Biquaternion Spectral Theory

Introduction

The biquaternion algebra $\mathbb{B} = \mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$ is unital and associative but neither commutative nor a division algebra, and both failures complicate the notion of a spectrum. Several inequivalent definitions of "spectrum" are in use, and the main source of error in the subject is the silent switching between them. We state, for every definition, exactly which set it defines; and "spectrum" without qualification always means the spectrum of the biquaternion defined by invertibility, that is, its set of eigenvalues in $\mathbb{C}$.

The treatment is mathematically honest: every claim is either proved or stated as a definition, and no new results or speculation are introduced. We assume the algebra, its conjugations, and its six distinguished subspaces from the article on biquaternion algebra; the norm and invertibility criterion from the article on biquaternion norm and invertibility; and the exponential and logarithm from the article on biquaternion elementary functions, where the eigenvalues $Q_0 \pm iB$ also occur, in the commutative subalgebra generated by one biquaternion.

The Spectrum of an Element of an Algebra

Let $A$ be a unital associative algebra over a field $F$, with unit $1_A$.

Definition 1.1. The spectrum of $a \in A$ is $\sigma_A(a) = \{\lambda \in F : a - \lambda 1_A \text{ is not invertible in } A\}$. It is also called the point spectrum when $A$ is a finite-dimensional algebra of operators. The definition is intrinsic: it uses only invertibility.

Proposition 1.2. If $A$ is a finite-dimensional unital algebra over $\mathbb{C}$ with $A \neq 0$, and $L_a(x) = ax$ is left multiplication, then $\sigma_A(a)$ is exactly the set of eigenvalues of $L_a$, and is nonempty.

Proof. $b$ is invertible iff $L_b$ is bijective. The characteristic polynomial of $L_a$ has a root $\mu \in \mathbb{C}$, and $L_a - \mu\,\mathrm{id} = L_{a-\mu 1_A}$ is singular, so $\mu \in \sigma_A(a)$; conversely $\lambda \in \sigma_A(a)$ makes $L_{a-\lambda 1_A}$ non-bijective, hence $\lambda$ an eigenvalue.

Corollary 1.3. For $A = M_n(\mathbb{C})$, $\sigma(M) = \{\lambda \in \mathbb{C} : \det(\lambda I - M) = 0\}$.

In finite dimension left and right invertibility coincide, since $ab = 1_A$ makes $L_a$ bijective; in infinite dimension they need not. This is the origin of the left/right distinction below, which cannot occur here.

Left and Right Eigenvalues

Let $D$ be a division ring and $M \in M_n(D)$, acting on column vectors $D^n$. For $\lambda \in D$ and $v \in D^n$ the products $\lambda v$ and $v\lambda$ differ in general.

Definition 2.1. A scalar $\lambda \in D$ is a right eigenvalue of $M$ if $Mv = v\lambda$ for some nonzero $v \in D^n$, and a left eigenvalue if $Mv = \lambda v$ for some nonzero $v$. The sets of such $\lambda$ are the right spectrum $\sigma_R(M)$ and the left spectrum $\sigma_L(M)$.

Proposition 2.2. $\lambda \in \sigma_L(M)$ iff $M - \lambda I$ is singular.

Proof. $Mv = \lambda v \iff (M-\lambda I)v = 0$, and a square matrix over a division ring is singular iff it has a nonzero kernel vector.

Thus $\sigma_L(M)$ is exactly the spectrum of Definition 1.1 with scalars in $D$. The right problem does not reduce this way, since $v\lambda \neq \lambda v$ in general; note that $\sigma_R(M)$ is a union of conjugacy classes, because $Mv = v\lambda$ gives $M(vw) = (vw)(w^{-1}\lambda w)$ for every nonzero $w$.

Example 2.3. For $D = \mathbb{H}$, $n = 1$, $M = [q]$, the equation $qv = \lambda v$ forces $\lambda = q$, so $\sigma_L([q]) = \{q\}$, while $\sigma_R([q]) = \{v^{-1}qv : v \in \mathbb{H}^\times\}$ is a point for real $q$ and a $2$-sphere otherwise. Thus $\sigma_L([q]) \neq \sigma_R([q])$ for non-real $q$.

Proposition 2.4. If $D$ is a field, then $\sigma_L(M) = \sigma_R(M) = \{\lambda : \det(\lambda I - M) = 0\}$.

Consequence for the biquaternions. Since the centre of $\mathbb{B}$ is $\mathbb{C}$, complex scalars commute with everything; if eigenvalues are required to lie in $\mathbb{C}$, the left and right problems for a biquaternion coincide and agree with the matrix spectrum below. If eigenvalues may be arbitrary elements of $\mathbb{B}$, the two spectra can differ, but that problem is not studied here.

The Biquaternion Spectrum

A biquaternion is written $\tilde{Q} = \sum_{\mu=0}^3 Q_\mu e_\mu$. Its trace and biquaternion norm functionals are $$ T(\tilde{Q}) = 2Q_0, \qquad D(\tilde{Q}) = N(\tilde{Q}) = Q_0^2 + Q_1^2 + Q_2^2 + Q_3^2 . $$ Write $\mathbf{Q} = Q_1 e_1 + Q_2 e_2 + Q_3 e_3$ and let $B = \sqrt{Q_1^2 + Q_2^2 + Q_3^2}$ be a fixed square root, the complex norm of the vector part; replacing $B$ by $-B$ changes nothing.

Definition 3.1. The spectrum of $\tilde{Q}$ is $\sigma(\tilde{Q}) = \{\lambda \in \mathbb{C} : \tilde{Q} - \lambda e_0 \text{ is not invertible in } \mathbb{B}\}$, the spectrum of Definition 1.1. Its characteristic polynomial is $$ p_{\tilde{Q}}(\lambda) = \lambda^2 - 2Q_0\lambda + N(\tilde{Q}) = \lambda^2 - 2Q_0\lambda + Q_0^2 + B^2 = (\lambda - Q_0 - iB)(\lambda - Q_0 + iB), $$ so $$ \sigma(\tilde{Q}) = \{Q_0 + iB, \, Q_0 - iB\}, $$ the two roots coinciding exactly when $B = 0$.

Consistency with invertibility. Since $N(\tilde{Q} - \lambda e_0) = (Q_0-\lambda)^2 + Q_1^2 + Q_2^2 + Q_3^2 = p_{\tilde{Q}}(\lambda)$, the spectrum contains $0$ exactly when $N(\tilde{Q}) = 0$, i.e. when $\tilde{Q}$ is zero or a zero divisor. The spectrum of left multiplication on $\mathbb{B} \cong \mathbb{C}^4$ is the same set, with doubled multiplicity (§Eigenspaces and Their Dimensions).

Cayley–Hamilton and the Trace and Determinant Functionals

Theorem 4.1 (Cayley–Hamilton). Every $\tilde{Q} \in \mathbb{B}$ satisfies $\tilde{Q}^2 - 2Q_0\tilde{Q} + N(\tilde{Q})e_0 = 0$, i.e. $\tilde{Q}^2 - T(\tilde{Q})\tilde{Q} + D(\tilde{Q})e_0 = 0$.

Proof. With $\tilde{Q} = Q_0e_0 + \mathbf{Q}$ and $\mathbf{Q}^2 = -B^2 e_0$, $$ \tilde{Q}^2 = Q_0^2 e_0 + 2Q_0\mathbf{Q} - B^2 e_0 = 2Q_0\tilde{Q} - (Q_0^2 + B^2)e_0 = 2Q_0\tilde{Q} - N(\tilde{Q})e_0. $$

Corollary 4.2. If $N(\tilde{Q}) \neq 0$, then $\tilde{Q}^{-1} = (2Q_0e_0 - \tilde{Q})/N(\tilde{Q}) = \tilde{Q}^{\natural}/N(\tilde{Q})$.

Proposition 4.3. The functionals $T, D$ satisfy $$ T(\tilde{P}+\tilde{Q}) = T(\tilde{P})+T(\tilde{Q}), \quad T(\lambda\tilde{Q}) = \lambda T(\tilde{Q}), \quad T(\tilde{P}\tilde{Q}) = T(\tilde{Q}\tilde{P}), \quad T(e_0) = 2, $$ $$ D(\tilde{P}\tilde{Q}) = D(\tilde{P})D(\tilde{Q}), \quad D(\lambda\tilde{Q}) = \lambda^2 D(\tilde{Q}), \quad D(e_0) = 1, $$ and both are invariant under similarity. The determinant $D = N$ is multiplicative, the trace is linear and cyclic but not multiplicative, and $D(\exp\tilde{Q}) = e^{T(\tilde{Q})}$.

Proof. Direct computation from the definitions, with $T(\tilde{Q}) = 2Q_0$ and $D(\tilde{Q}) = N(\tilde{Q})$.

The Commutative Subalgebra Generated by a Single Biquaternion

The functionals of the previous section are the coefficients of the vanishing polynomial, and this makes the algebra generated by a single element transparent. For $\tilde{Q} \in \mathbb{B}$, the commutative subalgebra generated by $\tilde{Q}$ is the $\mathbb{C}$-linear span of $\{e_0, \tilde{Q}, \tilde{Q}^2, \dots\}$, denoted $\mathbb{C}[\tilde{Q}]$; it is the smallest subalgebra of $\mathbb{B}$ containing $e_0$ and $\tilde{Q}$, and it is the image of the algebra homomorphism $$ \mathbb{C}[x] \longrightarrow \mathbb{B}, \qquad x \longmapsto \tilde{Q}. $$

The minimal polynomial. The kernel of this homomorphism is the principal ideal $(m_{\tilde{Q}})$ generated by the minimal polynomial $m_{\tilde{Q}}$, the monic polynomial of least degree with $m_{\tilde{Q}}(\tilde{Q}) = 0$. By Cayley–Hamilton $m_{\tilde{Q}}$ divides the characteristic polynomial $x^2 - 2Q_0x + N(\tilde{Q})$, so $\deg m_{\tilde{Q}} \in \{1, 2\}$, and $$ \mathbb{C}[\tilde{Q}] \cong \mathbb{C}[x]/(m_{\tilde{Q}}), \qquad \mathbb{C}[\tilde{Q}] = \operatorname{span}_{\mathbb{C}}\{e_0, \tilde{Q}\}. $$

The two cases. Let $\Delta = -4B^2$ be the discriminant of the characteristic polynomial.

$\Delta \neq 0$ ($B \neq 0$). $m_{\tilde{Q}} = x^2 - 2Q_0x + N(\tilde{Q})$ splits, with roots $Q_0 \pm iB$, and the Chinese remainder theorem gives $\mathbb{C}[\tilde{Q}] \cong \mathbb{C} \times \mathbb{C}$, evaluation at the two roots. This is the semisimple case; $\tilde{Q}$ is diagonalisable with eigenvalues $Q_0 \pm iB$.
$\Delta = 0$ and $\mathbf{Q} \neq 0$ ($B = 0$). $m_{\tilde{Q}} = (x - Q_0)^2$, and $\mathbb{C}[\tilde{Q}] \cong \mathbb{C}[\epsilon]/(\epsilon^2)$ with $\epsilon = \tilde{Q} - Q_0e_0$ nilpotent of order two. This is the local, non-semisimple case.
$\mathbf{Q} = 0$. $m_{\tilde{Q}} = x - Q_0$ and $\mathbb{C}[\tilde{Q}] = \mathbb{C}e_0$, of dimension one.

So $\dim_{\mathbb{C}} \mathbb{C}[\tilde{Q}]$ is $2$ in the first two cases and $1$ in the scalar case, and the isomorphism type is the same dichotomy as the conjugacy classification of §Similarity Classes and the Eigenvalue Dichotomy.

The functional calculus. Since $\mathbb{C}[\tilde{Q}] \cong \mathbb{C}[x]/(m_{\tilde{Q}})$, every power series $F(A) = \sum_n a_n A^n$ with complex coefficients has $F(\tilde{Q}) = \alpha e_0 + \beta \tilde{Q}$ for scalars $\alpha, \beta$ determined by the data of $F$ on the roots: the two values $F(Q_0 \pm iB)$ in the semisimple case, and $F(Q_0)$ together with $F'(Q_0)$ in the local case. Consequently $F(\tilde{Q})$ converges if and only if $F$ converges at the root or roots (in the local case, $F$ and its derivative). Every elementary function of a single biquaternion is an instance, and the standard identities of complex analysis hold in $\mathbb{C}[\tilde{Q}]$ because it is commutative; this is the structural fact behind Biquaternion Elementary Functions.

Similarity Classes and the Eigenvalue Dichotomy

Definition 5.1. Two biquaternions are conjugate (or similar) if $\tilde{Q}' = S\tilde{Q}S^{-1}$ for some invertible $S \in \mathbb{B}$. The functionals $T, D$ are the coefficients of the characteristic polynomial.

Theorem 5.2 (classification). Let $T = T(\tilde{Q})$, $D = D(\tilde{Q})$, and let $\Delta = T^2 - 4D = 4(Q_0^2 - N(\tilde{Q})) = -4B^2$ be the discriminant.

(1) Distinct eigenvalues ($\Delta \neq 0$, i.e. $B \neq 0$). $\tilde{Q}$ has the two distinct eigenvalues $Q_0 \pm iB$, and its conjugacy class is determined by the unordered pair of eigenvalues, equivalently by $(T,D)$.

Repeated eigenvalue ($\Delta = 0$, i.e. $B = 0$). The only eigenvalue is $Q_0$, and there are exactly two conjugacy classes with $(T,D) = (2Q_0, Q_0^2)$: the central class $\mathbf{Q} = 0$ ($\tilde{Q} = Q_0e_0$) and the non-semi-simple class $\mathbf{Q} \neq 0$, $\mathbf{Q}^2 = 0$ (conjugate to the Jordan block with diagonal $Q_0$).

Thus $(T,D)$ determines the conjugacy class except on the repeated locus, where the separating invariant is whether $\mathbf{Q} = 0$.

Proof. (1) $p_{\tilde{Q}}$ has two distinct roots, so $\tilde{Q}$ is diagonalizable with them, and any two such elements are similar. (2) $B=0$ gives $\mathbf{Q}^2 = 0$; if $\mathbf{Q}\neq 0$ then $\tilde{Q} - Q_0e_0$ is nonzero with square $0$, so $\tilde{Q}$ is non-semi-simple and hence similar to the Jordan block with diagonal $Q_0$.

In case 2, $N(\tilde{Q}) = Q_0^2$, so the non-semi-simple element is a zero divisor exactly when $Q_0 = 0$.

Eigenspaces and Their Dimensions

Definition 6.1. For $\lambda \in \sigma(\tilde{Q})$ the eigenspace is $E_\lambda = \{\tilde{P} \in \mathbb{B} : \tilde{Q}\tilde{P} = \lambda \tilde{P}\}$, the eigenspace of left multiplication by $\tilde{Q}$ on $\mathbb{B}$, and $\dim_\mathbb{C} E_\lambda$ is the geometric multiplicity.

Proposition 6.2. For $\lambda \in \sigma(\tilde{Q})$: (1) if $B \neq 0$ then $\dim_\mathbb{C} E_\lambda = 2$; (2) if $B = 0$ and $\mathbf{Q} = 0$ then $E_{Q_0} = \mathbb{B}$; (3) if $B = 0$ and $\mathbf{Q} \neq 0$ then $\dim_\mathbb{C} E_{Q_0} = 2$.

Proof. (1) Left multiplication by $\tilde{Q}$ is diagonalizable with the two distinct eigenvalues $Q_0 \pm iB$, each of algebraic multiplicity $2$, hence each eigenspace has dimension $2$. (2) $\tilde{Q} = Q_0e_0$ is central, so $\tilde{Q}\tilde{P} = Q_0\tilde{P}$ for every $\tilde{P}$. (3) Left multiplication by the nilpotent $\tilde{Q} - Q_0e_0$ has square $0$ and rank $2$, hence nullity $2$.

Corollary 6.3. In the regular module $\mathbb{B}$ the geometric multiplicity of an eigenvalue is $2$, except for a central element, where it is $4$; only for a nilpotent element does it fall below the algebraic multiplicity $4$.

Real and complex eigenvalues. A real eigenvalue does not force real eigenvectors: $E_\lambda \subseteq \mathbb{B}$ is still a complex subspace, of real dimension twice its complex dimension. In the Hermitian case both eigenvalues are real.

The Resolvent

Definition 7.1. The resolvent of $\tilde{Q}$ is $R(\lambda) = (\tilde{Q} - \lambda e_0)^{-1}$, defined for $\lambda \notin \sigma(\tilde{Q})$.

Proposition 7.2. For $\lambda \notin \sigma(\tilde{Q})$, $$ R(\lambda) = \frac{(Q_0-\lambda)e_0 - \mathbf{Q}}{(Q_0-\lambda)^2 + B^2} = \frac{(\tilde{Q} - \lambda e_0)^{\natural}}{p_{\tilde{Q}}(\lambda)}. $$ Proof. $\tilde{Q} - \lambda e_0$ has conjugate $(Q_0-\lambda)e_0 - \mathbf{Q}$ and norm $(Q_0-\lambda)^2 + B^2 = p_{\tilde{Q}}(\lambda)$; apply Corollary 4.2.

Corollary 7.3 (analyticity). $R$ is rational in $\lambda$ with denominator $p_{\tilde{Q}}(\lambda)$, hence holomorphic on $\mathbb{C}\setminus\sigma(\tilde{Q})$, with poles exactly at the two eigenvalues. The pole is simple when $\tilde{Q}$ is diagonalizable, and double when $B = 0$ with $\mathbf{Q} \neq 0$, where $p_{\tilde{Q}} = (\lambda-Q_0)^2$ but the numerator at $\lambda = Q_0$ is $-\mathbf{Q} \neq 0$.

Finally, the resolvent satisfies $R(\lambda) - R(\mu) = (\lambda-\mu)R(\lambda)R(\mu)$, since with $A = \tilde{Q}-\lambda e_0$ and $B = \tilde{Q}-\mu e_0$ one has $A^{-1}-B^{-1} = A^{-1}(B-A)B^{-1}$; consequently $R'(\lambda) = R(\lambda)^2$.

The Spectral Radius

Definition 8.1. The spectral radius is $\rho(\tilde{Q}) = \max\{|\lambda| : \lambda \in \sigma(\tilde{Q})\} = \max\{|Q_0+iB|, |Q_0-iB|\}$.

Proposition 8.2. $\rho(\tilde{Q}) = 0$ iff $\tilde{Q}$ is nilpotent, equivalently $\tilde{Q}^2 = 0$, equivalently $Q_0 = 0$ and $B = 0$.

Proof. $\rho = 0$ iff $Q_0 = B = 0$, and then $p_{\tilde{Q}}(\lambda) = \lambda^2$, so $\tilde{Q}^2 = 0$ by Cayley–Hamilton; conversely $\tilde{Q}^2 = 0$ makes $0$ the only eigenvalue. The nonzero pure nilpotents $\mathbf{Q}$ with $\mathbf{Q}^2 = 0$, which are zero divisors, are a special case.

Theorem 8.3 (Gelfand). For any norm on the finite-dimensional algebra $\mathbb{B}$, $\rho(\tilde{Q}) = \lim_{k\to\infty}\|\tilde{Q}^k\|^{1/k}$, independently of the biquaternion norm; with the Euclidean norm this reads $\lim_k\|\tilde{Q}^k\|_E^{1/k}$.

The Exponential and the Logarithm

The exponential $\exp(\tilde{Q}) = \sum_n \tilde{Q}^n/n!$ is entire and is the value of $e^A$ at the spectrum, in the holomorphic functional calculus on $\mathbb{C}[\tilde{Q}] = \mathrm{span}_\mathbb{C}\{e_0,\tilde{Q}\}$. The companion article gives $\exp(\tilde{Q}) = e^{Q_0}(\cos B\,e_0 + \sin B\,\hat{n})$ for $B \neq 0$ with $\hat{n} = \mathbf{Q}/B$, and $\exp(\tilde{Q}) = e^{Q_0}(e_0 + \mathbf{Q})$ for $B = 0$; the eigenvalues are $e^{Q_0 \pm iB}$ in the first case and the repeated $e^{Q_0}$ in the second.

Non-injectivity. Since $2\pi i n e_0$ is central, $\exp(\tilde{Q} + 2\pi i n e_0) = \exp(\tilde{Q})$ for every integer $n$; in particular $\exp(0) = \exp(2\pi i e_0) = e_0$, although $2\pi i e_0 \neq 0$.

The logarithm. A logarithm of $\tilde{Q}$ is any $\tilde{L}$ with $\exp(\tilde{L}) = \tilde{Q}$. The companion article proves that every invertible biquaternion has one, so $\exp : \mathbb{B} \to \mathbb{B}^\times$ is surjective; not being injective, it has a multivalued inverse, and a branch must be chosen. Non-uniqueness comes from adding a commuting element of the kernel, in particular $2\pi i n e_0$, and from choosing a complex logarithm of each eigenvalue. No continuous, and hence no holomorphic, logarithm can exist on the whole group of units: one applied to $\lambda e_0$ and composed with the determinant would give a continuous logarithm of $\lambda^2$ on $\mathbb{C}^\times$, impossible because $\lambda^2$ winds twice around the origin.

Hermitian and Normal Elements

Definition 10.1. $\tilde{Q}$ is Hermitian if $\tilde{Q}^{*} = \tilde{Q}$, unitary if $\tilde{Q}^{*}\tilde{Q} = \tilde{Q}\tilde{Q}^{*} = e_0$, and normal if $\tilde{Q}\tilde{Q}^{*} = \tilde{Q}^{*}\tilde{Q}$. Hermitian and unitary elements are normal.

The Hermitian condition is a subspace condition: writing $\tilde{Q} = \sum_\mu Q_\mu e_\mu$, it says exactly that $Q_0 \in \mathbb{R}$ and $Q_k = iq'_k$ with $q'_k \in \mathbb{R}$. So the Hermitian elements of $\mathbb{B}$ are precisely the elements of the Hermitian subspace $\mathbb{M}_+$, and the anti-Hermitian elements are precisely those of $\mathbb{M}_-$; the Hermitian idempotents $\tilde\Pi_1, \tilde\Pi_2$ of Theorem 11.2 lie in $\mathbb{M}_+$.

Proposition 10.3. If $\tilde{Q}$ is Hermitian then $T(\tilde{Q})$, $D(\tilde{Q})$, and both eigenvalues are real.

Proof. Hermitian elements have $Q_0 \in \mathbb{R}$ and $Q_k = iq'_k$ with $q'_k \in \mathbb{R}$, so $B^2 = -\sum_k(q'_k)^2$ is real and non-positive, $iB$ is real, and $Q_0 \pm iB \in \mathbb{R}$.

Proposition 10.4. $\tilde{Q}$ is normal iff left multiplication by $\tilde{Q}$ is unitarily diagonalizable, i.e. iff $\mathbb{B}$ has an orthonormal basis of eigenvectors with respect to the Hermitian form. In particular a normal element with $B = 0$ must be central, and $\tilde{Q}$ is Hermitian iff it is normal with real spectrum. A unitary element is normal and has all eigenvalues of modulus $1$.

The Spectral Theorem

Theorem 11.1. $M \in M_2(\mathbb{C})$ is normal iff $M = U\,\mathrm{diag}(\lambda_1,\lambda_2)\,U^\dagger$ with $U$ unitary and $\lambda_1,\lambda_2 \in \mathbb{C}$ its spectrum; if $M$ is Hermitian the $\lambda_i$ are real, and if $M$ is unitary they have modulus $1$.

Theorem 11.2 (biquaternion form). $\tilde{Q}$ is normal iff there are a unitary $\tilde{U} \in \mathbb{B}$ and $\lambda_1,\lambda_2 \in \mathbb{C}$ with $$ \tilde{U}\tilde{Q}\tilde{U}^{*} = \frac{\lambda_1+\lambda_2}{2}e_0 + \frac{i(\lambda_1-\lambda_2)}{2}e_3. $$ Equivalently $\tilde{Q} = \lambda_1\tilde\Pi_1 + \lambda_2\tilde\Pi_2$ with Hermitian idempotents $\tilde\Pi_1,\tilde\Pi_2$ satisfying $\tilde\Pi_1+\tilde\Pi_2 = e_0$, $\tilde\Pi_1\tilde\Pi_2 = 0$, $\tilde\Pi_i^\dagger = \tilde\Pi_i$, $\tilde\Pi_i^2 = \tilde\Pi_i$; for $\lambda_1 \neq \lambda_2$, $$ \tilde\Pi_1 = \frac{\tilde{Q}-\lambda_2e_0}{\lambda_1-\lambda_2}, \qquad \tilde\Pi_2 = \frac{\tilde{Q}-\lambda_1e_0}{\lambda_2-\lambda_1}. $$ Proof. This is Theorem 11.1 transported to $\mathbb{B}$, identifying a normal element with its diagonal form $\frac{\lambda_1+\lambda_2}{2}e_0 + \frac{i(\lambda_1-\lambda_2)}{2}e_3$ and using that Hermitian conjugation is a $*$-involution on $\mathbb{B}$.

The S-Spectrum and the Quaternionic Spectral Theorem

§The Spectral Theorem works for biquaternions because $\mathbb{C}$ is central. When the scalars do not commute, as over $\mathbb{H}$, the left/right ambiguity returns and a different spectrum is needed: the S-spectrum.

Definition 12.1. For $T \in M_n(\mathbb{H})$, the S-spectrum is $\sigma_S(T) = \{s \in \mathbb{H} : T^2 - 2\mathrm{Re}(s)T + |s|^2I \text{ is not invertible}\}$. It rests on the identity $s^2 - 2\mathrm{Re}(s)s + |s|^2 = 0$, which holds because $s - \mathrm{Re}(s)$ is pure imaginary.

Theorem 12.2. For $T \in M_n(\mathbb{H})$, $\sigma_S(T) = \sigma_R(T)$.

Proof (one direction). If $Tv = vs$ then $(T^2 - 2\mathrm{Re}(s)T + |s|^2I)v = v(s^2 - 2\mathrm{Re}(s)s + |s|^2) = 0$, so the operator is singular; the converse is proved in the references.

For $T = [q]$ this gives the $2$-sphere $\sigma_S([q]) = \{v^{-1}qv : v \in \mathbb{H}^\times\}$ for non-real $q$; compare Example 2.3.

Theorem 12.3. If $T \in M_n(\mathbb{H})$ is normal, then $T = U\,\mathrm{diag}(\lambda_1,\dots,\lambda_n)\,U^*$ for a unitary $U$ and complex $\lambda_i$ in a fixed slice, and $\sigma_S(T) = \bigcup_{i=1}^n [\lambda_i]$, where $[\lambda]$ is the conjugacy class $\{v^{-1}\lambda v : v \in \mathbb{H}^\times\}$. For Hermitian $T$ the $\lambda_i$ are real; for unitary $T$ they have modulus $1$.

Relation to the biquaternions. For a real quaternion $q$ viewed as a biquaternion, the complex spectrum $\{q_0 \pm i|\mathbf{q}|\}$ is the slice $\sigma(q) = [q] \cap \mathbb{C}_i$ of the S-sphere $[q]$. Thus the S-spectrum plays for quaternionic matrices the role the complex spectrum plays for biquaternion matrices.

Summary

The word "spectrum" is used in several inequivalent senses for $\mathbb{B}$, and the main source of error in the subject is the silent switching between them. Unqualified, the spectrum of a biquaternion means its spectrum under Definition 1.1: the two-element set

$$ \sigma(\tilde{Q}) = \{Q_0 + iB, \, Q_0 - iB\}, \qquad B = \sqrt{Q_1^2 + Q_2^2 + Q_3^2}, $$

the roots of the characteristic polynomial $\lambda^2 - 2Q_0\lambda + D(\tilde{Q})$, where $T(\tilde{Q}) = 2Q_0$ is the trace functional and $D(\tilde{Q}) = N(\tilde{Q})$ is the biquaternion norm. The two roots coincide exactly when $B = 0$; the set $\sigma(\tilde{Q})$ contains $0$ exactly when $\tilde{Q}$ is a zero divisor, by the invertibility criterion. Because $\mathbb{B}$ is non-commutative, the left and right spectra must be distinguished from it; they agree with $\sigma(\tilde{Q})$ as a set, with the multiplicities of the left spectrum (in the regular module) doubled.

The rest of the theory is intrinsic to $\mathbb{B}$: the commutative subalgebra generated by a single element and its functional calculus, Cayley–Hamilton, the similarity classification and the eigenvalue dichotomy, the eigenspace dimensions in the regular module, the resolvent and its expansion, the spectral radius, and the exponential and the logarithm. The spectral theorem takes the biquaternion form $\tilde{Q} = \lambda_1 \tilde\Pi_1 + \lambda_2 \tilde\Pi_2$ with Hermitian idempotents $\tilde\Pi_1, \tilde\Pi_2$ summing to $e_0$ and orthogonal; a Hermitian element has real spectrum, a unitary element spectrum on the unit circle.

When the scalars do not commute, as over $\mathbb{H}$, the left/right ambiguity requires a different notion, the S-spectrum $\sigma_S(T) = \{s \in \mathbb{H} : T^2 - 2\operatorname{Re}(s)T + |s|^2 I \text{ is not invertible}\}$, which coincides with the right spectrum and, for a real quaternion viewed as a biquaternion, restricts to the complex spectrum as $\sigma(q) = [q] \cap \mathbb{C}_i$. The S-spectrum thus plays for quaternionic matrices the role the complex spectrum plays for biquaternion matrices.

Summary of Notation

Symbol	Meaning
$\mathbb{B}$	Biquaternion algebra, $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{H}$
$\tilde{Q} = \sum_{\mu=0}^{3} Q_\mu e_\mu$	General biquaternion
$B = \sqrt{Q_1^2+Q_2^2+Q_3^2}$	Complex norm of the vector part
$T(\tilde{Q}) = 2Q_0$	Trace functional
$D(\tilde{Q}) = N(\tilde{Q})$	Determinant functional (norm)
$\sigma(\tilde{Q}) = \{Q_0 \pm iB\}$	Spectrum (complex spectrum)
$\sigma_L, \sigma_R$	Left and right spectra
$E_\lambda$	Eigenspace in the regular module $\mathbb{B}$
$R(\lambda)$	Resolvent
$\rho(\tilde{Q})$	Spectral radius
$\sigma_S(T)$	S-spectrum
$\tilde\Pi_1, \tilde\Pi_2$	Hermitian idempotents in the biquaternion spectral theorem, $\tilde\Pi_1+\tilde\Pi_2 = e_0$, $\tilde\Pi_1\tilde\Pi_2 = 0$