 means the y-coordinate is multiplied by 5. The transformation is $T(x, y) = (x/3, 5y)$. The standard matrix is formed by applying the transformation to the standard basis vectors. $T(\vec{e}_1) = T(1,0) = (1/3, 0)$ and $T(\vec{e}_2) = T(0,1) = (0, 5)$. These form the columns of the matrix, which is $\begin{pmatrix} 1/3 & 0 \\ 0 & 5 \end{pmatrix}$.
10. A transformation $T$ stretches any vector in $\mathbb{R}^2$ by a factor of 4 horizontally and compresses it to one-half its length vertically. Which transformation $T^{-1}$ maps the transformed vectors back to their original positions?
    Answer: A horizontal compression by $1/4$ and a vertical stretch by 2. The original transformation $T$ has matrix $A = \begin{pmatrix} 4 & 0 \\ 0 & 1/2 \end{pmatrix}$. The inverse transformation $T^{-1}$ is represented by the inverse matrix $A^{-1}$. For a diagonal matrix, the inverse is found by taking the reciprocal of each diagonal entry: $A^{-1} = \begin{pmatrix} 1/4 & 0 \\ 0 & 2 \end{pmatrix}$. This matrix corresponds to multiplying the x-coordinate by $1/4$ (a horizontal compression) and the y-coordinate by 2 (a vertical stretch).
11. A linear transformation $T(\vec{x}) = A\vec{x}$ uses the matrix $A = \begin{pmatrix} 3 & 0 \\ 0 & 1/2 \end{pmatrix}$. What is the equation of the image of the line $y = 2x + 4$ under this transformation?
    Answer: $y = \frac{1}{3}x + 2$ When you encounter a linear transformation applied to a curve, you need to understand how the transformation affects individual points, then reconstruct the transformed curve's equation. To find how the line $y = 2x + 4$ transforms under $T(\vec{x}) = A\vec{x}$ where $A = \begin{pmatrix} 3 & 0 \\ 0 & 1/2 \end{pmatrix}$, start with a general point $(x, y)$ on the original line. Since $y = 2x + 4$, any point can be written as $(x, 2x + 4)$. The transformation maps this point to: $A\begin{pmatrix} x \\ 2x + 4 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 1/2 \end{pmatrix}\begin{pmatrix} x \\ 2x + 4 \end{pmatrix} = \begin{pmatrix} 3x \\ \frac{1}{2}(2x + 4) \end{pmatrix} = \begin{pmatrix} 3x \\ x + 2 \end{pmatrix}$ So the transformed point is $(3x, x + 2)$. To find the equation relating these new coordinates, let $X = 3x$ and $Y = x + 2$. From $X = 3x$, we get $x = \frac{X}{3}$. Substituting into $Y = x + 2$: $Y = \frac{X}{3} + 2$, or $y = \frac{1}{3}x + 2$. Choice A ($y = \frac{1}{3}x + 4$) incorrectly keeps the original y-intercept unchanged, forgetting that the transformation scales the y-coordinates. Choice B ($y = 12x + 8$) mistakenly multiplies everything by factors from the matrix. Choice C ($y = 3x + 2$) confuses which coordinate gets which scaling factor. Remember: when transforming curves, apply the transformation to general points on the curve, then eliminate parameters to find the new equation. Don't just transform coefficients directly.
12. A linear transformation is represented by the matrix $A = \begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}$. This transformation is applied to a parallelogram defined by the vectors $\vec{u} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ originating from the origin. What is the area of the resulting parallelogram?
    Answer: 10 When you encounter a problem about linear transformations and areas, remember that the determinant of the transformation matrix tells you how areas scale under the transformation. To find the area of the transformed parallelogram, you need two pieces: the original area and the scaling factor. First, find the original area using the cross product formula. The area of a parallelogram formed by vectors $\vec{u} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ is the absolute value of their determinant: $|3 \cdot 4 - 1 \cdot 2| = |12 - 2| = 10$. Next, calculate how the transformation scales areas by finding $\det(A) = 2 \cdot \frac{1}{2} - 0 \cdot 0 = 1$. Since the determinant equals 1, areas are preserved under this transformation. Therefore, the transformed parallelogram has area $10 \times 1 = 10$. Looking at the wrong answers: Choice A (5) likely comes from forgetting to take the absolute value or making an arithmetic error in the cross product. Choice B (1) represents just the determinant of the transformation matrix, ignoring the original parallelogram entirely. Choice C (20) might result from incorrectly adding the determinants rather than multiplying the original area by the scaling factor. The key insight is that linear transformations scale all areas by the absolute value of the determinant. Always compute the original area first, then multiply by $|\det(A)|$ to find the transformed area.
13. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation that scales the x-coordinate by a factor of 6 and the y-coordinate by a factor of $1/2$. If a square with area 4 is transformed by $T$, what is the area of the resulting figure?
    Answer: 12 The standard matrix for this transformation is $A = \begin{pmatrix} 6 & 0 \\ 0 & 1/2 \end{pmatrix}$. The determinant of this matrix is $\det(A) = (6)(1/2) = 3$. The area of a transformed region is the original area multiplied by the absolute value of the determinant of the transformation matrix. Therefore, the new area is $4 \times |\det(A)| = 4 \times 3 = 12$.
14. Consider the non-uniform scaling transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ with standard matrix $A = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$. The shape of which of the following geometric figures is NOT preserved under $T$?
    Answer: A circle centered at the origin. A non-uniform scaling stretches or compresses the plane by different factors in different directions. A circle, defined by $x^2 + y^2 = r^2$, is characterized by all its points being equidistant from the center. Under this transformation, $x' = 4x$ and $y' = 2y$, so $x = x'/4$ and $y=y'/2$. The equation becomes $(\frac{x'}{4})^2 + (\frac{y'}{2})^2 = r^2$, or $\frac{x'^2}{16} + \frac{y'^2}{4} = r^2$. This is the equation of an ellipse, not a circle. Thus, the shape of a circle is not preserved. A square becomes a rectangle, a line segment becomes another line segment, and a parallelogram becomes another parallelogram.

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