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Concept 07 / 09Matrices and Determinants

Special Matrices: Orthogonal, Idempotent, Involutory and Nilpotent, and Properties of the Trace

Orthogonal matrices as rigid rotations and reflections, idempotent, involutory and nilpotent matrices and what they force on determinants and eigenvalues, the trace and its properties including tr(AB) = tr(BA), and why AB - BA can never equal I.

Figure 1
Several special classes of square matrices are defined by an algebraic identity, and each identity immediately restricts the determinant and the eigenvalues. Orthogonal: ATA=AAT=IA^{T}A = AA^{T} = I, so A−1=ATA^{-1} = A^{T}. Equivalently, the columns (and also the rows) of AA are mutually perpendicular unit vectors. Geometrically an orthogonal matrix is a rigid motion fixing the origin: it preserves lengths and angles, because ∣Ax∣2=xTATAx=xTx=∣x∣2|Ax|^{2} = x^{T}A^{T}Ax = x^{T}x = |x|^{2}. Taking determinants, ∣A∣2=1|A|^{2} = 1, so ∣A∣=±1|A| = \pm 1: +1+1 for rotations and −1-1 for reflections. The standard example is the rotation matrix R(θ)=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}, shown in the figure, which satisfies R(α)R(β)=R(α+β)R(\alpha)R(\beta) = R(\alpha + \beta) and R(θ)n=R(nθ)R(\theta)^{n} = R(n\theta), a matrix version of De Moivre's theorem. Products and inverses of orthogonal matrices are orthogonal. A common question type: find unknown entries so that a matrix is orthogonal, by imposing unit length on each column and zero dot product between columns. Idempotent: A2=AA^{2} = A. Then An=AA^{n} = A for every n≥1n \ge 1. Determinant: ∣A∣2=∣A∣|A|^{2} = |A|, so ∣A∣∈{0,1}|A| \in \{0, 1\}; the only invertible idempotent matrix is II (multiply A2=AA^{2} = A by A−1A^{-1}). Eigenvalues are 00 or 11. Projections are idempotent: projecting twice is the same as projecting once. If AA is idempotent, so is I−AI - A, and A(I−A)=OA(I - A) = O. Involutory: A2=IA^{2} = I, so A−1=AA^{-1} = A. Then ∣A∣=±1|A| = \pm 1 and eigenvalues are ±1\pm 1. Reflections are involutory. The matrix AA is involutory if and only if (I−A)(I+A)=O(I - A)(I + A) = O. Nilpotent: Ak=OA^{k} = O for some positive integer kk; the least such kk is the index. Then ∣A∣k=0|A|^{k} = 0, so ∣A∣=0|A| = 0, and every eigenvalue is 00, so tr A=0\text{tr}\,A = 0. A strictly upper triangular matrix (zeros on and below the diagonal) is nilpotent. If AA is nilpotent of index kk, then I−AI - A is invertible with (I−A)−1=I+A+A2+⋯+Ak−1(I - A)^{-1} = I + A + A^{2} + \cdots + A^{k-1}, a finite geometric series: multiply out and everything telescopes to I−Ak=II - A^{k} = I. Unitary and Hermitian (complex versions, occasionally tested): AθA^{\theta} denotes the conjugate transpose AˉT\bar{A}^{T}. Hermitian means Aθ=AA^{\theta} = A (the diagonal is real), skew-Hermitian means Aθ=−AA^{\theta} = -A (the diagonal is purely imaginary or zero), and unitary means AθA=IA^{\theta}A = I. Trace. The trace tr A=∑aii\text{tr}\,A = \sum a_{ii} is the sum of the diagonal entries. Properties: tr(A+B)=tr A+tr B\text{tr}(A + B) = \text{tr}\,A + \text{tr}\,B, tr(kA)=k tr A\text{tr}(kA) = k\,\text{tr}\,A, tr(AT)=tr A\text{tr}(A^{T}) = \text{tr}\,A, and the key one, tr(AB)=tr(BA)\text{tr}(AB) = \text{tr}(BA), even though AB≠BAAB \neq BA. Proof: tr(AB)=∑i∑kaikbki\text{tr}(AB) = \sum_{i}\sum_{k}a_{ik}b_{ki}, and tr(BA)=∑k∑ibkiaik\text{tr}(BA) = \sum_{k}\sum_{i}b_{ki}a_{ik}, the same double sum. It extends to cyclic permutations: tr(ABC)=tr(BCA)=tr(CAB)\text{tr}(ABC) = \text{tr}(BCA) = \text{tr}(CAB), but not to arbitrary reorderings. Also tr(ATA)=∑i,jaij2\text{tr}(A^{T}A) = \sum_{i,j}a_{ij}^{2}, which is 00 only if A=OA = O. And the trace equals the sum of the eigenvalues. Key Takeaways: - Orthogonal: ATA=IA^{T}A = I, A−1=ATA^{-1} = A^{T}, ∣A∣=±1|A| = \pm 1; columns are orthonormal. - Idempotent: A2=AA^{2} = A, ∣A∣∈{0,1}|A| \in \{0, 1\}, eigenvalues 00 or 11. - Involutory: A2=IA^{2} = I, A−1=AA^{-1} = A, ∣A∣=±1|A| = \pm 1. - Nilpotent: Ak=OA^{k} = O, ∣A∣=0|A| = 0, tr A=0\text{tr}\,A = 0; (I−A)−1=I+A+⋯+Ak−1(I - A)^{-1} = I + A + \cdots + A^{k-1}. - tr(AB)=tr(BA)\text{tr}(AB) = \text{tr}(BA); the trace is invariant under cyclic permutations only. - tr(ATA)=\text{tr}(A^{T}A) = sum of squares of all entries. - R(θ)n=R(nθ)R(\theta)^{n} = R(n\theta) for the rotation matrix. - Pitfall: assuming tr(ABC)=tr(ACB)\text{tr}(ABC) = \text{tr}(ACB). Only cyclic shifts are allowed. Brain-Teaser 1: Prove that there are no square matrices A,BA, B (with real or complex entries) such that AB−BA=IAB - BA = I. Solution: Take the trace of both sides. tr(AB−BA)=tr(AB)−tr(BA)=0\text{tr}(AB - BA) = \text{tr}(AB) - \text{tr}(BA) = 0, while tr(In)=n≥1\text{tr}(I_{n}) = n \ge 1. Contradiction. This short argument is a classic because it shows the power of an invariant: the trace ignores the order of a product, so any commutator AB−BAAB - BA has trace zero, and the identity matrix does not. Brain-Teaser 2: Find a,b,ca, b, c so that A=[02bcab−ca−bc]A = \begin{bmatrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \end{bmatrix} is orthogonal. Solution: The columns must be orthonormal. Column 1 is (0,a,a)(0, a, a): unit length gives 2a2=12a^{2} = 1. Column 2 is (2b,b,−b)(2b, b, -b): 6b2=16b^{2} = 1. Column 3 is (c,−c,c)(c, -c, c): 3c2=13c^{2} = 1. Dot products: column 1 with column 2 gives ab−ab=0ab - ab = 0; column 1 with column 3 gives −ac+ac=0-ac + ac = 0; column 2 with column 3 gives 2bc−bc−bc=02bc - bc - bc = 0. All automatically satisfied. So a=±12a = \pm\frac{1}{\sqrt{2}}, b=±16b = \pm\frac{1}{\sqrt{6}}, c=±13c = \pm\frac{1}{\sqrt{3}}, with any combination of signs. Checking the columns one by one is much faster than multiplying out ATAA^{T}A in full. Brain-Teaser 3: If AA is idempotent and I+AI + A is invertible, find (I+A)−1(I + A)^{-1} in terms of AA. What is (I+A)n(I + A)^{n}? Solution: Try (I+A)(I+kA)=I+(k+1)A+kA2=I+(2k+1)A(I + A)(I + kA) = I + (k + 1)A + kA^{2} = I + (2k + 1)A, using A2=AA^{2} = A. This equals II when k=−12k = -\frac{1}{2}. So (I+A)−1=I−12A(I + A)^{-1} = I - \frac{1}{2}A. For the power, since II and AA commute, the binomial theorem applies, and every power of AA equals AA: (I+A)n=I+[(n1)+(n2)+⋯+(nn)]A=I+(2n−1)A(I + A)^{n} = I + [\binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n}]A = I + (2^{n} - 1)A. Check at n=1n = 1: I+AI + A. The idempotent relation collapses all the higher powers, so the whole binomial sum reduces to a single coefficient.

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