Topic 3 Linear Transformations

This topic explores mapping vectors from one space to another. Key skills include finding the "standard matrix" $A$ for a transformation $T$ (often based on geometric descriptions or given vector mappings), determining the range (column space), and evaluating properties like injectivity (one-to-one) and surjectivity (onto).

Question 1: Standard Matrix from Geometric Transformations (Rotation & Reflection)

Source: Practice Exam 4, Question 9
Exact Question: Let $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be the linear transformation that first rotates a point around the origin by $-\frac{\pi }{3}$ radians and then reflects it through the line $x_2=0$. Find the standard matrix $A$ of $T$.

Methodology:

1. Objective: Construct the $2\times 2$ standard matrix $A=\left[\begin{array}{cc}T({\mathbf{e}}_1)& T({\mathbf{e}}_2)\end{array}\right]$.
2. Process: Apply the geometric sequence to the standard basis vectors ${\mathbf{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right]$ and ${\mathbf{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$ individually.
   • Step A (Rotation): A rotation by angle $\theta$ maps $(1,0)$ to $(\cos \theta ,\sin \theta )$ and $(0,1)$ to $(-\sin \theta ,\cos \theta )$.
   • Step B (Reflection): The line $x_2=0$ is the $x_1$-axis (the horizontal axis). Reflecting across the horizontal axis flips the sign of the vertical component (the $y$ or $x_2$ value): $(x,y)\to (x,-y)$.

Solution:

• Track ${\mathbf{e}}_1$:
  1. Rotate by $-\pi /3$ (which is $-{60}^{\circ }$): $\left[\begin{array}{c}\cos (-\pi /3)\\ \sin (-\pi /3)\end{array}\right]=\left[\begin{array}{c}1/2\\ -\sqrt{3}/2\end{array}\right]$.
  2. Reflect through $x_2=0$: Flip the second component. $⟹T({\mathbf{e}}_1)=\left[\begin{array}{c}1/2\\ \sqrt{3}/2\end{array}\right]$.
• Track ${\mathbf{e}}_2$:
  1. Rotate by $-\pi /3$: $\left[\begin{array}{c}-\sin (-\pi /3)\\ \cos (-\pi /3)\end{array}\right]=\left[\begin{array}{c}\sqrt{3}/2\\ 1/2\end{array}\right]$.
  2. Reflect through $x_2=0$: Flip the second component. $⟹T({\mathbf{e}}_2)=\left[\begin{array}{c}\sqrt{3}/2\\ -1/2\end{array}\right]$.
• Assemble the Matrix:
  Final Answer:$$A=\left[\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}\end{array}\right]$$

Question 2: Standard Matrix from Geometric Transformations (Projection & Rotation)

Source: Exam 2, Question 5
Exact Question: Let $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be the linear transformation that first applies the orthogonal projection onto the line $x_2=-x_1$ and then rotates the resulting vector $\frac{\pi }{2}$ radians (i.e., 90 degrees) counterclockwise around the origin. Find the standard matrix $A$ of $T$.

Methodology:

1. Objective: Again, find $A=\left[\begin{array}{cc}T({\mathbf{e}}_1)& T({\mathbf{e}}_2)\end{array}\right]$.
2. Projection Formula: The projection of a vector $\mathbf{x}$ onto a line spanned by a vector $\mathbf{v}$ is ${\text{proj}}_{\mathbf{v}}(\mathbf{x})=\frac{\mathbf{x}\cdot \mathbf{v}}{\mathbf{v}\cdot \mathbf{v}}\mathbf{v}$.
   • Find the direction vector $\mathbf{v}$ for the line $x_2=-x_1$. A simple choice is $\mathbf{v}=\left[\begin{array}{c}1\\ -1\end{array}\right]$.
3. Rotation Logic: A ${90}^{\circ }$ counterclockwise rotation maps any vector $(x,y)$ to $(-y,x)$.

Solution:

• Track ${\mathbf{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right]$:
  1. Project onto $\mathbf{v}=\left[\begin{array}{c}1\\ -1\end{array}\right]$: $\frac{(1)(1)+(0)(-1)}{1^2+(-1{)}^2}\left[\begin{array}{c}1\\ -1\end{array}\right]=\frac{1}{2}\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}1/2\\ -1/2\end{array}\right]$.
  2. Rotate ${90}^{\circ }$ ccw: Swap coordinates and negate the first. $⟹T({\mathbf{e}}_1)=\left[\begin{array}{c}-(-1/2)\\ 1/2\end{array}\right]=\left[\begin{array}{c}1/2\\ 1/2\end{array}\right]$.
• Track ${\mathbf{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$:
  1. Project onto $\mathbf{v}$: $\frac{(0)(1)+(1)(-1)}{2}\left[\begin{array}{c}1\\ -1\end{array}\right]=-\frac{1}{2}\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}-1/2\\ 1/2\end{array}\right]$.
  2. Rotate ${90}^{\circ }$ ccw: Swap and negate the first. $⟹T({\mathbf{e}}_2)=\left[\begin{array}{c}-(1/2)\\ -1/2\end{array}\right]=\left[\begin{array}{c}-1/2\\ -1/2\end{array}\right]$.
• Assemble the Matrix:
  Final Answer:$$A=\left[\begin{array}{cc}1/2& -1/2\\ 1/2& -1/2\end{array}\right]$$

Question 3: Standard Matrix from Non-Standard Mapping Images

Source: Exam 2, Question 9
Exact Question: Consider the linear transformation $T:{\mathbb{R}}^2\to {\mathbb{R}}^3$, which satisfies $T\left(\left[\begin{array}{c}3\\ -1\end{array}\right]\right)=\left[\begin{array}{c}3\\ 11\\ 5\end{array}\right]$ and $T\left(\left[\begin{array}{c}2\\ -1\end{array}\right]\right)=\left[\begin{array}{c}5\\ 5\\ 4\end{array}\right]$.
a. Find the standard matrix $A$ of $T$.
b. Is $T$ injective? Motivate your answer.

Methodology:

1. Part A: You don't know $T({\mathbf{e}}_1)$ or $T({\mathbf{e}}_2)$ directly. Because $T$ is linear, $T(c_1\mathbf{u}+c_2\mathbf{v})=c_{1}T(\mathbf{u})+c_{2}T(\mathbf{v})$.
   • Express ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ as linear combinations of the given inputs $\mathbf{u}=\left[\begin{array}{c}3\\ -1\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{c}2\\ -1\end{array}\right]$ by solving augmented matrices.
   • Apply the transformation $T$ to those combinations to get $T({\mathbf{e}}_1)$ and $T({\mathbf{e}}_2)$.
2. Part B: A transformation $T(\mathbf{x})=A\mathbf{x}$ is injective (one-to-one) if $A\mathbf{x}=\mathbf{0}$ has only the trivial solution. This happens if and only if every column of $A$ has a pivot.

Solution:
Part a:

• Find ${\mathbf{e}}_1$ in terms of $\mathbf{u},\mathbf{v}$: Solve $[\mathbf{u}\mathbf{v}\mid {\mathbf{e}}_1]⟹\left[\begin{array}{ccc}3& 2& 1\\ -1& -1& 0\end{array}\right]$.
  Row reduce: $R_1\to R_1+2R_2⟹\left[\begin{array}{ccc}1& 0& 1\\ -1& -1& 0\end{array}\right]$.
  $R_2\to R_2+R_1⟹\left[\begin{array}{ccc}1& 0& 1\\ 0& -1& 1\end{array}\right]⟹\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\end{array}\right]$.
  So, ${\mathbf{e}}_1=1\mathbf{u}-1\mathbf{v}$.
• Compute $T({\mathbf{e}}_1)$: $T({\mathbf{e}}_1)=T(\mathbf{u})-T(\mathbf{v})=\left[\begin{array}{c}3\\ 11\\ 5\end{array}\right]-\left[\begin{array}{c}5\\ 5\\ 4\end{array}\right]=\left[\begin{array}{c}-2\\ 6\\ 1\end{array}\right]$.
• Find ${\mathbf{e}}_2$ in terms of $\mathbf{u},\mathbf{v}$: Solve $[\mathbf{u}\mathbf{v}\mid {\mathbf{e}}_2]⟹\left[\begin{array}{ccc}3& 2& 0\\ -1& -1& 1\end{array}\right]$.
  Row reduce: $R_1\to R_1+2R_2⟹\left[\begin{array}{ccc}1& 0& 2\\ -1& -1& 1\end{array}\right]$.
  $R_2\to R_2+R_1⟹\left[\begin{array}{ccc}1& 0& 2\\ 0& -1& 3\end{array}\right]⟹\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& -3\end{array}\right]$.
  So, ${\mathbf{e}}_2=2\mathbf{u}-3\mathbf{v}$.
• Compute $T({\mathbf{e}}_2)$: $T({\mathbf{e}}_2)=2T(\mathbf{u})-3T(\mathbf{v})=2\left[\begin{array}{c}3\\ 11\\ 5\end{array}\right]-3\left[\begin{array}{c}5\\ 5\\ 4\end{array}\right]=\left[\begin{array}{c}6\\ 22\\ 10\end{array}\right]-\left[\begin{array}{c}15\\ 15\\ 12\end{array}\right]=\left[\begin{array}{c}-9\\ 7\\ -2\end{array}\right]$.
  Answer (a): $A=\left[\begin{array}{cc}-2& -9\\ 6& 7\\ 1& -2\end{array}\right]$

Part b:

• Row reduce $A$ to check for pivots.
  $\left[\begin{array}{cc}1& -2\\ 6& 7\\ -2& -9\end{array}\right]$ (swapped rows for easier math, $R_3\leftrightarrow R_1$).
  $R_2\to R_2-6R_1⟹\left[\begin{array}{cc}1& -2\\ 0& 19\\ -2& -9\end{array}\right]$.
  $R_3\to R_3+2R_1⟹\left[\begin{array}{cc}1& -2\\ 0& 19\\ 0& -13\end{array}\right]$.
  It's clear both columns will have pivots. (Echelon form reduces to $\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$).
  Answer (b): Since $A$ has a pivot in every column, the equation $A\mathbf{x}=\mathbf{0}$ only has the trivial solution. Thus, $T$ is injective.

Question 4: Range of a Transformation

Source: Practice Exam 3, Question 3
Exact Question: Let $T:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be a linear transformation with standard matrix $A=\left[\begin{array}{ccc}-5& -25& 2\\ -5& -25& 1\\ -5& -25& 4\end{array}\right]$. Is the range of $T$ equal to ${\mathbb{R}}^3$? If not, determine a vector $\mathbf{b}\in {\mathbb{R}}^3$ that is not in the range of $T$.

Methodology:

1. Objective: Determine if the columns of $A$ span ${\mathbb{R}}^3$. If they do not, find a vector $\mathbf{b}=\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right]$ that makes the system $A\mathbf{x}=\mathbf{b}$ inconsistent (i.e., creates a contradictory row like $0=1$).
2. Process: Set up the augmented matrix $[A\mid \mathbf{b}]$ and perform row reduction until you obtain a row of zeros on the left side. The corresponding algebraic expression on the right side will dictate the condition for $\mathbf{b}$ to be in the range.

Solution:

• Step 1: Set up augmented matrix $[A\mid \mathbf{b}]$.$$\left[\begin{array}{cccc}-5& -25& 2& b_1\\ -5& -25& 1& b_2\\ -5& -25& 4& b_3\end{array}\right]$$
• Step 2: Row reduce. $R_2\to R_2-R_1$, $R_3\to R_3-R_1$.$$\left[\begin{array}{cccc}-5& -25& 2& b_1\\ 0& 0& -1& b_2-b_1\\ 0& 0& 2& b_3-b_1\end{array}\right]$$
• Step 3: Eliminate row 3 column 3. $R_3\to R_3+2R_2$.$$\left[\begin{array}{cccc}-5& -25& 2& b_1\\ 0& 0& -1& b_2-b_1\\ 0& 0& 0& b_3-b_1+2(b_2-b_1)\end{array}\right]$$
• Step 4: Simplify the right side of the bottom row. $b_3-b_1+2b_2-2b_1=b_3-3b_1+2b_2$.
• Conclusion: The system is only consistent (and thus $\mathbf{b}$ is only in the range) if $b_3-3b_1+2b_2=0$. Since we are restricted by this equation, the range is not all of ${\mathbb{R}}^3$. To find a vector outside the range, we just need to pick numbers such that $b_3-3b_1+2b_2\ne 0$.
  Let $b_1=1,b_2=0,b_3=0$. Then $0-3(1)+2(0)=-3\ne 0$.
  Final Answer: No, the range is not ${\mathbb{R}}^3$. A vector not in the range is $\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$.

Question 5: Mapping Properties (Injectivity/Surjectivity/Bijectivity)

Source: Practice Exam 3, Question 10
Exact Question: Give an example of linear transformations $T:{\mathbb{R}}^2\to {\mathbb{R}}^3$ and $S:{\mathbb{R}}^3\to {\mathbb{R}}^2$ in such a way that $T$ is injective, $S$ is surjective, but $S\circ T$ is not bijective. Motivate your answer.

Methodology:

1. Objective: Construct standard matrices $A$ (for $T$) and $B$ (for $S$) that fulfill the required rules.
   • $T:{\mathbb{R}}^2\to {\mathbb{R}}^3$ means $A$ is $3\times 2$. For $T$ to be injective, $A$ must have a pivot in every column (2 pivots total).
   • $S:{\mathbb{R}}^3\to {\mathbb{R}}^2$ means $B$ is $2\times 3$. For $S$ to be surjective, $B$ must have a pivot in every row (2 pivots total).
   • $S\circ T$ is represented by the matrix product $BA$. For it to be bijective (both injective and surjective), the resulting $2\times 2$ matrix must be invertible (have 2 pivots). We need to arrange $A$ and $B$ so that $BA$ has less than 2 pivots (rank $<2$).
2. Process: Pick very simple matrices (like parts of the identity matrix) and test the product.

Solution:

• Step 1: Construct Injective $T$. Let $A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$. Every column has a pivot.
• Step 2: Construct Surjective $S$. Let $B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\end{array}\right]$. Every row has a pivot.
• Step 3: Compute $S\circ T⟹BA$.$$BA=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$$
• Conclusion: The matrix $BA$ represents $S\circ T$. It is in echelon form and only has 1 pivot. Since a $2\times 2$ matrix needs 2 pivots to be invertible/bijective, this transformation is not bijective.
  Final Answer: Let $A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\end{array}\right]$. $T$ is injective (pivot in every column), $S$ is surjective (pivot in every row), but $BA=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$ which lacks a pivot in the second row/column, making it not bijective.