Mathematics / Algebra Basic & Higher Algebra 100% Free Open Access
Chapter 6 • Theory & Derivations

Matrix Algebra & Determinants

Matrix Rings, Special Matrix Classes, Multilinear Determinants, Cauchy-Binet Multiplicativity & Adjugate Inverses

§6.1 Algebra of Matrices: Rings, Transposition & Trace

1. The Matrix Ring

The collection of $m \times n$ matrices with entries in a field $\mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$), denoted $M_{m \times n}(\mathbb{F})$, forms a vector space under entrywise addition and scalar multiplication.
For $A \in M_{m \times p}(\mathbb{F})$ and $B \in M_{p \times n}(\mathbb{F})$, their matrix product $C = AB \in M_{m \times n}(\mathbb{F})$ has entries: $$\mathbf{c_{ij} = \sum_{k=1}^p a_{ik} b_{kj}}$$ The product is associative ($A(BC) = (AB)C$) and distributes over addition, but is strictly non-commutative ($AB \ne BA$ in general). The square matrices $M_n(\mathbb{F})$ form a non-commutative ring with identity $I_n$.

2. The Transpose & Conjugate Transpose

The transpose $A^T$ of $A = [a_{ij}]$ is defined by $(A^T)_{ij} = a_{ji}$. Key properties: $$(A + B)^T = A^T + B^T, \quad (cA)^T = c A^T, \quad (A^T)^T = A, \quad \mathbf{(AB)^T = B^T A^T}$$ For complex matrices, the Hermitian adjoint (conjugate transpose) is $A^\dagger \equiv (\bar{A})^T$, satisfying $(AB)^\dagger = B^\dagger A^\dagger$.

3. The Trace Function

For a square matrix $A \in M_n(\mathbb{F})$, the trace is the sum of its diagonal elements: $$\mathbf{\text{tr}(A) \equiv \sum_{i=1}^n a_{ii}}$$ Cyclic Invariance Theorem: For any $A \in M_{m \times n}$ and $B \in M_{n \times m}$: $$\mathbf{\text{tr}(AB) = \text{tr}(BA)}$$ Proof: $\text{tr}(AB) = \sum_{i=1}^m (AB)_{ii} = \sum_{i=1}^m \sum_{j=1}^n a_{ij} b_{ji} = \sum_{j=1}^n \sum_{i=1}^m b_{ji} a_{ij} = \sum_{j=1}^n (BA)_{jj} = \text{tr}(BA)$. $\blacksquare$

§6.2 Taxonomy of Special Classes of Matrices

1. Real Special Matrices

Let $A \in M_n(\mathbb{R})$:

  • Symmetric: $A^T = A \iff a_{ij} = a_{ji}$.
  • Skew-Symmetric: $A^T = -A \iff a_{ij} = -a_{ji}$ (diagonal entries must be zero: $a_{ii} = 0$).
  • Orthogonal: $A^T A = A A^T = I \iff A^{-1} = A^T$. Rows (and columns) form an orthonormal basis of $\mathbb{R}^n$. Orthogonal transformations preserve Euclidean lengths and angles: $\|Ax\| = \|x\|$.

2. Complex Special Matrices

Let $A \in M_n(\mathbb{C})$:

  • Hermitian: $A^\dagger = A \iff a_{ij} = \bar{a}_{ji}$ (diagonal entries must be purely real: $a_{ii} \in \mathbb{R}$).
  • Skew-Hermitian: $A^\dagger = -A \iff a_{ij} = -\bar{a}_{ji}$ (diagonal entries must be purely imaginary).
  • Unitary: $A^\dagger A = I \iff A^{-1} = A^\dagger$. Unitary matrices preserve the complex inner product: $\langle Ux, Uy \rangle = \langle x, y \rangle$.

3. Algebraic Operational Classes

  • Idempotent: $A^2 = A$ (represents projection operators).
  • Nilpotent: $A^k = 0$ for some integer $k \ge 1$ (the smallest such $k$ is the index of nilpotency).
  • Involutory: $A^2 = I \iff A^{-1} = A$ (represents reflection operators).

§6.3 Axiomatic Characterization & Laplace Expansion of Determinants

1. Axiomatic Definition of the Determinant

The determinant is the unique function $\det: M_n(\mathbb{F}) \to \mathbb{F}$ satisfying three fundamental axioms:

  1. Multilinearity: $\det$ is a linear function of each row when all other rows are held fixed.
  2. Alternating Property: Swapping any two rows negates the determinant: $\det(R_1, \dots, R_i, \dots, R_j, \dots, R_n) = -\det(R_1, \dots, R_j, \dots, R_i, \dots, R_n)$. Consequently, if two rows are identical, $\det A = 0$.
  3. Normalization: $\det(I_n) = 1$.

2. The Leibniz Permutation Formula

From the axioms, the explicit determinant formula is: $$\mathbf{\det(A) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}}$$ where the sum ranges over all $n!$ permutations $\sigma$ of the symmetric group $S_n$, and $\text{sgn}(\sigma) \in \{+1, -1\}$ is the permutation parity.

3. Laplace's Cofactor Expansion Theorem

The $(i, j)$-th minor $M_{ij}$ is the determinant of the $(n-1) \times (n-1)$ submatrix formed by deleting row $i$ and column $j$. The $(i, j)$-th cofactor is: $$C_{ij} \equiv (-1)^{i+j} M_{ij}$$ Theorem (Laplace): The determinant can be evaluated by expanding along any arbitrary row $i$ or column $j$: $$\mathbf{\det(A) = \sum_{j=1}^n a_{ij} C_{ij} \quad (\text{Expansion along row } i)}$$ $$\mathbf{\det(A) = \sum_{i=1}^n a_{ij} C_{ij} \quad (\text{Expansion along column } j)}$$

§6.4 Fundamental Determinant Properties & Multiplicativity

1. Invariance and Operational Properties

  • Transpose Invariance: $\det(A^T) = \det(A)$. Row operations and column operations have identical effects on determinants!
  • Triangular Matrices: If $A$ is upper-triangular, lower-triangular, or diagonal, its determinant is simply the product of its diagonal entries: $$\det(A) = a_{11} a_{22} \cdots a_{nn}$$
  • Type III Row Operations: Adding a scalar multiple of one row to another preserves the determinant strictly unchanged: $\det(A) = \det(E A)$ where $\det(E) = 1$.
  • Scalar Multiplication: For an $n \times n$ matrix, $\det(c A) = c^n \det(A)$.

2. The Multiplicative Theorem

Theorem (Cauchy–Binet): For any two $n \times n$ matrices $A$ and $B$: $$\mathbf{\det(AB) = \det(A) \det(B)}$$ Immediate Corollaries:

  • A matrix $A$ is invertible if and only if $\det(A) \ne 0$.
  • If $A$ is invertible: $\det(A^{-1}) = \frac{1}{\det(A)}$.
  • If $A$ and $B$ are similar ($B = P^{-1} A P$): $\det(B) = \det(P^{-1})\det(A)\det(P) = \det(A)$.
  • If $Q$ is orthogonal: $Q^T Q = I \implies [\det(Q)]^2 = 1 \implies \det(Q) = \pm 1$.

§6.5 The Classical Adjugate Matrix, Matrix Inversion & Cramer's Rule

1. The Classical Adjugate (Adjoint)

The adjugate $\text{adj}(A)$ of an $n \times n$ matrix $A$ is the transpose of its cofactor matrix $C = [C_{ij}]$: $$\mathbf{\text{adj}(A) \equiv C^T \implies [\text{adj}(A)]_{ij} = C_{ji} = (-1)^{i+j} M_{ji}}$$

2. The Fundamental Inversion Theorem

Theorem: For any $n \times n$ matrix $A$: $$\mathbf{A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = (\det A) I_n}$$ Proof: The $(i, k)$-th entry of $A \cdot \text{adj}(A)$ is $\sum_{j=1}^n a_{ij} [\text{adj}(A)]_{jk} = \sum_{j=1}^n a_{ij} C_{kj}$. If $i = k$, this is the Laplace expansion of $\det(A)$ along row $i$. If $i \ne k$, this represents the expansion of a matrix with two identical rows ($i$ and $k$), which vanishes identically (alien cofactors). Thus $[A \cdot \text{adj}(A)]_{ik} = \delta_{ik} \det(A)$. $\blacksquare$

Consequently, whenever $\det(A) \ne 0$, the unique matrix inverse is: $$\mathbf{A^{-1} = \frac{1}{\det(A)} \text{adj}(A)}$$

3. Cramer’s Rule for Linear Systems

For an $n \times n$ system $AX = B$ with $\det(A) \ne 0$, $X = A^{-1} B = \frac{1}{\det A}\text{adj}(A) B$. Entrywise: $$\mathbf{x_i = \frac{\det(A_i)}{\det(A)}}$$ where $A_i$ is the matrix obtained by replacing the $i$-th column of $A$ with the constant vector $B$.

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.

Foundational Mechanics Example 6.1: Determinant Evaluation via Row Operations & Cofactor Inversion

Given the $3 \times 3$ matrix $A = \begin{pmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 4 & 2 & 1 \end{pmatrix}$: (a) Compute $\det(A)$. (b) Construct the adjugate matrix $\text{adj}(A)$ and find $A^{-1}$.

Step 1: Compute det(A) via Row Operations
$$\det(A) = \begin{vmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 4 & 2 & 1 \end{vmatrix} \xrightarrow{R_3 \to R_3 - 2R_1} \begin{vmatrix} 2 & 1 & 3 \\ 1 & 0 & 2 \\ 0 & 0 & -5 \end{vmatrix} \\ \text{Expand along row 3: } \det(A) = (-5) \cdot (-1)^{3+3} \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = (-5)(1)(0 - 1) = 5$$

Subtracting $2R_1$ from $R_3$ produces two zeros in row 3, making cofactor expansion trivial.

Step 2: Compute All 9 Cofactors
$$C_{11} = +\begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix} = -4, \quad C_{12} = -\begin{vmatrix} 1 & 2 \\ 4 & 1 \end{vmatrix} = 7, \quad C_{13} = +\begin{vmatrix} 1 & 0 \\ 4 & 2 \end{vmatrix} = 2 \\ C_{21} = -\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = 5, \quad C_{22} = +\begin{vmatrix} 2 & 3 \\ 4 & 1 \end{vmatrix} = -10, \quad C_{23} = -\begin{vmatrix} 2 & 1 \\ 4 & 2 \end{vmatrix} = 0 \\ C_{31} = +\begin{vmatrix} 1 & 3 \\ 0 & 2 \end{vmatrix} = 2, \quad C_{32} = -\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = -1, \quad C_{33} = +\begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = -1$$

Evaluate the signed minors for each position.

Step 3: Transpose to Form adj(A) and Compute A⁻¹
$$\text{adj}(A) = C^T = \begin{pmatrix} -4 & 5 & 2 \\ 7 & -10 & -1 \\ 2 & 0 & -1 \end{pmatrix} \\ A^{-1} = \frac{1}{5} \begin{pmatrix} -4 & 5 & 2 \\ 7 & -10 & -1 \\ 2 & 0 & -1 \end{pmatrix} = \begin{pmatrix} -0.8 & 1.0 & 0.4 \\ 1.4 & -2.0 & -0.2 \\ 0.4 & 0.0 & -0.2 \end{pmatrix}$$

Divide the transposed cofactor matrix by $\det(A) = 5$.

Final Answer & Physical Insight

\det(A) = 5; \quad \mathbf{A^{-1} = \frac{1}{5}\begin{pmatrix} -4 & 5 & 2 \\ 7 & -10 & -1 \\ 2 & 0 & -1 \end{pmatrix}}.

Intermediate University Exam Example 6.2: Symmetric-Skew Decomposition & Vandermonde Determinant

(a) Prove that every square matrix $A$ can be uniquely decomposed as $A = S + K$, where $S$ is symmetric and $K$ is skew-symmetric. (b) Evaluate the $3 \times 3$ Vandermonde determinant $V(a, b, c) = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$.

Step 1: Construct the Unique Decomposition
$$S \equiv \frac{A + A^T}{2}, \quad K \equiv \frac{A - A^T}{2} \\ S^T = \frac{A^T + (A^T)^T}{2} = \frac{A^T + A}{2} = S \quad (\text{Symmetric}) \\ K^T = \frac{A^T - (A^T)^T}{2} = \frac{A^T - A}{2} = -K \quad (\text{Skew-symmetric}) \\ S + K = \frac{A + A^T + A - A^T}{2} = A$$

Construct $S$ and $K$ explicitly, demonstrating existence.

Step 2: Prove Uniqueness
$$\text{Suppose } A = S' + K'. \text{ Then } A^T = S'^T + K'^T = S' - K' \\ A + A^T = 2S' \implies S' = \frac{A + A^T}{2} = S \\ A - A^T = 2K' \implies K' = \frac{A - A^T}{2} = K \quad \blacksquare$$

Taking transposes and adding/subtracting establishes uniqueness.

Step 3: Evaluate Vandermonde Determinant via Row Operations
$$V(a, b, c) = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \xrightarrow{\substack{R_2 \to R_2 - R_1 \\ R_3 \to R_3 - R_1}} \begin{vmatrix} 1 & a & a^2 \\ 0 & b - a & b^2 - a^2 \\ 0 & c - a & c^2 - a^2 \end{vmatrix} \\ = \begin{vmatrix} b - a & (b - a)(b + a) \\ c - a & (c - a)(c + a) \end{vmatrix} = (b - a)(c - a) \begin{vmatrix} 1 & b + a \\ 1 & c + a \end{vmatrix} \\ = (b - a)(c - a) [(c + a) - (b + a)] = (b - a)(c - a)(c - b) = (c - b)(b - a)(c - a)$$

Factoring $(b-a)$ and $(c-a)$ leaves a simple $2 \times 2$ determinant.

Final Answer & Physical Insight

\mathbf{A = \frac{A + A^T}{2} + \frac{A - A^T}{2}} \text{ is unique; } \quad \mathbf{V(a, b, c) = (b - a)(c - a)(c - b)}.

Honors / Proof Challenge Example 6.3: Closed Formula for Circulant Determinants via Roots of Unity

For the $3 \times 3$ circulant matrix $C = \begin{pmatrix} a & b & c \\ c & a & b \\ b & c & a \end{pmatrix}$, prove that $\det(C) = (a + b + c)(a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c) = a^3 + b^3 + c^3 - 3abc$, where $\omega = e^{i 2\pi/3}$ is the cube root of unity.

Step 1: Direct Column Transformation using Roots of Unity
$$\text{Let } \omega \text{ satisfy } \omega^3 = 1, \; 1 + \omega + \omega^2 = 0 \\ \text{Add } \omega \cdot \text{Col}_2 + \omega^2 \cdot \text{Col}_3 \text{ to } \text{Col}_1: \\ \text{Row 1: } a + \omega b + \omega^2 c \\ \text{Row 2: } c + \omega a + \omega^2 b = \omega(a + \omega b + \omega^2 c) \quad (\text{since } \omega^3 c = c) \\ \text{Row 3: } b + \omega c + \omega^2 a = \omega^2(a + \omega b + \omega^2 c)$$

The linear combination produces a common factor $(a + \omega b + \omega^2 c)$ down the entire first column!

Step 2: Factor Out Eigenvalues
$$\text{For each } k \in \{0, 1, 2\}, \text{ substituting the root } \omega^k \text{ proves that } \lambda_k = a + \omega^k b + \omega^{2k} c \text{ is an eigenvalue of } C! \\ \det(C) = \prod_{k=0}^2 \lambda_k = (a + b + c)(a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c)$$

The determinant of any matrix equals the product of its eigenvalues.

Step 3: Expand the Product
$$(a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c) = a^2 + \omega^2 ab + \omega ac + \omega ab + b^2 + \omega^2 bc + \omega^2 ac + \omega bc + c^2 \\ = a^2 + b^2 + c^2 + (\omega + \omega^2)(ab + bc + ca) = a^2 + b^2 + c^2 - (ab + bc + ca) \\ \det(C) = (a + b + c)[a^2 + b^2 + c^2 - ab - bc - ca] = a^3 + b^3 + c^3 - 3abc \quad \blacksquare$$

Using $1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1$ recovers Euler's classical cubic product factorization.

Final Answer & Physical Insight

\mathbf{\det(C) = a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c)}.