Spectral Regularity and Degeneracy in the Biquaternion Algebra

Introduction

The biquaternion algebra carries a second quadratic condition, the square of the vector part,

$$ B^2=Q_x^2+Q_y^2+Q_z^2, $$

which involves no scalar term. It alone decides whether the two eigenvalues of an element can be told apart: the characteristic polynomial is $X^2-2Q_tX+Q_t^2+B^2$, and its discriminant is $-4B^2$. This article defines the eigenvalue problem, proves the spectral criterion, and separates the elements of $\mathbb{B}$ into the two cases of the value of $B^2$: the regular case $B^2\neq0$, where the two eigenvalues are distinct and the element has a unique spectral decomposition, and the degeneracy case $B^2=0$, the spectral-degenerate cone, on which the two eigenvalues coalesce. This article is the study of this one condition and of its two cases.

Notation. An element is $\tilde Q=Q_te_t+\vec Q$, its natural conjugation is $\tilde Q^{\natural}=Q_t-\vec Q$, and the product is the plain product of the algebra, associative with the two-sided unit $e_t$. The vector part is $\vec Q=Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z$, and

$$ B^2=Q_x^2+Q_y^2+Q_z^2,\qquad \vec Q^2=-B^2e_t. $$

The number $B$ is a fixed complex square root of $B^2$; the other root only exchanges the two eigenvalues. The companion article Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra studies the multiplicative norm and its two cases; this article studies $B^2$ and does not use that norm. The matrix realization $\mathsf{M}_2$ of Introduction to the 2×2 Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions is used only to compute the determinant $\det\mathsf{M}_2(\tilde Q)=Q_t^2+B^2$. The algebra and its conventions are The Biquaternion Algebra and Its Vector Spaces; the eigenvalue decomposition, the normality criterion and the spectral theorem are The Biquaternion Eigenvalue Decomposition.

The Spectral Criterion

An eigenvalue is a scalar for which a left ideal is proper, and the criterion turns that definition into a single polynomial equation.

Definition (eigenvalue). A scalar $\lambda\in\mathbb{C}$ is an eigenvalue of $\tilde Q$ if the left ideal $\mathbb{B}(\tilde Q-\lambda e_t)$ is proper, that is $\mathbb{B}(\tilde Q-\lambda e_t)\neq\mathbb{B}$.

Definition (eigenvector). A non-zero $v\in\mathbb{B}$ is an eigenvector of $\tilde Q$ for $\lambda$ if $(\tilde Q-\lambda e_t)v=0$. The eigenvectors for $\lambda$, with the origin, are the left annihilator

$$ \operatorname{Ann}_{\ell}(\tilde Q-\lambda e_t)=\{v:\ (\tilde Q-\lambda e_t)v=0\}, $$

which is a right ideal of $\mathbb{B}$.

Definition (spectral idempotent). A non-zero idempotent $e$ is a spectral idempotent of $\tilde Q$ if $e\tilde Q=\tilde Qe=\lambda e$ for some scalar $\lambda$.

Theorem (the spectral criterion). For $\tilde Q\in\mathbb{B}$ and $\lambda\in\mathbb{C}$, the following are equivalent:

$\lambda$ is an eigenvalue of $\tilde Q$;
$\tilde Q-\lambda e_t$ is not a unit of the plain product;
$\det\mathsf{M}_2(\tilde Q-\lambda e_t)=0$;
$(\lambda-Q_t)^2+B^2=0$.

Proof. Write $a=\tilde Q-\lambda e_t$.

(1)$\iff$(2). If $a$ is a unit then $\mathbb{B}a=\mathbb{B}$, so the left ideal is not proper. Conversely, if $\mathbb{B}a=\mathbb{B}$ then $e_t=xa$ for some $x$. Left multiplication by $a$ is injective, since $ay=0$ gives $y=xay=0$. In finite dimension an injective endomorphism is bijective, so there is $b$ with $ab=e_t$. Then $a(ba-e_t)=0$, hence $ba=e_t$ by injectivity, and $b$ is a two-sided inverse.

(2)$\iff$(3). The isomorphism $\mathsf{M}_2$ turns the plain product into the matrix product, so $a$ is a unit exactly when $\mathsf{M}_2(a)$ is invertible, and a square matrix over a field is invertible exactly when its determinant is non-zero.

(3)$\iff$(4). For $a=a_te_t+\vec a$ the matrix is

$$ \mathsf{M}_2(a)=\begin{pmatrix} a_t-ia_z & -ia_x-a_y\\ -ia_x+a_y & a_t+ia_z\end{pmatrix}, $$

whose determinant is $a_t^2+a_x^2+a_y^2+a_z^2$. With $a_t=Q_t-\lambda$ and $\vec a=\vec Q$ this is $(\lambda-Q_t)^2+B^2$.

Corollary (the characteristic polynomial). For every $\lambda$,

$$ \det\mathsf{M}_2(\tilde Q-\lambda e_t)=(\lambda-Q_t)^2+B^2, $$

so the characteristic polynomial is

$$ p_{\tilde Q}(X)=X^2-2Q_tX+Q_t^2+B^2, $$

of discriminant

$$ \operatorname{disc}p_{\tilde Q}=4Q_t^2-4(Q_t^2+B^2)=-4B^2. $$

The eigenvalues are

$$ \lambda_\pm=Q_t\pm iB, $$

and they coincide exactly when $B=0$, that is exactly when $B^2=0$.

Corollary (the eigenspaces). For a non-scalar $\tilde Q$ and each eigenvalue $\lambda$, the eigenspace $\operatorname{Ann}_{\ell}(\tilde Q-\lambda e_t)$ is a right ideal of complex dimension $2$. For $\tilde Q=Q_te_t$ the only eigenvalue is $Q_t$, and its eigenspace is all of $\mathbb{B}$.

Proof. A non-zero non-unit of $\mathbb{B}$ is a matrix of rank one in $\mathsf{M}_2$, and left multiplication by a rank-one matrix has rank $2$; hence the kernel has complex dimension $4-2=2$. At the scalar the difference $\tilde Q-Q_te_t$ is zero.

Corollary (Cayley–Hamilton and the inverse). Every $\tilde Q$ satisfies

$$ \tilde Q^2-2Q_t\tilde Q+(Q_t^2+B^2)e_t=0, $$

and if $Q_t^2+B^2\neq0$ then $\tilde Q^{-1}=(2Q_te_t-\tilde Q)/(Q_t^2+B^2)=\tilde Q^{\natural}/(Q_t^2+B^2)$.

Proof. With $\tilde Q=Q_te_t+\vec Q$ and $\vec Q^2=-B^2e_t$, $\tilde Q^2=Q_t^2e_t+2Q_t\vec Q-B^2e_t=2Q_t\tilde Q-(Q_t^2+B^2)e_t$. Multiplying by $2Q_te_t-\tilde Q=\tilde Q^{\natural}$ on either side gives $(Q_t^2+B^2)e_t$.

Remark (left and right eigenvalues). The centre of $\mathbb{B}$ is $\mathbb{C}$, so complex scalars commute with every element. A left eigenvalue, $\tilde Qv=\lambda v$, and a right eigenvalue, $\tilde Qv=v\lambda$, therefore define the same set when the scalar is required to lie in $\mathbb{C}$, and that set is $\sigma(\tilde Q)$. The distinction between the two eigenvalue problems, which is genuine over a noncommutative division ring (Quaternion Spectral Theory, §The Spectrum of a Quaternion), collapses here.

The Two Cases

The discriminant of the characteristic polynomial is $-4B^2$, and its vanishing, and nothing else, separates the algebra into two spectral cases.

Definition (spectral decomposition). A spectral decomposition of $\tilde Q$ is either the one-term writing $\tilde Q=Q_te_t$ when $\tilde Q$ is scalar, or a two-term writing

$$ \tilde Q=\lambda_+e_++\lambda_-e_-, $$

where $\lambda_+\neq\lambda_-$ and $e_+,e_-$ are non-zero idempotents with

$$ e_+^2=e_+,\qquad e_-^2=e_-,\qquad e_+e_-=e_-e_+=0,\qquad e_++e_-=e_t. $$

The scalars $\lambda_\pm$ are the eigenvalues and the idempotents $e_\pm$ are the spectral idempotents.

Criterion (four descriptions). For a non-scalar $\tilde Q$ the following are equivalent:

$B^2\neq0$;
the eigenvalues are distinct;
$\tilde Q$ is semisimple;
$L_{\tilde Q}$ is diagonalisable.

Proof. (1)$\iff$(2): the eigenvalues are $Q_t\pm iB$, and they are distinct exactly when $B\neq0$. (2)$\iff$(3): for a non-scalar element the minimal polynomial is the characteristic polynomial $(X-Q_t)^2+B^2$, of degree two, since a degree-one minimal polynomial would make the element scalar; it is squarefree exactly when its two roots are distinct. (3)$\iff$(4): the left regular representation is faithful and carries $\mathbb{C}[\tilde Q]$ isomorphically onto $\mathbb{C}[L_{\tilde Q}]$, so $\tilde Q$ and $L_{\tilde Q}$ have the same minimal polynomial, and an operator is diagonalisable exactly when that polynomial is squarefree.

The two cases are collected in the table; the last two columns are proved in §The Spectral Decomposition and §The Eigenbasis of Left Multiplication.

case	condition	spectrum	spectral decomposition	eigenbasis
regular	$B^2\neq0$	$\{\lambda_+,\lambda_-\}$, distinct	two terms, unique	exists
degenerate	$B^2=0$	$\{Q_t\}$	one term iff $\vec Q=0$, none otherwise	every basis iff $\vec Q=0$, none otherwise

Remark (the words). Unqualified regular means spectrally regular, $B^2\neq0$, and degenerate means spectrally degenerate, $B^2=0$; both words are fixed by the discriminant, as norm-regular and norm-degenerate are fixed by the value of the norm in the companion article. Degenerate here does not mean defective: the vertex is semisimple. The comparison with the norm is Relations between Quaternionic Norm and Spectral Regularity and Degeneracy.

Remark (the condition is not multiplicative). Neither case is stable under the product. Two elements of $B^2=0$ can multiply to an element of $B^2\neq0$: both factors of

$$ (\vec e_x+i\vec e_y)(\vec e_x-i\vec e_y)=-2i\vec e_z $$

have $B^2=0$, and the product has $B^2=-4$. Two elements of $B^2\neq0$ can multiply to an element of $B^2=0$: both factors of

$$ (e_t+\vec e_x)(e_t-\vec e_x)=2e_t $$

have $B^2=1$, and the product has $B^2=0$. So the regular case is not a subgroup, and the spectral condition is not preserved by multiplication in either direction.

The Regular Case: Distinct Eigenvalues

When $B^2\neq0$ the two eigenvalues are distinct, the spectral idempotents are the Lagrange idempotents, and the decomposition is unique.

The Spectral Decomposition

Theorem (existence). Let $B^2\neq0$ and put

$$ \vec\Xi=-\frac{i\vec Q}{B},\qquad \tilde\Pi_\pm=\frac{e_t\pm\vec\Xi}{2},\qquad \lambda_\pm=Q_t\pm iB. $$

Then $\tilde\Pi_\pm$ are non-zero idempotents with $\tilde\Pi_+\tilde\Pi_-=\tilde\Pi_-\tilde\Pi_+=0$ and $\tilde\Pi_++\tilde\Pi_-=e_t$, and

$$ \tilde Q=\lambda_+\tilde\Pi_++\lambda_-\tilde\Pi_-,\qquad \tilde Q\tilde\Pi_\pm=\lambda_\pm\tilde\Pi_\pm. $$

Proof. Since $\vec Q^2=-B^2e_t$, one has $\vec\Xi^2=e_t$ and $\vec Q=iB\vec\Xi$. Hence $\tilde\Pi_\pm^2=(e_t\pm2\vec\Xi+e_t)/4=\tilde\Pi_\pm$ and $\tilde\Pi_+\tilde\Pi_-=(e_t-\vec\Xi^2)/4=0$, and the sum is $e_t$. The idempotents are non-zero because $\vec\Xi$ has zero scalar part, so $\vec\Xi\neq\pm e_t$. For the action, $\vec Q\vec\Xi=(-i/B)\vec Q^2=iBe_t$, hence

$$ \tilde Q\tilde\Pi_+=\frac{(Q_t e_t+\vec Q)(e_t+\vec\Xi)}{2} =\frac{(Q_t+iB)(e_t+\vec\Xi)}{2}=\lambda_+\tilde\Pi_+, $$

and $\tilde Q\tilde\Pi_-=\lambda_-\tilde\Pi_-$ the same way. Adding the two identities multiplied by the eigenvalues gives the decomposition.

Corollary (Lagrange form). The idempotents are the Lagrange idempotents,

$$ \tilde\Pi_\pm=\frac{\tilde Q-\lambda_\mp e_t}{\lambda_\pm-\lambda_\mp}, $$

and the spectral decomposition is unique.

Proof. The first identity is the computation $\tilde Q-\lambda_-=iB(e_t+\vec\Xi)$, and its analogue with the signs exchanged. For uniqueness, the minimal polynomial $(X-\lambda_+)(X-\lambda_-)$ has distinct roots, so the Chinese remainder theorem gives $\mathbb{C}[\tilde Q]\cong\mathbb{C}[X]/((X-\lambda_+)(X-\lambda_-))\cong\mathbb{C}\times\mathbb{C}$; the idempotents of that product are $0,e_t,\tilde\Pi_+,\tilde\Pi_-$, and every spectral idempotent lies in the centraliser $\mathbb{C}[\tilde Q]$.

Corollary (spectral projectors). $L_{\tilde\Pi_+}$ is the projector onto the $\lambda_+$-eigenspace: it sends $X$ to $\tilde\Pi_+X$, with image $\tilde\Pi_+\mathbb{B}$ and kernel $\tilde\Pi_-\mathbb{B}$. The same statement holds with the two signs exchanged.

The Eigenbasis of Left Multiplication

Theorem (the eigenspaces). For $B^2\neq0$ and each sign,

$$ \ker(L_{\tilde Q}-\lambda_\pm)=\tilde\Pi_\pm\mathbb{B}, $$

a right ideal of complex dimension $2$. Each eigenvalue therefore has multiplicity two.

Proof. Since $\tilde Q-\lambda_+=iB(\vec\Xi-e_t)=-2iB\tilde\Pi_-$, the equation $(\tilde Q-\lambda_+)X=0$ is $\tilde\Pi_-X=0$, that is $X=\tilde\Pi_+X$, that is $X\in\tilde\Pi_+\mathbb{B}$. In a basis in which $\tilde\Pi_+=\operatorname{diag}(1,0)$, the ideal $\tilde\Pi_+\mathbb{B}$ is the set of matrices whose second row vanishes, of complex dimension $2$.

Corollary (eigenbasis). Choose non-zero $\vec\Upsilon_-\in\tilde\Pi_+\mathbb{B}\tilde\Pi_-$ and $\vec\Upsilon_+\in\tilde\Pi_-\mathbb{B}\tilde\Pi_+$. Then

$$ (\tilde\Pi_+,\ \vec\Upsilon_-,\ \vec\Upsilon_+,\ \tilde\Pi_-) $$

is an eigenbasis of $L_{\tilde Q}$, with diagonal values $(\lambda_+,\lambda_+,\lambda_-,\lambda_-)$. An eigenbasis is any pair of bases of the two eigenspaces, so it is determined only up to an invertible change of basis on each.

Proof. The mixed corners are one-dimensional, and $\vec\Upsilon_-\in\tilde\Pi_+\mathbb{B}$ while $\vec\Upsilon_+\in\tilde\Pi_-\mathbb{B}$, so the four vectors form a basis of the two eigenspaces and hence of $\mathbb{B}$.

Remark (the two sides). The minimal left ideals $S_\pm=\mathbb{B}\tilde\Pi_\pm$ are the eigenspaces of the right multiplication $R_{\tilde Q}$, with eigenvalues $\lambda_\pm$. They are not eigenspaces of $L_{\tilde Q}$: $S_+$ contains $\tilde\Pi_+$ and $\vec\Upsilon_+$, with eigenvalues $\lambda_+$ and $\lambda_-$, and $S_-$ contains $\tilde\Pi_-$ and $\vec\Upsilon_-$, with eigenvalues $\lambda_-$ and $\lambda_+$. The eigenbasis above is an eigenbasis of left multiplication, not of right multiplication.

Remark (the degenerate side). If $B^2=0$ and $\vec Q\neq0$, the minimal polynomial is $(X-Q_t)^2$, the eigenspace has dimension $2$, and no eigenbasis exists. If $\vec Q=0$ the element is scalar and $L_{\tilde Q}=Q_t\,\mathrm{id}$; every basis is then an eigenbasis.

The Degenerate Case: The Cone

On the cone $B^2=0$ the eigenvalues have coalesced, and one further question decides the spectral behaviour of a point: whether the vector part vanishes.

The Spectral-Degenerate Cone

Definition. The spectral-degenerate cone is

$$ \mathcal S=\{B^2=0\}=\{\vec Q^2=0\}=\{\operatorname{disc}p_{\tilde Q}=0\}. $$

Theorem (three readings of one equation). The equation $B^2=0$ is simultaneously:

geometric: the vector part is isotropic, $Q_x^2+Q_y^2+Q_z^2=0$;
algebraic: the vector part is square-zero, $\vec Q^2=0$, hence nilpotent of index two when it is non-zero;
spectral: the discriminant vanishes, and the two eigenvalues coalesce at $Q_t$.

Theorem (the geometry). The cone is the product of the scalar line and the isotropic cone of the vector subspace,

$$ \mathcal S=\mathbb{C}e_t\times\{\vec Q\in\operatorname{Vect}(\mathbb{B}):\vec Q^2=0\}. $$

It is a complex cone of complex dimension $3$ and real dimension $6$. Its singular locus is the vertex $\{\vec Q=0\}$, the whole scalar line; off the vertex it is a smooth hypersurface.

Proof. The equation $B^2=0$ is one complex equation in the four complex coefficients, homogeneous of degree two. Its gradient with respect to $(Q_t,Q_x,Q_y,Q_z)$ is

$$ (0,\,2Q_x,\,2Q_y,\,2Q_z), $$

which vanishes exactly when $\vec Q=0$. Off the vertex the gradient is non-zero and the equation cuts a smooth hypersurface, of complex dimension $4-1=3$ and real dimension $8-2=6$. On the vertex the cone is $\mathbb{C}e_t\times C$, with $C$ the isotropic cone of $\mathbb{C}^3$, which is singular at its apex; the product is singular along the whole line.

Warning (what the cone is not). The equation involves only $\vec Q$: the cone is blind to the scalar part, and the whole scalar line lies on it. It is not a subspace, since $\vec e_x+i\vec e_y$ and $\vec e_x-i\vec e_y$ lie on it while their sum $2\vec e_x$ does not. It is not the set of defective elements: a point of the vertex is semisimple with minimal polynomial $X-Q_t$. The cone is exactly the set of elements with a repeated eigenvalue.

The Behaviour on the Cone

Theorem (the vertex and its complement). Let $\tilde Q=Q_te_t+\vec Q\in\mathcal S$.

At the vertex, $\vec Q=0$: the element is the scalar $Q_te_t$, its minimal polynomial is $X-Q_t$, its spectrum is $\{Q_t\}$ and it is semisimple. Every idempotent commutes with it, the spectral decomposition is the one-term writing $\tilde Q=Q_te_t$, and every basis is an eigenbasis.
Off the vertex, $\vec Q\neq0$: the vector part is square-zero, the minimal polynomial is $(X-Q_t)^2$, the spectrum is $\{Q_t\}$ and the element is defective. Its eigenspace is $\ker(L_{\vec Q})$, of complex dimension $2$, and no two-term spectral decomposition or eigenbasis exists.

Proof. At the vertex the difference $\tilde Q-Q_te_t$ vanishes. Off it, $\vec Q^2=0$ and $\vec Q\neq0$, so $(\tilde Q-Q_te_t)^2=0$ while $\tilde Q-Q_te_t\neq0$; the minimal polynomial therefore divides $(X-Q_t)^2$ and is not $X-Q_t$, hence equals $(X-Q_t)^2$. A minimal polynomial with a repeated root is not squarefree, so the element is defective. Since $Q_te_t$ is central, the centraliser of $\tilde Q$ is the centraliser of $\vec Q$, namely the dual numbers $\mathbb{C}[\vec Q]=\{ae_t+b\vec Q\}$, whose only idempotents are $0$ and $e_t$; hence there is no non-trivial spectral idempotent and no two-term spectral decomposition. The eigenspace is $\ker(L_{\vec Q})$, and left multiplication by a non-zero square-zero element has rank $2$, so the kernel has complex dimension $2$; since $2<4$, no eigenbasis exists.

The two behaviours are collected in the table.

point of the cone	condition	spectrum	minimal polynomial	spectral decomposition	eigenspace
vertex	$B^2=0$, $\vec Q=0$	$\{Q_t\}$	$X-Q_t$	one term	all of $\mathbb{B}$, every basis
off the vertex	$B^2=0$, $\vec Q\neq0$	$\{Q_t\}$	$(X-Q_t)^2$	none	$\ker(L_{\vec Q})$, complex dimension $2$, no eigenbasis

Remark (no further spectral split). The distinguishing condition of the table is $\vec Q$, not $Q_t$: the vertex is the whole scalar line, and every point of it has the same spectral behaviour whatever the value of $Q_t$; off the vertex the minimal polynomial is $(X-Q_t)^2$ for every $Q_t$. The type of the minimal polynomial distinguishes the vertex from its complement and nothing finer.

The Similarity Classification

Definition. Two biquaternions $\tilde Q_1,\tilde Q_2$ are similar if $\tilde Q_2=S\tilde Q_1S^{-1}$ for some unit $S\in\mathbb{B}$.

Theorem (the classification).

If $B^2\neq0$, the similarity class is determined by the unordered pair of eigenvalues $\{\lambda_+,\lambda_-\}$.
If $B^2=0$, the spectrum is the singleton $\{Q_t\}$, and exactly two similarity classes share it: the central class $\tilde Q=Q_te_t$, and the non-semisimple class $\vec Q\neq0$ with $\vec Q^2=0$, which is similar to the Jordan block $J_2(Q_t)$.

Proof. (1) The characteristic polynomial has two distinct roots, so $\tilde Q$ is diagonalisable with those roots; two diagonalisable matrices with the same spectrum are conjugate in $M_2(\mathbb{C})\cong\mathbb{B}$, and the spectrum is unordered because the two eigenvalues are exchanged by the choice of $B$. (2) On the cone $\vec Q^2=0$. If $\vec Q=0$ the element is central. If $\vec Q\neq0$, then $\tilde Q-Q_te_t$ is non-zero with square $0$, so the minimal polynomial is $(X-Q_t)^2$ and the Jordan form is a single block of size $2$; all non-zero square-zero elements are rank-one nilpotents, hence conjugate.

The unordered spectrum therefore determines the similarity class off the cone, and carries no information on it beyond $Q_t$: the discriminant is what separates the central class from the non-semisimple one.

The Two Cases on the Remarkable Subspaces

The condition $B^2$ depends only on the vector part, so its restriction to a remarkable subspace is read from the vector coordinates of that subspace.

remarkable subspace	real basis	restriction of $B^2$	elements with $B^2=0$
$\mathbb{C}_{\mathbb{B}}$, the centre	$e_t,\,ie_t$	$0$	the whole centre
$\operatorname{Vect}(\mathbb{B})$, the vector subspace	$\vec e_k,\,i\vec e_k$	$\sum_kQ_k^2$, realified signature $(3,3)$	the isotropic cone, complex dimension $2$
$\mathbb{H}_{\mathbb{B}}$, the quaternion subspace	$e_t,\vec e_x,\vec e_y,\vec e_z$	$q_x^2+q_y^2+q_z^2$	$\mathbb{R}e_t$
$i\mathbb{H}_{\mathbb{B}}$, the anti-quaternion subspace	$ie_t,i\vec e_x,i\vec e_y,i\vec e_z$	$-(q_x^2+q_y^2+q_z^2)$	$\mathbb{R}ie_t$
$\mathbb{M}_{+}$, the Hermitian subspace	$e_t,i\vec e_x,i\vec e_y,i\vec e_z$	$-(a_x^2+a_y^2+a_z^2)$	$\mathbb{R}e_t$
$\mathbb{M}_{-}$, the anti-Hermitian subspace	$ie_t,\vec e_x,\vec e_y,\vec e_z$	$a_x^2+a_y^2+a_z^2$	$\mathbb{R}ie_t$

Corollary. Every element of the centre is spectrally degenerate, and the centre is the only remarkable subspace contained in $\mathcal S$. On the quaternion and anti-Hermitian subspaces the condition is positive semidefinite, on the anti-quaternion and Hermitian subspaces negative semidefinite; in each case the zero set is the real or the imaginary scalar line. Only on the vector subspace is the condition a genuine cone, the isotropic cone of complex dimension $2$ and real dimension $4$.

Proof. The values are read from the displayed real bases: on the centre all three vector coefficients vanish; on $\mathbb{H}_{\mathbb{B}}$ and $i\mathbb{H}_{\mathbb{B}}$ the coefficients $Q_k$ are real or purely imaginary; on $\mathbb{M}_{+}$ and $\mathbb{M}_{-}$ the scalar coefficient is real or purely imaginary and the vector coefficients are purely imaginary or real. Each non-central restriction is a sum of three squares with a sign, so its zero set is the scalar line of the subspace.

Summary

The biquaternion algebra carries a second quadratic condition, the square of the vector part, $B^2=Q_x^2+Q_y^2+Q_z^2$, which involves no scalar term. It decides whether the two eigenvalues of an element can be told apart: the characteristic polynomial is $X^2-2Q_tX+Q_t^2+B^2$, of discriminant $-4B^2$, with eigenvalues $Q_t\pm iB$.

The eigenvalue problem is stated without a module and without an acting operator: an eigenvalue is a scalar for which the left ideal $\mathbb{B}(\tilde Q-\lambda e_t)$ is proper, an eigenvector lies in the left annihilator, and a spectral idempotent is an idempotent commuting with the element. The criterion proves that $\lambda$ is an eigenvalue exactly when $\tilde Q-\lambda e_t$ is a non-unit, exactly when $\det\mathsf{M}_2(\tilde Q-\lambda e_t)=0$, exactly when $(\lambda-Q_t)^2+B^2=0$. The eigenspace of a non-scalar element has complex dimension $2$; Cayley–Hamilton gives the inverse whenever the determinant does not vanish.

The regular case is $B^2\neq0$. The eigenvalues are distinct, the element is semisimple, and its left multiplication is diagonalisable. The spectral idempotents are the Lagrange idempotents $\tilde\Pi_\pm=(\tilde Q-\lambda_\mp e_t)/(\lambda_\pm-\lambda_\mp)$, which coincide with $(e_t\pm\vec\Xi)/2$ for $\vec\Xi=-i\vec Q/B$; the decomposition is unique, and an eigenbasis exists with diagonal values $(\lambda_+,\lambda_+,\lambda_-,\lambda_-)$.

The degeneracy case is the cone $B^2=0$, of complex dimension $3$ and real dimension $6$, the product of the scalar line and the isotropic cone of the vector subspace, smooth off the scalar vertex. On it the two eigenvalues have coalesced at $Q_t$. The spectral behaviour is uniform along the whole vertex: the element is scalar and semisimple, every idempotent commutes with it, and the decomposition has one term. Off the vertex the minimal polynomial is $(X-Q_t)^2$, the element is defective, the eigenspace has complex dimension $2$, and no two-term spectral decomposition or eigenbasis exists. The unordered spectrum determines the similarity class off the cone; on the cone it distinguishes only the central class from the non-semisimple one. The condition is blind to the scalar part, so the whole centre is degenerate, and no case is stable under multiplication. The comparison with the norm is Relations between Quaternionic Norm and Spectral Regularity and Degeneracy.

Summary of Notation

symbol	meaning
$\mathbb{B}=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$	the biquaternion algebra
$\tilde Q=Q_te_t+\vec Q$	a general element
$\vec Q=Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z$	the vector part
$B^2=Q_x^2+Q_y^2+Q_z^2$	the square of the vector part, the discriminant up to $-4$
$p_{\tilde Q}(X)=X^2-2Q_tX+Q_t^2+B^2$	the characteristic polynomial, equal to $(X-Q_t)^2+B^2$
$\det\mathsf{M}_2(a)=a_t^2+a_x^2+a_y^2+a_z^2$, for $a=a_te_t+\vec a$	the determinant of the matrix realization, the invertibility test
$\operatorname{disc}p_{\tilde Q}=-4B^2$	the discriminant
$\lambda_\pm=Q_t\pm iB$	the eigenvalues
$\mathcal S=\{B^2=0\}$	the spectral-degenerate cone
$\operatorname{Ann}_{\ell}(a)=\{v:av=0\}$	the left annihilator, a right ideal
$\vec\Xi=-i\vec Q/B$	the pure root, with $\vec\Xi^2=e_t$
$\tilde\Pi_\pm=(e_t\pm\vec\Xi)/2$	the spectral idempotents, equal to $(\tilde Q-\lambda_\mp e_t)/(\lambda_\pm-\lambda_\mp)$