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Concept 06 / 09Matrices and Determinants

Characteristic Equation, Cayley-Hamilton Theorem and Powers of Matrices

Eigenvalues as roots of |A - lambda I| = 0, the trace and determinant as their sum and product, the Cayley-Hamilton theorem, and using it to find inverses and high powers of a matrix without repeated multiplication.

Figure 1
For a square matrix AA, a non-zero vector vv is an eigenvector with eigenvalue λ\lambda if Av=λvAv = \lambda v: the transformation AA does not turn vv, it only scales it. Rewriting, (A−λI)v=O(A - \lambda I)v = O with v≠Ov \neq O, which is a homogeneous system with a non-trivial solution. By the previous concept, this happens exactly when ∣A−λI∣=0|A - \lambda I| = 0. This polynomial equation in λ\lambda is the characteristic equation of AA, and its roots are the eigenvalues. For a 2×22 \times 2 matrix it reads λ2−(tr A)λ+∣A∣=0\lambda^{2} - (\text{tr}\,A)\lambda + |A| = 0, where the trace tr A\text{tr}\,A is the sum of the diagonal entries. For a 3×33 \times 3 matrix, λ3−(tr A)λ2+(sum of the principal 2×2 minors)λ−∣A∣=0\lambda^{3} - (\text{tr}\,A)\lambda^{2} + (\text{sum of the principal } 2 \times 2 \text{ minors})\lambda - |A| = 0. By Vieta's formulas, the sum of the eigenvalues equals the trace and the product equals the determinant. So a matrix is singular exactly when 00 is an eigenvalue. Other quick facts: the eigenvalues of a triangular matrix are its diagonal entries; AA and ATA^{T} have the same eigenvalues; if λ\lambda is an eigenvalue of AA, then λk\lambda^{k} is an eigenvalue of AkA^{k}, 1λ\frac{1}{\lambda} of A−1A^{-1}, and λ+c\lambda + c of A+cIA + cI. Real symmetric matrices have only real eigenvalues. The figure shows A=[2112]A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}: the characteristic equation is λ2−4λ+3=0\lambda^{2} - 4\lambda + 3 = 0, so λ=3\lambda = 3 and λ=1\lambda = 1. Vectors along (1,1)(1, 1) are stretched by 33, vectors along (1,−1)(1, -1) are unchanged, and any other vector is turned towards the (1,1)(1, 1) direction. The Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation. If ∣A−λI∣=λn+cn−1λn−1+⋯+c0|A - \lambda I| = \lambda^{n} + c_{n-1}\lambda^{n-1} + \cdots + c_{0}, then An+cn−1An−1+⋯+c0I=OA^{n} + c_{n-1}A^{n-1} + \cdots + c_{0}I = O. For 2×22 \times 2 matrices, A2−(tr A)A+∣A∣I=OA^{2} - (\text{tr}\,A)A + |A|I = O, which you can verify directly by multiplying out. A tempting but wrong 'proof' is to substitute λ=A\lambda = A into ∣A−λI∣|A - \lambda I| and get ∣A−A∣=0|A - A| = 0; this confuses the number 00 with the zero matrix. The real proof uses the adjoint of A−λIA - \lambda I. Uses of Cayley-Hamilton. Inverse: if ∣A∣≠0|A| \neq 0, the constant term c0=±∣A∣c_{0} = \pm|A| is non-zero, so rearrange An+⋯+c1A=−c0IA^{n} + \cdots + c_{1}A = -c_{0}I and factor out AA. For 2×22 \times 2: A−1=1∣A∣((tr A)I−A)A^{-1} = \frac{1}{|A|}((\text{tr}\,A)I - A). High powers: every power AkA^{k} can be reduced to a combination of I,A,…,An−1I, A, \ldots, A^{n-1}. For 2×22 \times 2, Ak=pkA+qkIA^{k} = p_{k}A + q_{k}I. One way to find pk,qkp_{k}, q_{k}: the same relation holds for the eigenvalues, λk=pkλ+qk\lambda^{k} = p_{k}\lambda + q_{k} for each eigenvalue, giving two linear equations (when the eigenvalues are distinct). Special cases: if A2=AA^{2} = A (idempotent) then Ak=AA^{k} = A for all k≥1k \ge 1; if A2=IA^{2} = I (involutory) then AkA^{k} alternates between AA and II; if A2=kAA^{2} = kA then An=kn−1AA^{n} = k^{n-1}A. Key Takeaways: - Eigenvalues: roots of ∣A−λI∣=0|A - \lambda I| = 0. Sum =tr A= \text{tr}\,A, product =∣A∣= |A|. - 2×22 \times 2 characteristic equation: λ2−(tr A)λ+∣A∣=0\lambda^{2} - (\text{tr}\,A)\lambda + |A| = 0. - Cayley-Hamilton: A2−(tr A)A+∣A∣I=OA^{2} - (\text{tr}\,A)A + |A|I = O for 2×22 \times 2. - A−1=(tr A)I−A∣A∣A^{-1} = \frac{(\text{tr}\,A)I - A}{|A|} for 2×22 \times 2. - AA singular   ⟺  0\iff 0 is an eigenvalue. - Triangular matrix: eigenvalues are the diagonal entries. - If A2=kAA^{2} = kA, then An=kn−1AA^{n} = k^{n-1}A. - Pitfall: 'proving' Cayley-Hamilton by substituting λ=A\lambda = A into a determinant. Brain-Teaser 1: For A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, express A3A^{3} as pA+qIpA + qI and find A−1A^{-1}. Solution: tr A=5\text{tr}\,A = 5 and ∣A∣=−2|A| = -2, so Cayley-Hamilton gives A2=5A+2IA^{2} = 5A + 2I. Then A3=A⋅A2=5A2+2A=5(5A+2I)+2A=27A+10IA^{3} = A \cdot A^{2} = 5A^{2} + 2A = 5(5A + 2I) + 2A = 27A + 10I. For the inverse, A2−5A=2IA^{2} - 5A = 2I, so A(A−5I)=2IA(A - 5I) = 2I and A−1=12(A−5I)=12[−423−1]A^{-1} = \frac{1}{2}(A - 5I) = \frac{1}{2}\begin{bmatrix} -4 & 2 \\ 3 & -1 \end{bmatrix}. Check against the 2×22 \times 2 formula: 1−2[4−2−31]\frac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}, which is the same. Repeatedly replacing A2A^{2} reduces any power to a linear expression. Brain-Teaser 2: AA is a 3×33 \times 3 matrix with eigenvalues 1,2,31, 2, 3. Find ∣A∣|A|, tr(A2)\text{tr}(A^{2}), ∣A−1+I∣|A^{-1} + I| and ∣adj A∣|\text{adj}\,A|. Solution: ∣A∣=1⋅2⋅3=6|A| = 1 \cdot 2 \cdot 3 = 6. The eigenvalues of A2A^{2} are 1,4,91, 4, 9, so tr(A2)=14\text{tr}(A^{2}) = 14. The eigenvalues of A−1+IA^{-1} + I are 1λ+1\frac{1}{\lambda} + 1, that is 2,32,432, \frac{3}{2}, \frac{4}{3}, whose product is 44. And ∣adj A∣=∣A∣2=36|\text{adj}\,A| = |A|^{2} = 36. Working with eigenvalues turns questions about complicated matrices into arithmetic on three numbers. Brain-Teaser 3: If A=[111111111]A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}, find AnA^{n} and the eigenvalues of AA. Solution: Every entry of A2A^{2} is a row of ones dotted with a column of ones, which is 33, so A2=3AA^{2} = 3A. Then An=3n−1AA^{n} = 3^{n-1}A by induction: An+1=AnA=3n−1A2=3nAA^{n+1} = A^{n}A = 3^{n-1}A^{2} = 3^{n}A. For eigenvalues: AA has rank 11, so ∣A∣=0|A| = 0, and 00 is an eigenvalue of multiplicity 22 (any vector with entries summing to 00 is sent to OO). The trace is 33, so the remaining eigenvalue is 33, with eigenvector (1,1,1)(1, 1, 1). The relation A2=3AA^{2} = 3A also says that every eigenvalue satisfies λ2=3λ\lambda^{2} = 3\lambda, so λ∈{0,3}\lambda \in \{0, 3\}, which is consistent.