Linear Transformation: Definition, Example, and Solution

STEP 1 - Given Data

$$A = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}, \quad v = \begin{bmatrix} -2 \ 1 \end{bmatrix}, \quad b = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}, \quad x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}, \quad T(x) = Ax$$

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Definition of Linear Transformation

A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation if for all $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ and all scalars $c$:

1. Additivity: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. Homogeneity: $T(c\mathbf{v}) = c,T(\mathbf{v})$

Example: For any $m \times n$ matrix $A$, the map $T(\mathbf{x}) = A\mathbf{x}$ is a linear transformation (a matrix transformation).

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STEP 2 - Solve

Part (a): Find $T(v)$

$$T(v) = Av = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix} \begin{bmatrix} -2 \ 1 \end{bmatrix}$$

• Row 1: $(1)(-2) + (-3)(1) = -2 - 3 = -5$
• Row 2: $(3)(-2) + (5)(1) = -6 + 5 = -1$
• Row 3: $(-1)(-2) + (7)(1) = 2 + 7 = 9$

$$\boxed{T(v) = \begin{bmatrix} -5 \ -1 \ 9 \end{bmatrix}}$$

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Part (b): Find $x \in \mathbb{R}^2$ with $T(x) = b$

Solve $Ax = b$:

$$\begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$$

Augmented matrix:

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 3 & 5 & 2 \ -1 & 7 & 1 \end{array}\right]$$

$R_2 \to R_2 - 3R_1$: $(0,\ 5+9,\ 2-9) = (0, 14, -7)$

$R_3 \to R_3 + R_1$: $(0,\ 7-3,\ 1+3) = (0, 4, 4)$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 14 & -7 \ 0 & 4 & 4 \end{array}\right]$$

$R_2 \to \tfrac{1}{14}R_2$: $(0, 1, -0.5)$

$R_3 \to R_3 - 4R_2$:

• Col 2: $4 - 4(1) = 0$
• Col 3 (RHS): $4 - 4(-0.5) = 4 + 2 = 6$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 1 & -0.5 \ 0 & 0 & 6 \end{array}\right]$$

The last row states $0 = 6$, which is impossible.

Verification of consistency using $R_3 - \frac{4}{14}R_2$ (before scaling): $$4 - \tfrac{4}{14}(-7) = 4 + 2 = 6 \neq 0$$

Same contradiction.

Conclusion: The system $Ax = b$ is inconsistent. There is no $x \in \mathbb{R}^2$ whose image under $T$ is $b = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$.

Geometric reason: The columns of $A$ span at most a 2-dimensional subspace (plane) of $\mathbb{R}^3$. The vector $b$ does not lie in $\text{Col}(A)$, so $b$ is not in the range of $T$.

$$\boxed{\text{No solution: } b \text{ is not in the range of } T.}$$

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State the conclusion: part (a) gives $T(v) = [-5, -1, 9]^T$, and the row reduction in part (b) reaches the contradiction $0 = 6$, so no such $x$ exists and the system is inconsistent.

STEP 1 - EXTRACT

Given data:

$$\mathbf{u} = \begin{bmatrix} 1 \ -3 \ 2 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix}$$

Tasks:

1. State the condition for equality of two column vectors in $\mathbb{R}^2$, with an example.
2. Compute $\mathbf{u} + 3\mathbf{v}$ and $\mathbf{u} - 2\mathbf{v}$.

Note: The vectors given are actually in $\mathbb{R}^3$ (three components), though the theory question refers to $\mathbb{R}^2$. The equality definition generalizes; I answer the theory for $\mathbb{R}^2$ as asked and compute using the given $\mathbb{R}^3$ vectors.

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STEP 2 - SOLVE

Part 1: Equality of two column vectors in $\mathbb{R}^2$

Two column vectors in $\mathbb{R}^2$ are equal if and only if their corresponding components are equal.

If $$\mathbf{u} = \begin{bmatrix} u_1 \ u_2 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} v_1 \ v_2 \end{bmatrix}$$ then $$\mathbf{u} = \mathbf{v} \iff u_1 = v_1 \ \text{and}\ u_2 = v_2.$$

Example: $$\begin{bmatrix} 3 \ -2 \end{bmatrix} = \begin{bmatrix} 3 \ -2 \end{bmatrix} \quad \text{(equal, matching entries)}$$ but $$\begin{bmatrix} 3 \ -2 \end{bmatrix} \neq \begin{bmatrix} -2 \ 3 \end{bmatrix} \quad \text{(order matters, entries differ)}.$$

Part 2: Compute $\mathbf{u} + 3\mathbf{v}$

$$3\mathbf{v} = 3\begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix} = \begin{bmatrix} 3 \ -3 \ 9 \end{bmatrix}$$

$$\mathbf{u} + 3\mathbf{v} = \begin{bmatrix} 1 \ -3 \ 2 \end{bmatrix} + \begin{bmatrix} 3 \ -3 \ 9 \end{bmatrix} = \begin{bmatrix} 4 \ -6 \ 11 \end{bmatrix}$$

Compute $\mathbf{u} - 2\mathbf{v}$

$$2\mathbf{v} = \begin{bmatrix} 2 \ -2 \ 6 \end{bmatrix}$$

$$\mathbf{u} - 2\mathbf{v} = \begin{bmatrix} 1 \ -3 \ 2 \end{bmatrix} - \begin{bmatrix} 2 \ -2 \ 6 \end{bmatrix} = \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix}$$

Final Results

$$\mathbf{u} + 3\mathbf{v} = \begin{bmatrix} 4 \ -6 \ 11 \end{bmatrix}, \qquad \mathbf{u} - 2\mathbf{v} = \begin{bmatrix} -1 \ -1 \ -4 \end{bmatrix}$$

Least Squares Line: $y = a_0 + a_1x$

Given data

Points: $(2,1), (5,2), (7,3), (8,3)$, with $n = 4$.

Normal Equations

$$\sum y = na_0 + a_1\sum x$$ $$\sum xy = a_0\sum x + a_1\sum x^2$$

Computing Sums

Substituting

$$9 = 4a_0 + 22a_1 \quad (1)$$ $$57 = 22a_0 + 142a_1 \quad (2)$$

Solving

Multiply (1) by 22, (2) by 4:

$$198 = 88a_0 + 484a_1 \quad (3)$$ $$228 = 88a_0 + 568a_1 \quad (4)$$

Subtract (3) from (4):

$$30 = 84a_1 \implies a_1 = \frac{30}{84} = \frac{5}{14} \approx 0.357$$

Substitute into (1):

$$9 = 4a_0 + 22\cdot\frac{5}{14} = 4a_0 + \frac{55}{7}$$ $$4a_0 = 9 - \frac{55}{7} = \frac{8}{7} \implies a_0 = \frac{2}{7} \approx 0.286$$

Result

$$\boxed{y = \frac{2}{7} + \frac{5}{14}x \approx 0.286 + 0.357x}$$

Definition of a Group

Definition: A non-empty set G together with a binary operation * is called a group, denoted (G, *), if the following axioms are satisfied:

1. Closure: For all a, b ∈ G, a * b ∈ G
2. Associativity: For all a, b, c ∈ G, (a * b) * c = a * (b * c)
3. Identity: There exists an element e ∈ G such that a * e = e * a = a for all a ∈ G
4. Inverse: For every a ∈ G, there exists an element a⁻¹ ∈ G such that a * a⁻¹ = a⁻¹ * a = e

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Proof: (Z, +) is a Group

Claim: The set of integers Z = {..., -2, -1, 0, 1, 2, ...} forms a group under the addition operation (+).

We verify all four group axioms:

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Axiom 1: Closure

For all a, b ∈ Z, the sum a + b is also an integer.

Example: 3 + (-5) = -2 ∈ Z, 7 + 4 = 11 ∈ Z

Therefore, + is a binary operation on Z, and Z is closed under addition. ✓

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Axiom 2: Associativity

For all a, b, c ∈ Z:

$$(a + b) + c = a + (b + c)$$

Example: (2 + 3) + 4 = 5 + 4 = 9 and 2 + (3 + 4) = 2 + 7 = 9

Since addition of integers is always associative, this holds for all a, b, c ∈ Z. ✓

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Axiom 3: Existence of Identity

There exists 0 ∈ Z such that for all a ∈ Z:

$$a + 0 = 0 + a = a$$

Example: 5 + 0 = 0 + 5 = 5

Therefore, 0 is the additive identity in Z. ✓

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Axiom 4: Existence of Inverse

For every a ∈ Z, there exists (-a) ∈ Z such that:

$$a + (-a) = (-a) + a = 0$$

Example: For a = 7, the inverse is -7 ∈ Z, and 7 + (-7) = 0

Since for every integer a, its negative (-a) is also an integer, every element has an inverse in Z. ✓

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Conclusion

Since all four group axioms -- Closure, Associativity, Identity, and Inverse -- are satisfied, we conclude that:

$$\boxed{(\mathbb{Z},\ +) \text{ is a group.}}$$

Note: Since a + b = b + a for all a, b ∈ Z (commutativity also holds), (Z, +) is in fact an Abelian (commutative) group.

Definition of Ring and Proof that (Z⁺, +, ×) is Not a Ring

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Definition of Ring

An algebraic structure (R, +, ×) with two binary operations, addition (+) and multiplication (×), is called a ring if it satisfies the following conditions:

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Proof: (Z⁺, +, ×) is NOT a Ring

Let Z⁺ = {1, 2, 3, 4, ...} be the set of positive integers.

To show that (Z⁺, +, ×) is not a ring, it is sufficient to show that at least one condition of a ring is violated.

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Checking Condition 3: Existence of Additive Identity

For (Z⁺, +, ×) to be a ring, there must exist an element 0 ∈ Z⁺ such that:

a + 0 = 0 + a = a, for all a ∈ Z⁺

But 0 ∉ Z⁺, since Z⁺ = {1, 2, 3, ...} contains only positive integers.

Therefore, no additive identity exists in Z⁺.

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Checking Condition 4: Existence of Additive Inverse

For (Z⁺, +, ×) to be a ring, for every a ∈ Z⁺, there must exist −a ∈ Z⁺ such that:

a + (−a) = 0

Example: Take a = 3 ∈ Z⁺. Then we need −3 ∈ Z⁺.

But −3 ∉ Z⁺, since Z⁺ contains no negative integers.

Therefore, additive inverses do not exist in Z⁺.

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Conclusion

Since the set of positive integers Z⁺ fails to satisfy:

• the existence of an additive identity (0 ∉ Z⁺), and
• the existence of additive inverses (−a ∉ Z⁺ for any a ∈ Z⁺),

the algebraic structure (Z⁺, +, ×) is NOT a ring. $\blacksquare$