Chapter 1 • Theory & Derivations

Unit 1: Vectors in R^n & C^n, Systems of Linear Equations & Applications

Foundations of Euclidean and complex vector spaces, geometry of R^n and C^n, Cauchy-Schwarz and Minkowski inequalities, Gaussian elimination and Gauss-Jordan reduction to Reduced Row Echelon Form (RREF), consistency criteria, and applied network flow, electrical networks, chemical balancing, and polynomial interpolation.

§1.1Vectors in Euclidean R^n and Complex C^n Spaces

### 1. Vectors in Real Euclidean Space $\mathbb{R}^n$ Let $n$ be a positive integer. The set of all ordered $n$-tuples of real numbers is denoted by $\mathbb{R}^n$: $$\mathbb{R}^n = \left\{ \vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} : x_i \in \mathbb{R}, \; i = 1, 2, \dots, n \right\}$$ Vectors in $\mathbb{R}^n$ admit two fundamental algebraic operations: 1. **Vector Addition:** For $\vec{x}, \vec{y} \in \mathbb{R}^n$, $$\vec{x} + \vec{y} = \begin{pmatrix} x_1 + y_1 \\ x_2 + y_2 \\ \vdots \\ x_n + y_n \end{pmatrix}$$ 2. **Scalar Multiplication:** For $c \in \mathbb{R}$ and $\vec{x} \in \mathbb{R}^n$, $$c\vec{x} = \begin{pmatrix} c x_1 \\ c x_2 \\ \vdots \\ c x_n \end{pmatrix}$$ #### The Standard Euclidean Inner Product (Dot Product) The standard inner product of $\vec{x}, \vec{y} \in \mathbb{R}^n$ is the scalar: $$\vec{x} \cdot \vec{y} = \langle \vec{x}, \vec{y} \rangle = \sum_{i=1}^n x_i y_i = \vec{x}^T \vec{y}$$ The **Euclidean Norm (Length)** of $\vec{x}$ is defined by: $$\|\vec{x}\| = \sqrt{\vec{x} \cdot \vec{x}} = \sqrt{\sum_{i=1}^n x_i^2}$$ The **Euclidean Distance Metric** between two points $\vec{x}, \vec{y} \in \mathbb{R}^n$ is: $$d(\vec{x}, \vec{y}) = \|\vec{x} - \vec{y}\| = \sqrt{\sum_{i=1}^n (x_i - y_i)^2}$$ --- ### 2. Fundamental Inequalities in $\mathbb{R}^n$ #### Theorem 1.1 (The Cauchy-Schwarz Inequality): For any vectors $\vec{x}, \vec{y} \in \mathbb{R}^n$: $$|\vec{x} \cdot \vec{y}| \le \|\vec{x}\| \|\vec{y}\|$$ *Rigorous Proof:* If $\vec{y} = \vec{0}$, both sides are $0$ and the equality holds trivially. Assume $\vec{y} \ne \vec{0}$. For any real scalar $t \in \mathbb{R}$, consider the squared norm: $$0 \le \|\vec{x} - t\vec{y}\|^2 = (\vec{x} - t\vec{y}) \cdot (\vec{x} - t\vec{y}) = \|\vec{x}\|^2 - 2t(\vec{x} \cdot \vec{y}) + t^2 \|\vec{y}\|^2$$ This is a non-negative quadratic polynomial $q(t) = A t^2 + B t + C$ where $A = \|\vec{y}\|^2 > 0$, $B = -2(\vec{x} \cdot \vec{y})$, and $C = \|\vec{x}\|^2$. Since $q(t) \ge 0$ for all $t \in \mathbb{R}$, its discriminant $\Delta = B^2 - 4AC$ must be non-positive: $$\Delta = 4(\vec{x} \cdot \vec{y})^2 - 4 \|\vec{y}\|^2 \|\vec{x}\|^2 \le 0 \implies (\vec{x} \cdot \vec{y})^2 \le \|\vec{x}\|^2 \|\vec{y}\|^2$$ Taking the positive square root of both sides yields $|\vec{x} \cdot \vec{y}| \le \|\vec{x}\| \|\vec{y}\|$. Equality holds if and only if $\vec{x}$ and $\vec{y}$ are linearly dependent. $\blacksquare$ #### Theorem 1.2 (The Triangle / Minkowski Inequality): For any $\vec{x}, \vec{y} \in \mathbb{R}^n$: $$\|\vec{x} + \vec{y}\| \le \|\vec{x}\| + \|\vec{y}\|$$ *Proof:* Expanding the squared norm using the Cauchy-Schwarz inequality: $$\|\vec{x} + \vec{y}\|^2 = (\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) = \|\vec{x}\|^2 + 2(\vec{x} \cdot \vec{y}) + \|\vec{y}\|^2 \le \|\vec{x}\|^2 + 2\|\vec{x}\|\|\vec{y}\| + \|\vec{y}\|^2 = (\|\vec{x}\| + \|\vec{y}\|)^2$$ Taking square roots on both sides establishes the inequality. $\blacksquare$ --- ### 3. Vectors in Complex Space $\mathbb{C}^n$ The set of ordered $n$-tuples of complex numbers is denoted by $\mathbb{C}^n$: $$\mathbb{C}^n = \left\{ \vec{z} = \begin{pmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{pmatrix} : z_k = a_k + i b_k \in \mathbb{C}, \; a_k, b_k \in \mathbb{R} \right\}$$ For $\vec{z} \in \mathbb{C}^n$, its **complex conjugate** is $\overline{\vec{z}} = (\overline{z}_1, \dots, \overline{z}_n)^T$, and its **conjugate transpose (Hermitian adjoint)** is: $$\vec{z}^* = \vec{z}^H = \overline{\vec{z}}^T = (\overline{z}_1, \overline{z}_2, \dots, \overline{z}_n)$$ #### Standard Inner Product in $\mathbb{C}^n$ To ensure that the norm of a non-zero complex vector is always a positive real number, the inner product in $\mathbb{C}^n$ is defined with complex conjugation: $$\langle \vec{u}, \vec{v} \rangle = \vec{v}^* \vec{u} = \sum_{k=1}^n u_k \overline{v}_k$$ This inner product satisfies: 1. **Conjugate Symmetry:** $\langle \vec{v}, \vec{u} \rangle = \overline{\langle \vec{u}, \vec{v} \rangle}$ 2. **Linearity in First Argument:** $\langle \alpha \vec{u}_1 + \beta \vec{u}_2, \vec{v} \rangle = \alpha \langle \vec{u}_1, \vec{v} \rangle + \beta \langle \vec{u}_2, \vec{v} \rangle$ 3. **Conjugate Linearity in Second Argument:** $\langle \vec{u}, \alpha \vec{v}_1 + \beta \vec{v}_2 \rangle = \overline{\alpha} \langle \vec{u}, \vec{v}_1 \rangle + \overline{\beta} \langle \vec{u}, \vec{v}_2 \rangle$ 4. **Positive Definiteness:** $\langle \vec{z}, \vec{z} \rangle = \sum_{k=1}^n |z_k|^2 \ge 0$, and $\langle \vec{z}, \vec{z} \rangle = 0 \iff \vec{z} = \vec{0}$ The **Complex Euclidean Norm** is $\|\vec{z}\| = \sqrt{\langle \vec{z}, \vec{z} \rangle} = \sqrt{\sum_{k=1}^n |z_k|^2}$, and the distance metric is $d(\vec{u}, \vec{v}) = \|\vec{u} - \vec{v}\| = \sqrt{\sum_{k=1}^n |u_k - v_k|^2}$.

§1.2Systems of Linear Equations, Gaussian Elimination & RREF

### 1. General System of Linear Equations A general system of $m$ linear equations in $n$ unknowns $x_1, x_2, \dots, x_n$ over a field $F$ ($\mathbb{R}$ or $\mathbb{C}$) is formulated as: $$\begin{aligned} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n &= b_1 \\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n &= b_2 \\ &\vdots \\ a_{m1} x_1 + a_{m2} x_2 + \cdots + a_{mn} x_n &= b_m \end{aligned}$$ In compact matrix notation: $$A\vec{x} = \vec{b}$$ where $A \in M_{m \times n}(F)$ is the **coefficient matrix**, $\vec{x} \in F^n$ is the **unknown vector**, and $\vec{b} \in F^m$ is the **constant vector**: $$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}, \quad \vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix}$$ The **Augmented Matrix** is denoted by $[A \mid \vec{b}] \in M_{m \times (n+1)}(F)$. - If $\vec{b} = \vec{0}$, the system $A\vec{x} = \vec{0}$ is **homogeneous**. It always possesses the trivial solution $\vec{x} = \vec{0}$. - If $\vec{b} \ne \vec{0}$, the system is **non-homogeneous**. --- ### 2. Elementary Row Operations and Row Equivalence Two augmented matrices are **row equivalent** if one can be transformed into the other via a finite sequence of **Elementary Row Operations (EROs)**: 1. **Row Swap ($R_i \leftrightarrow R_j$):** Interchange rows $i$ and $j$. 2. **Row Scaling ($R_i \to c R_i$):** Multiply row $i$ by a non-zero scalar $c \ne 0$. 3. **Row Addition ($R_i \to R_i + c R_j$):** Add $c$ times row $j$ to row $i$ ($i \ne j$). Each ERO corresponds to left multiplication by an invertible **elementary matrix** $E \in M_{m \times m}(F)$. Because elementary matrices are invertible, EROs preserve the solution set of the linear system. --- ### 3. Reduced Row Echelon Form (RREF) A matrix is in **Row Echelon Form (REF)** if: 1. All zero rows are at the bottom of the matrix. 2. The leading entry (first non-zero entry from the left, called the **pivot**) of each non-zero row is strictly to the right of the leading entry of the row above it. 3. All entries in a column below a leading entry are zero. A matrix is in **Reduced Row Echelon Form (RREF)** if, in addition: 4. The leading entry (pivot) in each non-zero row is strictly equal to $1$. 5. Each column containing a leading $1$ has zeros in all other entries (both above and below the pivot). #### Theorem 1.3 (Uniqueness of RREF): Every matrix $A \in M_{m \times n}(F)$ is row-equivalent to one and only one matrix $R$ in Reduced Row Echelon Form: $R = \text{rref}(A)$. #### Classification of Variables and Solutions: Let $R = [R_A \mid \vec{d}] = \text{rref}([A \mid \vec{b}])$ have $r$ non-zero rows (pivots): - **Pivot Variables (Basic Variables):** Unknowns $x_j$ corresponding to columns of $R_A$ containing a pivot $1$. - **Free Variables:** Unknowns $x_j$ corresponding to columns without pivots. Total free variables = $n - r$. #### Theorem 1.4 (The Rouché-Capelli Consistency Theorem): A linear system $A\vec{x} = \vec{b}$ is **consistent** (has at least one solution) if and only if the augmented matrix $[A \mid \vec{b}]$ has the same rank as the coefficient matrix $A$: $$\text{rank}(A) = \text{rank}([A \mid \vec{b}])$$ Equivalently, the system is consistent if and only if the last column of $\text{rref}([A \mid \vec{b}])$ does not contain a pivot row of the form $(0, 0, \dots, 0 \mid 1)$. - **Unique Solution:** Consistent and $\text{rank}(A) = n$ (zero free variables). - **Infinitely Many Solutions:** Consistent and $\text{rank}(A) = r < n$ ($n - r$ free variables). - **Inconsistent (No Solution):** $\text{rank}([A \mid \vec{b}]) = \text{rank}(A) + 1$.

§1.3Applied Linear Systems: Networks, Circuits, Chemistry & Interpolation

### 1. Network Flow Analysis A directed network consists of a set of junctions (nodes) connected by branches (edges) carrying flow rates $x_1, x_2, \dots, x_k$. The governing physical principle is **Kirchhoff's Flow Conservation Law**: $$\sum \text{Flow In} = \sum \text{Flow Out} \quad \text{at every junction node}$$ For an urban traffic grid or pipeline network with $m$ nodes and external net inputs/outputs $b_i$: $$\sum_{j \in \text{incoming}(i)} x_j - \sum_{k \in \text{outgoing}(i)} x_k = b_i \quad (i = 1, \dots, m)$$ This yields a linear system $A\vec{x} = \vec{b}$. Overall network feasibility requires total inflow equals total outflow: $\sum_{i=1}^m b_i = 0$. --- ### 2. Electrical Resistor Networks Consider an electrical circuit containing resistors and DC voltage sources: 1. **Kirchhoff's Current Law (KCL):** The algebraic sum of currents at any node is zero: $\sum I_{\text{in}} = \sum I_{\text{out}}$. 2. **Kirchhoff's Voltage Law (KVL):** The algebraic sum of voltage changes around any closed loop is zero: $\sum V_{\text{sources}} = \sum I R$. By defining loop currents $I_1, I_2, \dots, I_L$ (Mesh Current Analysis), KVL produces a symmetric, diagonally dominant linear system: $$\begin{pmatrix} R_{11} & -R_{12} & \cdots \\ -R_{21} & R_{22} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} \begin{pmatrix} I_1 \\ I_2 \\ \vdots \end{pmatrix} = \begin{pmatrix} \mathcal{E}_1 \\ \mathcal{E}_2 \\ \vdots \end{pmatrix}$$ Solving via Gauss-Jordan elimination determines all branch currents and power dissipations $P_k = I_k^2 R_k$. --- ### 3. Balancing Chemical Reaction Equations A chemical reaction preserves the number of atoms of each chemical element. Consider balancing the combustion of propane: $$x_1 \text{C}_3\text{H}_8 + x_2 \text{O}_2 \to x_3 \text{CO}_2 + x_4 \text{H}_2\text{O}$$ Equating atom counts for Carbon ($\text{C}$), Hydrogen ($\text{H}$), and Oxygen ($\text{O}$): - $\text{C}: 3x_1 - 1x_3 = 0$ - $\text{H}: 8x_1 - 2x_4 = 0$ - $\text{O}: 2x_2 - 2x_3 - 1x_4 = 0$ This is a homogeneous linear system $A\vec{x} = \vec{0}$ with coefficient matrix: $$A = \begin{pmatrix} 3 & 0 & -1 & 0 \\ 8 & 0 & 0 & -2 \\ 0 & 2 & -2 & -1 \end{pmatrix}$$ Row reduction to RREF yields: $$\begin{pmatrix} 1 & 0 & 0 & -1/4 \\ 0 & 1 & 0 & -5/4 \\ 0 & 0 & 1 & -3/4 \end{pmatrix} \implies \begin{cases} x_1 = \frac{1}{4}x_4 \\ x_2 = \frac{5}{4}x_4 \\ x_3 = \frac{3}{4}x_4 \end{cases}$$ Choosing the smallest positive integer for the free variable $x_4 = 4$ gives the unique stoichiometric solution $\vec{x} = (1, 5, 3, 4)^T$: $$\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}$$ --- ### 4. Polynomial Curve Interpolation and the Vandermonde Matrix Given $n+1$ distinct data points $(t_0, y_0), (t_1, y_1), \dots, (t_n, y_n)$ with $t_i \ne t_j$ for $i \ne j$, there exists a unique polynomial of degree at most $n$: $$p(t) = c_0 + c_1 t + c_2 t^2 + \cdots + c_n t^n$$ satisfying the interpolation conditions $p(t_i) = y_i$ for all $i = 0, \dots, n$. Setting up the linear system $V\vec{c} = \vec{y}$ reveals the **Vandermonde Matrix** $V$: $$\begin{pmatrix} 1 & t_0 & t_0^2 & \cdots & t_0^n \\ 1 & t_1 & t_1^2 & \cdots & t_1^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & t_n & t_n^2 & \cdots & t_n^n \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \\ \vdots \\ c_n \end{pmatrix} = \begin{pmatrix} y_0 \\ y_1 \\ \vdots \\ y_n \end{pmatrix}$$ #### Theorem 1.5 (Vandermonde Determinant Identity): $$\det(V) = \prod_{0 \le j < i \le n} (t_i - t_j)$$ Since the interpolation nodes $t_i$ are pairwise distinct, $t_i - t_j \ne 0$ for all $i > j$. Consequently, $\det(V) \ne 0$, proving that $V$ is invertible and the interpolating polynomial $p(t)$ exists and is strictly unique.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1 • Foundational Complete Gauss-Jordan Reduction & Parametric Vector Solution

Solve the following non-homogeneous system of linear equations in four variables over $\mathbb{R}$ using elementary row operations and Gauss-Jordan elimination: $$\begin{aligned} x_1 - 2x_2 + x_3 + 3x_4 &= 2 \\ 2x_1 - 4x_2 + 3x_3 + 8x_4 &= 7 \\ -x_1 + 2x_2 + 2x_3 + 3x_4 &= 7 \end{aligned}$$ Find the Reduced Row Echelon Form (RREF) of the augmented matrix, identify the pivot and free variables, and express the general solution in parametric vector form.

**Step 1: Set up the augmented matrix $[A \mid \vec{b}]$:** $$[A \mid \vec{b}] = \begin{pmatrix} 1 & -2 & 1 & 3 & 2 \\ 2 & -4 & 3 & 8 & 7 \\ -1 & 2 & 2 & 3 & 7 \end{pmatrix}$$ **Step 2: Eliminate entries below pivot in Column 1:** Apply $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 + R_1$: - Row 2: $(2 - 2, \; -4 - 2(-2), \; 3 - 2(1), \; 8 - 2(3), \; 7 - 2(2)) = (0, 0, 1, 2, 3)$ - Row 3: $(-1 + 1, \; 2 - 2, \; 2 + 1, \; 3 + 3, \; 7 + 2) = (0, 0, 3, 6, 9)$ The matrix becomes: $$\begin{pmatrix} 1 & -2 & 1 & 3 & 2 \\ 0 & 0 & 1 & 2 & 3 \\ 0 & 0 & 3 & 6 & 9 \end{pmatrix}$$ **Step 3: Eliminate entries in Column 3 to achieve RREF:** Apply $R_3 \to R_3 - 3R_2$: - Row 3: $(0, 0, 3 - 3, \; 6 - 6, \; 9 - 9) = (0, 0, 0, 0, 0)$ Apply $R_1 \to R_1 - R_2$: - Row 1: $(1, -2, \; 1 - 1, \; 3 - 2, \; 2 - 3) = (1, -2, 0, 1, -1)$ The resulting matrix is in Reduced Row Echelon Form: $$\text{rref}([A \mid \vec{b}]) = \begin{pmatrix} 1 & -2 & 0 & 1 & -1 \\ 0 & 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$ **Step 4: Identify pivots and free variables:** - Pivots are located in columns 1 and 3. Therefore, $x_1$ and $x_3$ are **basic (pivot) variables**. - Columns 2 and 4 contain no pivots. Therefore, $x_2$ and $x_4$ are **free variables**. Let $x_2 = s$ and $x_4 = t$ where $s, t \in \mathbb{R}$. From Row 1: $x_1 - 2s + t = -1 \implies x_1 = -1 + 2s - t$. From Row 2: $x_3 + 2t = 3 \implies x_3 = 3 - 2t$. **Step 5: Write the solution in parametric vector form:** $$\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} -1 + 2s - t \\ s \\ 3 - 2t \\ t \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 3 \\ 0 \end{pmatrix} + s \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ 0 \\ -2 \\ 1 \end{pmatrix}$$

Final Answer & Analytical Insight

$\text{rref}([A \mid \vec{b}]) = \begin{pmatrix} 1 & -2 & 0 & 1 & -1 \\ 0 & 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$. The general solution is $\vec{x} = \vec{x}_p + s\vec{v}_1 + t\vec{v}_2 = \begin{pmatrix} -1 \\ 0 \\ 3 \\ 0 \end{pmatrix} + s \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} -1 \\ 0 \\ -2 \\ 1 \end{pmatrix}$ for all $s, t \in \mathbb{R}$.

Tier 2 • Intermediate Exam Network Flow & Resistor Loop Conservation

A traffic network has four intersections $A, B, C, D$ connected by one-way streets with traffic flows $x_1, x_2, x_3, x_4, x_5$ (measured in vehicles/hour): - Node $A$: Inflow of $400$ enters; flows $x_1$ (to $B$) and $x_2$ (to $C$) leave. - Node $B$: Inflow $x_1$ arrives; flows $x_3$ (to $D$) and external outflow of $250$ leave. - Node $C$: Inflow $x_2$ arrives along with external inflow of $150$; flow $x_4$ (to $D$) and $x_5$ leave. - Node $D$: Inflows $x_3$ and $x_4$ arrive; external outflow of $300$ leaves. (a) Set up the linear system balancing inflow and outflow at every intersection. (b) Solve the system to determine the general flow vector. (c) If the road segment $CD$ is closed ($x_4 = 0$), determine the required flows on all remaining streets to avoid traffic congestion.

Tier 3 • Honors Challenge Vandermonde Determinant & Polynomial Interpolation Uniqueness

Prove rigorously by induction on $n \ge 1$ that the determinant of the $(n+1) \times (n+1)$ Vandermonde matrix: $$V_n = \begin{pmatrix} 1 & t_0 & t_0^2 & \cdots & t_0^n \\ 1 & t_1 & t_1^2 & \cdots & t_1^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & t_n & t_n^2 & \cdots & t_n^n \end{pmatrix}$$ satisfies the product formula $\det(V_n) = \prod_{0 \le j < i \le n} (t_i - t_j)$. Then, deduce that if the nodes $t_0, t_1, \dots, t_n \in F$ are distinct, there exists a unique polynomial $p(t) \in F[t]$ of degree $\le n$ passing through any prescribed points $(t_i, y_i)$. Explain the connection to the Lagrange interpolation formula.

**Step 1: Base Case $n = 1$:** $$V_1 = \begin{pmatrix} 1 & t_0 \\ 1 & t_1 \end{pmatrix} \implies \det(V_1) = 1 \cdot t_1 - 1 \cdot t_0 = t_1 - t_0 = \prod_{0 \le j < i \le 1} (t_i - t_j)$$ The base case holds. **Step 2: Inductive Step:** Assume the formula holds for size $n$, i.e., for any $n$ distinct nodes, $\det(V_{n-1}) = \prod_{0 \le j < i \le n-1} (t_i - t_j)$. To compute $\det(V_n)$, perform elementary column operations from right to left: subtract $t_0$ times column $k-1$ from column $k$, for $k = n+1, n, \dots, 2$: $$C_k \to C_k - t_0 C_{k-1}$$ These column operations do not alter the determinant ($\det(V_n)$ is invariant). In the first row, every entry from column $2$ onward becomes: $$t_0^k - t_0 t_0^{k-1} = 0$$ Thus, the first row becomes $(1, 0, 0, \dots, 0)$. For row $i$ ($i \ge 1$), entry in column $k$ becomes: $$t_i^k - t_0 t_i^{k-1} = (t_i - t_0) t_i^{k-1}$$ Expanding $\det(V_n)$ along the first row: $$\det(V_n) = 1 \cdot \det \begin{pmatrix} (t_1 - t_0) & (t_1 - t_0)t_1 & \cdots & (t_1 - t_0)t_1^{n-1} \\ (t_2 - t_0) & (t_2 - t_0)t_2 & \cdots & (t_2 - t_0)t_2^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ (t_n - t_0) & (t_n - t_0)t_n & \cdots & (t_n - t_0)t_n^{n-1} \end{pmatrix}$$ Factor out the common scalar $(t_i - t_0)$ from each row $i = 1, 2, \dots, n$: $$\det(V_n) = \left( \prod_{i=1}^n (t_i - t_0) \right) \cdot \det \begin{pmatrix} 1 & t_1 & \cdots & t_1^{n-1} \\ 1 & t_2 & \cdots & t_2^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & t_n & \cdots & t_n^{n-1} \end{pmatrix}$$ The remaining matrix is precisely $V_{n-1}$ for the nodes $(t_1, t_2, \dots, t_n)$. By the induction hypothesis: $$\det(V_{n-1}) = \prod_{1 \le j < i \le n} (t_i - t_j)$$ Multiplying this by $\prod_{i=1}^n (t_i - t_0)$ includes all terms where $j = 0$: $$\det(V_n) = \prod_{0 \le j < i \le n} (t_i - t_j)$$ The induction is complete. **Step 3: Uniqueness of Interpolating Polynomial:** Since nodes $t_0, \dots, t_n$ are pairwise distinct, $t_i - t_j \ne 0$ for all $i > j$, so $\det(V_n) \ne 0$. Hence $V_n$ is non-singular and invertible. The linear system $V_n \vec{c} = \vec{y}$ has the unique solution $\vec{c} = V_n^{-1} \vec{y}$, establishing the existence and uniqueness of $p(t) \in \mathbb{P}_n$. In the Lagrange basis $\ell_i(t) = \prod_{j \ne i} \frac{t - t_j}{t_i - t_j}$, the unique polynomial is expressed directly as $p(t) = \sum_{i=0}^n y_i \ell_i(t)$, avoiding explicit matrix inversion.

Final Answer & Analytical Insight

$\det(V_n) = \prod_{0 \le j < i \le n} (t_i - t_j)$. Since $t_i \ne t_j$ for $i \ne j$, $\det(V_n) \ne 0$, which guarantees $V_n$ is invertible, proving the interpolating polynomial $p(t)$ of degree $\le n$ is strictly unique.