Chapter 10

Cayley-Hamilton Theorem

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Problem 10.1.1

Let AA be an n×nn \times n matrix over C\mathbb {C} and

p(λ)=det⁡(A−λIn). p(\lambda ) = \operatorname {det}\left(A - \lambda I_{n}\right).

Show that p(A)=0np(A) = 0_{n}, i.e. the matrix AA satisfies its characteristic equation (Cayley-Hamilton theorem).

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Problem 10.1.2

Let AA be an arbitrary 2×22 \times 2 matrix. Show that

A2−Atr⁡(A)+I2det⁡(A)=0 A^{2} - A \operatorname {tr}(A) + I_{2} \operatorname {det}\left(A\right) = 0

and therefore (tr⁡(A))2=tr⁡(A2)+2det⁡(A)(\operatorname {tr}(A))^{2} = \operatorname {tr}(A^{2}) + 2 \operatorname {det}\left(A\right).

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Problem 10.1.3

Consider the 2×22 \times 2 nonnormal matrix

A=(1101). A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.

Calculate f(A)=A3f(A) = A^{3} using the Cayley-Hamilton theorem.

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Problem 10.1.4

Apply the Cayley-Hamilton theorem to the 3×33 \times 3 matrix A=(ajk)A = (a_{jk}) and express the result using the trace and determinant of AA.

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Problem 10.1.5

Let AA be an n×nn \times n matrix. Assume that the inverse matrix of AA exists. The inverse matrix can be calculated as follows (Csanky’s algorithm). Let

p(x):=det⁡(xIn−A). p(x) := \operatorname {det}\left(xI_{n} - A\right).

The roots are, by definition, the eigenvalues λ1,λ2,…,λn\lambda_{1}, \lambda_{2}, \ldots , \lambda_{n} of AA. We write

p(x)=xn+c1xn−1+⋯+cn−1x+cn p(x) = x^{n} + c_{1}x^{n-1} + \cdots + c_{n-1}x + c_{n}

where cn=(−1)ndet⁡(A)c_{n} = (-1)^{n} \operatorname {det}\left(A\right). Since AA is nonsingular we have cn≠0c_{n} \neq 0 and vice versa. The Cayley-Hamilton theorem states that

p(A)=An+c1An−1+⋯+cn−1A+cnIn=0n. p(A) = A^{n} + c_{1}A^{n-1} + \cdots + c_{n-1}A + c_{n}I_{n} = 0_{n}.

Multiplying this equation with A−1A^{-1} we obtain

A−1=1−cn(An−1+c1An−2+⋯+cn−1In). A^{-1} = \frac{1}{-c_{n}}(A^{n-1} + c_{1}A^{n-2} + \cdots + c_{n-1}I_{n}).

If we have the coefficients cjc_{j} we can calculate the inverse matrix AA. Let

sk:=∑j=1nλjk. s_{k} := \sum _{j=1}^{n} \lambda _{j}^{k}.

Then the sjs_{j} and cjc_{j} satisfy the following n×nn \times n lower triangular system of linear equations

(100…0s120…0s2s13…0⋮⋮⋮⋱⋮sn−1sn−2…s1n)(c1c2c3⋮cn)=(−s1−s2−s3⋮−sn). \begin{pmatrix} 1 & 0 & 0 & \ldots & 0 \\ s_{1} & 2 & 0 & \ldots & 0 \\ s_{2} & s_{1} & 3 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ s_{n-1} & s_{n-2} & \ldots & s_{1} & n \end{pmatrix} \begin{pmatrix} c_{1} \\ c_{2} \\ c_{3} \\ \vdots \\ c_{n} \end{pmatrix} = \begin{pmatrix} -s_{1} \\ -s_{2} \\ -s_{3} \\ \vdots \\ -s_{n} \end{pmatrix}.

Since

tr⁡(Ak)=λ1k+λ2k+⋯+λnk=sk \operatorname {tr}(A^{k}) = \lambda _{1}^{k} + \lambda _{2}^{k} + \cdots + \lambda _{n}^{k} = s_{k}

we find sks_{k} for k=1,…,nk = 1, \ldots , n. Thus we can solve the linear equation for cjc_{j}. Finally, using the inverse-matrix formula above we obtain the inverse matrix of AA. Apply Csanky's algorithm to the 3×33 \times 3 matrix

A=(101100011). A = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix}.
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Problem 10.1.6

Let AA be an n×nn \times n matrix. Let

c(z):=det⁡(zIn−A)=zn−∑k=0n−1ckzk c(z) := \operatorname {det}\left(zI_{n} - A\right) = z^{n} - \sum _{k=0}^{n-1} c_{k}z^{k}

be the characteristic polynomial of AA. Apply the Cayley-Hamilton theorem c(A)=0nc(A) = 0_{n} to calculate exp⁡(A)\exp (A).

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Problem 10.1.7
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(i)

Let AA be an n×nn \times n matrix with A3=InA^{3} = I_{n}. Calculate exp⁡(A)\exp (A) using

exp⁡(A)=∑j=0∞Ajj!. \exp (A) = \sum _{j=0}^{\infty } \frac{A^{j}}{j!}.
(ii)

Let

B=(010001100) B = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}

with B3=I3B^{3} = I_{3}. Calculate exp⁡(B)\exp (B) applying the Cayley-Hamilton theorem.

Problem 10.1.8

Let AA be an n×nn \times n matrix over C\mathbb {C}. Let ff be an entire function, i.e. an analytic function on the whole complex plane, for example exp⁡(z)\exp (z), sin⁡(z)\sin (z), cos⁡(z)\cos (z). Using the Cayley-Hamilton theorem we can write

f(A)=an−1An−1+an−2An−2+⋯+a2A2+a1A+a0In f(A) = a_{n-1}A^{n-1} + a_{n-2}A^{n-2} + \cdots + a_{2}A^{2} + a_{1}A + a_{0}I_{n}

where the complex numbers a0,a1,…,an−1a_{0}, a_{1}, \ldots , a_{n-1} are determined as follows: Let

r(λ):=an−1λn−1+an−2λn−2+⋯+a2λ2+a1λ+a0 r(\lambda ) := a_{n-1}\lambda ^{n-1} + a_{n-2}\lambda ^{n-2} + \cdots + a_{2}\lambda ^{2} + a_{1}\lambda + a_{0}

which is the right-hand side of the previous equation with AjA^{j} replaced by λj\lambda^{j}, where j=0,1,…,n−1j = 0, 1, \ldots , n-1 of each distinct eigenvalue λj\lambda_{j} of the matrix AA, we consider the equation

f(λj)=r(λj). f(\lambda _{j}) = r(\lambda _{j}).

If λj\lambda_{j} is an eigenvalue of multiplicity kk, for k>1k > 1, then we consider also the following equations

f′(λ)∣λ=λj=r′(λ)∣λ=λjf′′(λ)∣λ=λj=r′′(λ)∣λ=λj⋯=⋯f(k−1)(λ)∣λ=λj=r(k−1)(λ)∣λ=λj. \begin{align} f'(\lambda )\big|_{\lambda = \lambda _{j}} & = r'(\lambda )\big|_{\lambda = \lambda _{j}} \\ f''(\lambda )\big|_{\lambda = \lambda _{j}} & = r''(\lambda )\big|_{\lambda = \lambda _{j}} \\ \cdots & = \cdots \\ f^{(k-1)}(\lambda )\big|_{\lambda = \lambda _{j}} & = r^{(k-1)}(\lambda )\big|_{\lambda = \lambda _{j}}. \end{align}

Apply this technique to find exp⁡(A)\exp (A) with

A=(cccc),c∈R,c≠0. A = \begin{pmatrix} c & c \\ c & c \end{pmatrix}, \qquad c \in \mathbb {R}, \quad c \neq 0.
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Problem 10.1.9

Let z1,z2,z3∈Cz_{1}, z_{2}, z_{3} \in \mathbb {C}. Assume that at least two of the three complex numbers are nonzero. Consider the 2×22 \times 2 matrix

A=(z1z2z30). A = \begin{pmatrix} z_{1} & z_{2} \\ z_{3} & 0 \end{pmatrix}.

Calculate exp⁡(A)\exp (A) applying the Cayley-Hamilton theorem. The characteristic equation for AA is given by λ2−z1λ−z2z3=0\lambda^{2} - z_{1}\lambda - z_{2}z_{3} = 0 with the eigenvalues

λ+=12z1+z12+4z2z3,λ−=12z1−z12+4z2z3. \lambda _{+} = \frac{1}{2}z_{1} + \sqrt{z_{1}^{2} + 4z_{2}z_{3}}, \qquad \lambda _{-} = \frac{1}{2}z_{1} - \sqrt{z_{1}^{2} + 4z_{2}z_{3}}.

Cayley-Hamilton theorem then states that A2−z1A−z1z3I2=02A^{2} - z_{1}A - z_{1}z_{3}I_{2} = 0_{2}.

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Problem 10.1.10

Let α∈R\alpha \in \mathbb {R} and α≠0\alpha \neq 0. Consider the 2×22 \times 2 matrix

A(α)=(0αα0). A(\alpha ) = \begin{pmatrix} 0 & \alpha \\ \alpha & 0 \end{pmatrix}.

Calculate sin⁡(A(α))\sin (A(\alpha )) applying the Cayley-Hamilton theorem.

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Problem 10.1.11

Calculate

sec⁡(π200π20π2π200π2−π20π200−π2). \sec \begin{pmatrix} \frac{\pi }{\sqrt{2}} & 0 & 0 & \frac{\pi }{\sqrt{2}} \\ 0 & \frac{\pi }{\sqrt{2}} & \frac{\pi }{\sqrt{2}} & 0 \\ 0 & \frac{\pi }{\sqrt{2}} & -\frac{\pi }{\sqrt{2}} & 0 \\ \frac{\pi }{\sqrt{2}} & 0 & 0 & -\frac{\pi }{\sqrt{2}} \end{pmatrix}.
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Problem 10.1.12

Let AA, BB be n×nn \times n matrices such that ABAB=0nABAB = 0_{n}. Can we conclude that BABA=0nBABA = 0_{n}?

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Problem 10.2.1

Consider the nonnormal invertible 2×22 \times 2 matrix

A=(110−1). A = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}.

Find A2A^{2}, A3A^{3} and A−1A^{-1} utilizing the Cayley-Hamilton theorem.

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Problem 10.2.2

Let ϵ∈R\epsilon \in \mathbb {R} and ϵ≠0\epsilon \neq 0. Let

A(ϵ)=(0ϵϵ0). A(\epsilon ) = \begin{pmatrix} 0 & \epsilon \\ \epsilon & 0 \end{pmatrix}.
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(i)

Calculate cos⁡(A(ϵ))\cos (A(\epsilon )) and cos⁡(A(ϵ)⊗A(ϵ))\cos (A(\epsilon ) \otimes A(\epsilon )) applying the Cayley-Hamilton theorem.

(ii)

Calculate tanh⁡(A(ϵ))\tanh (A(\epsilon )) applying the Cayley-Hamilton theorem.

Problem 10.2.3

Let ϵ∈R\epsilon \in \mathbb {R}. Consider the 4×44 \times 4 matrix

A(ϵ)=(0ϵϵϵ)⊗(0ϵϵϵ). A(\epsilon ) = \begin{pmatrix} 0 & \epsilon \\ \epsilon & \epsilon \end{pmatrix} \otimes \begin{pmatrix} 0 & \epsilon \\ \epsilon & \epsilon \end{pmatrix}.

Find exp⁡(A(ϵ))\exp (A(\epsilon )) applying the Cayley-Hamilton theorem.

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Problem 10.2.4

Consider the Bell matrix

B=12(1001011001−10100−1) B = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{pmatrix}

which is a unitary matrix. Apply the Cayley-Hamilton theorem to find the skew-hermitian matrix KK such that B=eKB = e^{K}.

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