Linear Transformation

Definition

A mapping $T: V \to W$ (where $V, W$ are vector spaces) is a linear transformation if for all vectors $\mathbf{u}, \mathbf{v}$ in $V$ and all scalars $c$:

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

Example: For any $m \times n$ matrix $A$, define $T(\mathbf{x}) = A\mathbf{x}$. This is linear since $A(\mathbf{u}+\mathbf{v}) = A\mathbf{u}+A\mathbf{v}$ and $A(c\mathbf{u}) = c(A\mathbf{u})$.

Given Data

$$A = \begin{bmatrix} 1 & -3 \ 3 & 5 \ -1 & 7 \end{bmatrix},\quad v = \begin{bmatrix} 2 \ -1 \end{bmatrix},\quad b = \begin{bmatrix} 3 \ 2 \ 4 \end{bmatrix},\quad x = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}$$

Note: Since $A$ is $3\times 2$, the transformation is actually $T:\mathbb{R}^2 \to \mathbb{R}^3$ (the problem statement's $\mathbb{R}^2 \to \mathbb{R}^2$ is a typo).

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: $(-1)(2)+(7)(-1) = -2-7 = -9$

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

Part (b): Find $x$ with $T(x) = b$

Solve $Ax = b$ via the augmented matrix:

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 3 & 5 & 2 \ -1 & 7 & 4 \end{array}\right]$$

$R_2 \to R_2 - 3R_1$: $[0,\ 14,\ -7]$ $R_3 \to R_3 + R_1$: $[0,\ 4,\ 7]$

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

$R_2 \to \tfrac{1}{14}R_2$: $[0,\ 1,\ -\tfrac12]$

$R_3 \to R_3 - 4R_2$:

• Column 2: $4 - 4(1) = 0$
• RHS: $7 - 4(-\tfrac12) = 7 + 2 = 9$

$$\left[\begin{array}{cc|c} 1 & -3 & 3 \ 0 & 1 & -\frac{1}{2} \ 0 & 0 & 9 \end{array}\right]$$

The last row reads $0 = 9$, which is a contradiction.

Conclusion: The system $Ax = b$ is inconsistent. There is no $x \in \mathbb{R}^2$ whose image under $T$ equals $b = \begin{bmatrix}3\2\4\end{bmatrix}$; that is, $b$ is not in the range (column space) of $T$.

(Remark: A textbook-consistent version of this classic problem uses $b = \begin{bmatrix}3\2\-5\end{bmatrix}$, giving $x_1 = \tfrac32,\ x_2 = -\tfrac12$. With the stated $b = \begin{bmatrix}3\2\4\end{bmatrix}$, the system has no solution.)

Equality of Column Vectors and Vector Operations

STEP 1 - Given Data

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

Required: definition of vector equality with example; compute $\mathbf{u} + 3\mathbf{v}$ and $\mathbf{u} - 2\mathbf{v}$.

Note: The header says $\mathbb{R}^2$, but the given vectors have 3 components, so they lie in $\mathbb{R}^3$. The equality definition is stated generally; the arithmetic is done component-wise.

STEP 2 - Solution

Part 1: When are two column vectors equal?

Two column vectors are equal if and only if they have the same number of components and their corresponding components are equal.

For $\mathbb{R}^2$: if $\mathbf{a} = \begin{bmatrix} a_1 \ a_2 \end{bmatrix}$, $\mathbf{b} = \begin{bmatrix} b_1 \ b_2 \end{bmatrix}$, then

$$\mathbf{a} = \mathbf{b} \iff a_1 = b_1 \text{ and } a_2 = b_2$$

Example:

$$\begin{bmatrix} 2 \ 7 \end{bmatrix} = \begin{bmatrix} 2 \ 7 \end{bmatrix}$$

since $2 = 2$ and $7 = 7$. But $\begin{bmatrix} 2 \ 7 \end{bmatrix} \neq \begin{bmatrix} 7 \ 2 \end{bmatrix}$, since order matters.

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

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

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

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

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

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

Final Results

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

Eigenvalues of the Matrix

STEP 1 - EXTRACT: Given Data

$$A = \begin{bmatrix} 6 & 3 & -8 \ 0 & -2 & 0 \ 1 & 0 & -3 \end{bmatrix}$$

All entries are readable and complete.

STEP 2 - SOLVE

Characteristic Equation

$$\det(A - \lambda I) = 0$$

$$A - \lambda I = \begin{bmatrix} 6-\lambda & 3 & -8 \ 0 & -2-\lambda & 0 \ 1 & 0 & -3-\lambda \end{bmatrix}$$

Determinant (expand along second row, two zeros)

$$\det(A - \lambda I) = (-2-\lambda)\det\begin{bmatrix} 6-\lambda & -8 \ 1 & -3-\lambda \end{bmatrix}$$

Compute the 2×2 determinant:

$$(6-\lambda)(-3-\lambda) - (-8)(1) = (-18 - 6\lambda + 3\lambda + \lambda^2) + 8 = \lambda^2 - 3\lambda - 10$$

So:

$$\det(A - \lambda I) = (-2-\lambda)(\lambda^2 - 3\lambda - 10) = 0$$

Factor

$$\lambda^2 - 3\lambda - 10 = (\lambda - 5)(\lambda + 2)$$

Thus:

$$(-2-\lambda)(\lambda - 5)(\lambda + 2) = 0$$ $$-(\lambda + 2)^2(\lambda - 5) = 0$$

Eigenvalues

$$\boxed{\lambda = -2 \text{ (multiplicity 2)}, \quad \lambda = 5}$$

Verification (trace and determinant check):

• Sum of eigenvalues: $-2 - 2 + 5 = 1$; trace of $A = 6 + (-2) + (-3) = 1$ ✓
• Product of eigenvalues: $(-2)(-2)(5) = 20$; $\det(A)$: expand along row 2 gives $(-2)[(6)(-3) - (-8)(1)] = (-2)(-18+8) = (-2)(-10) = 20$ ✓

Both checks confirm the result.

Null Space of a Matrix

The null space of an $m \times n$ matrix $A$, written $N(A)$ or $\text{Null}(A)$, is the set of all vectors $\mathbf{x}$ in $\mathbb{R}^n$ that satisfy the homogeneous equation $A\mathbf{x} = \mathbf{0}$:

$$N(A) = {, \mathbf{x} \in \mathbb{R}^n \mid A\mathbf{x} = \mathbf{0} ,}$$

so the null space is exactly the solution set of $A\mathbf{x} = \mathbf{0}$. Its vectors carry as many components as $A$ has columns, while the zero vector on the right carries as many components as $A$ has rows.

The null space is always a subspace of $\mathbb{R}^n$. It contains the zero vector because $A\mathbf{0} = \mathbf{0}$, it is closed under addition because $A\mathbf{u} = \mathbf{0}$ and $A\mathbf{v} = \mathbf{0}$ give

$$A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} = \mathbf{0}$$

and it is closed under scalar multiplication because $A\mathbf{u} = \mathbf{0}$ gives

$$A(c\mathbf{u}) = c(A\mathbf{u}) = \mathbf{0}$$

Its dimension is called the nullity of $A$, and by the rank-nullity theorem the rank and the nullity add up to $n$, the number of columns.

Showing that $v$ lies in $\text{Null}(A)$

We are given

$$A = \begin{bmatrix} 1 & -3 & -2 \ -5 & 9 & 1 \end{bmatrix}, \qquad v = \begin{bmatrix} 5 \ 3 \ -2 \end{bmatrix}$$

Here $A$ has three columns, so the vectors of its null space live in $\mathbb{R}^3$, and $v$ has three components, which makes the test meaningful. Membership needs no row reduction at all, because it is settled by the single defining condition $Av = \mathbf{0}$. Forming the product row by row,

$$Av = \begin{bmatrix} 1 & -3 & -2 \ -5 & 9 & 1 \end{bmatrix} \begin{bmatrix} 5 \ 3 \ -2 \end{bmatrix}$$

The first row gives

$$(1)(5) + (-3)(3) + (-2)(-2) = 5 - 9 + 4 = 0$$

and the second row gives

$$(-5)(5) + (9)(3) + (1)(-2) = -25 + 27 - 2 = 0$$

so that

$$Av = \begin{bmatrix} 0 \ 0 \end{bmatrix} = \mathbf{0}$$

Since $Av = \mathbf{0}$, the vector $v$ satisfies the defining condition of the null space. Therefore $v \in \text{Null}(A)$, which is what the question asked us to show.

Verifying Linear Independence of $1^k$, $(-2)^k$, $3^k$

STEP 1 - Given Data

Three discrete-time signals:

• $f_1(k) = 1^k$
• $f_2(k) = (-2)^k$
• $f_3(k) = 3^k$

Task: verify they are linearly independent.

STEP 2 - Solve

Definition

Signals are linearly independent if $$C_1 f_1(k) + C_2 f_2(k) + C_3 f_3(k) = 0 \quad \text{for all } k$$ forces $C_1 = C_2 = C_3 = 0$.

Set up the equation

$$C_1(1)^k + C_2(-2)^k + C_3(3)^k = 0 \quad \forall k$$

Substitute $k = 0, 1, 2$:

$k=0$: $\quad C_1 + C_2 + C_3 = 0 \quad (1)$

$k=1$: $\quad C_1 - 2C_2 + 3C_3 = 0 \quad (2)$

$k=2$: $\quad C_1 + 4C_2 + 9C_3 = 0 \quad (3)$

Coefficient matrix and determinant

$$A = \begin{bmatrix} 1 & 1 & 1 \ 1 & -2 & 3 \ 1 & 4 & 9 \end{bmatrix}$$

This is a Vandermonde-type matrix in the bases $1, -2, 3$. Compute:

$$\det(A) = 1\begin{vmatrix} -2 & 3 \ 4 & 9 \end{vmatrix} - 1\begin{vmatrix} 1 & 3 \ 1 & 9 \end{vmatrix} + 1\begin{vmatrix} 1 & -2 \ 1 & 4 \end{vmatrix}$$

$$= 1(-18 - 12) - 1(9 - 3) + 1(4 + 2)$$

$$= -30 - 6 + 6 = -30$$

Cross-check (Vandermonde formula): for bases $a=1, b=-2, c=3$, $$\det = (b-a)(c-a)(c-b) = (-3)(2)(5) = -30 \checkmark$$

Conclusion

Since $\det(A) = -30 \neq 0$, the homogeneous system has only the trivial solution: $$C_1 = C_2 = C_3 = 0$$

Therefore the signals $1^k$, $(-2)^k$, and $3^k$ are linearly independent. $\blacksquare$

(Note: this is expected, since distinct exponential bases $1, -2, 3$ always yield linearly independent geometric signals.)

Step 1 - Extract: Given Data

• Vector $\mathbf{u} = (0, -2, 2, -3)$ in $\mathbb{R}^4$
• Task: Define unit vector; find unit vector in the direction of $\mathbf{u}$.

All data present.

Step 2 - Solve

Definition of Unit Vector

A unit vector is a vector whose length (norm) is equal to $1$. For any nonzero vector $\mathbf{v} \in \mathbb{R}^n$, the unit vector in its direction is obtained by dividing the vector by its norm:

$$\hat{v} = \frac{\mathbf{v}}{|\mathbf{v}|}, \qquad |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2}$$

Finding the Unit Vector of $\mathbf{u} = (0, -2, 2, -3)$

Step 1: Compute the norm

$$|\mathbf{u}| = \sqrt{(0)^2 + (-2)^2 + (2)^2 + (-3)^2} = \sqrt{0 + 4 + 4 + 9} = \sqrt{17}$$

Step 2: Divide by the norm

$$\hat{u} = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{1}{\sqrt{17}}(0, -2, 2, -3)$$

$$\boxed{\hat{u} = \left(0,\ \frac{-2}{\sqrt{17}},\ \frac{2}{\sqrt{17}},\ \frac{-3}{\sqrt{17}}\right)}$$

Step 3: Verification

$$|\hat{u}| = \sqrt{0 + \frac{4}{17} + \frac{4}{17} + \frac{9}{17}} = \sqrt{\frac{17}{17}} = 1 \checkmark$$

The vector $\hat{u}$ has norm $1$ and points in the direction of $\mathbf{u}$, so it is the required unit vector.

Definition of Ring and Proof

Definition of Ring

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

---

Proof: (Z⁺, +, ×) is NOT a Ring

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

We check the ring axioms one by one.

---

Checking R1 (Closure for addition)

For any a, b ∈ Z⁺, a + b ∈ Z⁺.

Example: 2 + 3 = 5 ∈ Z⁺. ✓

Closure under addition is satisfied.

---

Checking R3 (Existence of Additive Identity)

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

a + 0 = 0 + a = a, ∀ a ∈ Z⁺

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

Therefore, no additive identity exists in Z⁺. ✗

---

Checking R4 (Existence of Additive Inverse)

For any a ∈ Z⁺, we need (−a) ∈ Z⁺ such that a + (−a) = 0.

Example: For a = 3, we need −3 ∈ Z⁺. But −3 ∉ Z⁺. ✗

Therefore, no additive inverse exists for any element in Z⁺. ✗

---

Conclusion

The set of positive integers Z⁺ fails to satisfy:

• Axiom R3: Additive identity (0) does not belong to Z⁺.
• Axiom R4: Additive inverse (−a) does not belong to Z⁺ for any a ∈ Z⁺.

Since at least two fundamental ring axioms are violated, (Z⁺, +, ×) is NOT a ring. $\blacksquare$