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:
- Multilinearity: $\det$ is a linear function of each row when all other rows are held fixed.
- 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, $\det A = 0$.
- 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 $B$.
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.
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 $A^{-1}$.
Subtracting $2R_1$ from $R_3$ produces two zeros in row 3, making cofactor expansion trivial.
Evaluate the signed minors for each position.
Divide the transposed cofactor matrix by $\det(A) = 5$.
\det(A) = 5; \quad \mathbf{A^{-1} = \frac{1}{5}\begin{pmatrix} -4 & 5 & 2 \\ 7 & -10 & -1 \\ 2 & 0 & -1 \end{pmatrix}}.
(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 $V(a, b, c) = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$.
Construct $S$ and $K$ explicitly, demonstrating existence.
Taking transposes and adding/subtracting establishes uniqueness.
Factoring $(b-a)$ and $(c-a)$ leaves a simple $2 \times 2$ determinant.
\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)}.
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.
The linear combination produces a common factor $(a + \omega b + \omega^2 c)$ down the entire first column!
The determinant of any matrix equals the product of its eigenvalues.
Using $1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1$ recovers Euler's classical cubic product factorization.
\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)}.