Eigenbasis of a Biquaternion in Matrix Elements Representations

Introduction

The eigenbasis of a biquaternion $\tilde Q=Q_te_t+\vec Q$ is the basis $$ \tilde\Pi_+,\qquad \vec\Upsilon_-,\qquad \vec\Upsilon_+,\qquad \tilde\Pi_- $$ of the Peirce decomposition of its eigenprojections, in which multiplication by $\tilde Q$ is diagonal with eigenvalues $\lambda_+,\lambda_+,\lambda_-,\lambda_-$. The basis itself and the construction of the nilpotents $\vec\Upsilon_\pm$ from the two mixed Peirce corners are the subject of The Quaternionic Norm-Regular, Spectrally Regular Sector, §The Peirce Corners and the Nilpotents and §The Eigenbasis; the eigenvalue decomposition is that of The Biquaternion Eigenvalue Decomposition, §The Eigenprojectors. This article reads the same objects in the two matrix element representations of the algebra.

The first is the $2\times2$ matrix element representation $\mathsf{M}_2:\mathbb{B}\to M_2(\mathbb{C})$, the fixed isomorphism of Introduction to the $2\times2$ Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions. Every biquaternion becomes a $2\times2$ complex matrix, the two minimal left ideals become the two columns, and the eigenbasis becomes a basis of matrix units. The second is the $4\times4$ left regular representation $\mathsf{M}_4^{L}:\mathbb{B}\to M_4(\mathbb{C})$ of Introduction to the $4\times4$ Matrix Element Representation $M_4(\mathbb{C})_L$ of Biquaternions, with its right companion $\mathsf{M}_4^{R}$. The $4\times4$ matrices carry the multiplication on the four-vector space; in the eigenbasis order they are Kronecker products of the $2\times2$ matrix with an identity.

The article is organised around three questions. Which matrices are the eigenprojections and the nilpotents? They are the spectral projectors $\rho_\pm=\mathsf{M}_2(\tilde\Pi_\pm)$ and the two off-diagonal matrix units, and the eigenbasis model $\mathsf{M}_2^{Q}$ is the map that sends them to the four matrix units. What does a change of basis do? The fixed model, the eigenbasis of the standard root $i\vec e_z$, and the eigenbasis are three bases of the same four-dimensional space; the passage between them is a conjugation, and the diagonalising matrix $\mathsf{C}_Q$ is determined by the eigenvectors up to the rescaling of the nilpotents. What happens in the $4\times4$ representation? The left and right regular matrices are block matrices built from the $2\times2$ Peirce matrix, and the eigenbasis order diagonalises both.

Conventions. The base and the elements are those of Conventions in Mathematics: $\tilde Q=Q_te_t+Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z$, $B=\sqrt{Q_x^2+Q_y^2+Q_z^2}$, $\vec\Xi=-i\vec Q/B$, $\tilde\Pi_\pm=\tfrac12(e_t\pm\vec\Xi)$, $\lambda_\pm=Q_t\pm iB$. A general element distinct from $\tilde Q$ is written $\tilde Q_{\mathrm{sp}}$ or $\tilde Q_3$. The fixed $2\times2$ model is $\mathsf{M}_2$; the eigenbasis model is $\mathsf{M}_2^{Q}$; the regular matrices are $\mathsf{M}_4^{L}$ and $\mathsf{M}_4^{R}$, with the shorter $\mathsf{M}_4=\mathsf{M}_4^{L}$. The two matrix assignments are those of the two introduction articles cited above and are not redefined here.

Boundaries. The eigenvalue decomposition is The Biquaternion Eigenvalue Decomposition, §The Eigenprojectors; the Peirce decomposition and the construction of the nilpotents are The Quaternionic Norm-Regular, Spectrally Regular Sector, §The Eigenbasis, the Nilpotents, and the Peirce Corners. The $2\times2$ assignment, its invertibility and its invariants are Introduction to the $2\times2$ Matrix Element Representation $M_2(\mathbb{C})$ of Biquaternions; the $4\times4$ assignment, its multiplicativity and its remarkable subspaces are Introduction to the $4\times4$ Matrix Element Representation $M_4(\mathbb{C})_L$ of Biquaternions. The matrix algebra is Matrix Algebras, the spectral projections of a general matrix are The Eigenvalue Decomposition of a Matrix, and the change of basis of an endomorphism is Matrix Algebras, §Change of Basis.

The $2\times2$ Matrix Element Representation

The fixed model sends the four units to four matrices and the four coefficients to a $2\times2$ complex matrix.

The assignment is fixed by its values on the basis, $$ \mathsf{M}_2(e_t)=\mathrm I_2,\qquad \mathsf{M}_2(\vec e_x)=\begin{pmatrix}0&-i\\-i&0\end{pmatrix},\qquad \mathsf{M}_2(\vec e_y)=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad \mathsf{M}_2(\vec e_z)=\begin{pmatrix}-i&0\\0&i\end{pmatrix}, $$ and extended by $\mathbb{C}$-linearity. A general element therefore has the matrix $$ \mathsf{M}_2(\tilde Q_{\mathrm{sp}})= \begin{pmatrix} Q_{\mathrm{sp},t}-iQ_{\mathrm{sp},z}&-iQ_{\mathrm{sp},x}-Q_{\mathrm{sp},y}\\ -iQ_{\mathrm{sp},x}+Q_{\mathrm{sp},y}&Q_{\mathrm{sp},t}+iQ_{\mathrm{sp},z} \end{pmatrix}. $$ The map is an isomorphism of $\mathbb{C}$-algebras, so products, inverses, powers and conjugates are read in the matrices without loss. The inverse map is $$ \mathsf{M}_2^{-1}:\begin{pmatrix}a&b\\c&d\end{pmatrix}\longmapsto Q_te_t+Q_x\vec e_x+Q_y\vec e_y+Q_z\vec e_z, \qquad Q_t=\tfrac{a+d}{2},\quad Q_x=\tfrac{i(b+c)}{2},\quad Q_y=\tfrac{c-b}{2},\quad Q_z=\tfrac{i(a-d)}{2}. $$ The invariants are $$ \boxed{\; \operatorname{tr}\mathsf{M}_2(\tilde Q_{\mathrm{sp}})=2Q_{\mathrm{sp},t},\qquad \det\mathsf{M}_2(\tilde Q_{\mathrm{sp}})=\langle\tilde Q_{\mathrm{sp}}\rangle_{\natural}=Q_{\mathrm{sp},t}^2+Q_{\mathrm{sp},x}^2+Q_{\mathrm{sp},y}^2+Q_{\mathrm{sp},z}^2\;}. $$ The four basis matrices are collected in the table.

A general element is not a matrix unit. The entries of $\mathsf{M}_2(\tilde Q_{\mathrm{sp}})$ are linear combinations of two coefficients each, so the four matrix entries and the four biquaternion coefficients carry the same data in two different arrangements. The arrangement is fixed and is not a choice of basis: it is the assignment above. A different isomorphism differs from this one by conjugation by an invertible matrix, which is the subject of §Base Changes.

The image of the algebra is everything. Since $\mathbb{B}$ and $M_2(\mathbb{C})$ have complex dimension four and $\mathsf{M}_2$ is injective, the image is the whole of $M_2(\mathbb{C})$. This is the structural difference with the split-complex and reduced biquaternion cases, where the image is a proper commutative subalgebra. The whole matrix algebra is available: every $2\times2$ complex matrix is a biquaternion, and every matrix identity is a biquaternion identity.

The Eigenprojections in the Model

The eigenprojections become the spectral projectors of the matrix, and the two eigenvalues become its two diagonal entries in the eigenbasis.

For the standard root $\vec\Xi_0=i\vec e_z$ the idempotents are $\tilde\Pi_{\pm,0}=\tfrac12(e_t\pm i\vec e_z)$, and their matrices are the two diagonal matrix units: $$ \rho_+^{(0)}=\mathsf{M}_2(\tilde\Pi_{+,0})=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad \rho_-^{(0)}=\mathsf{M}_2(\tilde\Pi_{-,0})=\begin{pmatrix}0&0\\0&1\end{pmatrix}. $$ For a general element, with $\vec\Xi=-i\vec Q/B$, $$ \rho_\pm=\mathsf{M}_2(\tilde\Pi_\pm)=\frac{1}{2}\left(\mathrm I_2\pm\mathsf{M}_2(\vec\Xi)\right), \qquad \mathsf{M}_2(\vec\Xi)=\frac{1}{B} \begin{pmatrix} -Q_z&-Q_x+iQ_y\\ -Q_x-iQ_y&Q_z \end{pmatrix}. $$ Written out, $$ \rho_+=\frac{1}{2B} \begin{pmatrix} B-Q_z&-Q_x+iQ_y\\ -Q_x-iQ_y&B+Q_z \end{pmatrix}, \qquad \rho_-=\frac{1}{2B} \begin{pmatrix} B+Q_z&Q_x-iQ_y\\ Q_x+iQ_y&B-Q_z \end{pmatrix}. $$ The two matrices are the spectral projectors of $\mathsf{M}_2(\tilde Q)$: $$ \rho_\pm^2=\rho_\pm,\qquad \rho_+\rho_-=\rho_-\rho_+=0,\qquad \rho_++\rho_-=\mathrm I_2,\qquad \operatorname{tr}\rho_\pm=1,\qquad \mathsf{M}_2(\tilde Q)\rho_\pm=\lambda_\pm\rho_\pm. $$ Each has rank one; its image is the line of the corresponding eigenvector. The columns of $\rho_+$ are eigenvectors of eigenvalue $\lambda_+$, and the columns of $\rho_-$ eigenvectors of eigenvalue $\lambda_-$. Every nonzero column is one, and the two columns of one projector are proportional.

The nilpotents in the model. The symmetric closed form of The Quaternionic Norm-Regular, Spectrally Regular Sector, §The Peirce Corners and the Nilpotents, is, with $\Sigma=Q_x+Q_y+Q_z$, $$ \vec\Upsilon_\pm=\frac{B\,\vec w-\dfrac{\Sigma}{B}\,\vec Q\pm\dfrac{i}{2}\,[\vec Q,\vec w]}{2\sqrt{\Sigma^2-3B^2}}, \qquad \vec w=\vec e_x+\vec e_y+\vec e_z, $$ and its image is $$ \mathsf{M}_2(\vec\Upsilon_\pm)=\frac{N_\pm}{2\sqrt{\Sigma^2-3B^2}}, $$ where $$ N_\pm= \begin{pmatrix} i\left(\dfrac{\Sigma Q_z}{B}-B\right)\pm(Q_x-Q_y) & -B(1+i)+\dfrac{\Sigma}{B}(iQ_x+Q_y)\pm\bigl[(Q_y-Q_z)-i(Q_z-Q_x)\bigr]\\[6pt] B(1-i)+\dfrac{\Sigma}{B}(iQ_x-Q_y)\pm\bigl[(Q_y-Q_z)+i(Q_z-Q_x)\bigr] & -i\left(\dfrac{\Sigma Q_z}{B}-B\right)\mp(Q_x-Q_y) \end{pmatrix}. $$ Both matrices are rank one and trace zero, and $$ \mathsf{M}_2(\vec\Upsilon_+)\mathsf{M}_2(\vec\Upsilon_-)=\rho_-,\qquad \mathsf{M}_2(\vec\Upsilon_-)\mathsf{M}_2(\vec\Upsilon_+)=\rho_+,\qquad \mathsf{M}_2(\vec\Upsilon_\pm)^2=0, $$ with $$ \rho_+\mathsf{M}_2(\vec\Upsilon_-)\rho_-=\mathsf{M}_2(\vec\Upsilon_-),\qquad \rho_-\mathsf{M}_2(\vec\Upsilon_+)\rho_+=\mathsf{M}_2(\vec\Upsilon_+). $$ The entries are not symmetric in $x,y,z$ even though the elements are, because the fixed model $\mathsf{M}_2$ itself singles out $\vec e_z$: it is the representation in which $\mathsf{M}_2(\vec e_z)$ is diagonal. The symmetric formula is a statement in the algebra; its matrix image inherits the asymmetry of the representation.

The four basis elements of the eigenbasis therefore have the following matrices in the fixed model.

On the cone $\Sigma^2=3B^2$ the symmetric formula is singular; there the chart of The Quaternionic Norm-Regular, Spectrally Regular Sector, §The Peirce Corners and the Nilpotents, applies, and the matrices are its $\mathsf{M}_2$.

The standard case. For $\vec Q=Q_z\vec e_z$ the matrix is diagonal, $\mathsf{M}_2(\tilde Q)=\operatorname{diag}(Q_t-iQ_z,Q_t+iQ_z)$, and the projectors are $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\0&1\end{pmatrix}$. The general formula reduces to them when the sign of $B$ is chosen so that $B=-Q_z$; the other sign exchanges the two labels. This is the sign freedom of the root, read in the model.

The spectral reading. The matrices $\rho_\pm$ are the images of the eigenprojections, so the eigenvalue decomposition $\tilde Q=\lambda_+\tilde\Pi_++\lambda_-\tilde\Pi_-$ becomes $$ \mathsf{M}_2(\tilde Q)=\lambda_+\rho_++\lambda_-\rho_-. $$ The trace is $\lambda_++\lambda_-=2Q_t$, the determinant is $\lambda_+\lambda_-=\langle\tilde Q\rangle_{\natural}$, and the characteristic polynomial of the matrix is $(X-\lambda_+)(X-\lambda_-)$. The two facts are the matrix form of the eigenvalue decomposition of the element.

The Eigenbasis in the Model

The four elements of the eigenbasis become the four matrix units, and the Peirce matrix of $\tilde Q$ becomes diagonal.

The eigenbasis is $$ \tilde\Pi_+,\qquad \vec\Upsilon_-,\qquad \vec\Upsilon_+,\qquad \tilde\Pi_-, \qquad \vec\Upsilon_+\in\mathbb{B}_{-+},\quad \vec\Upsilon_-\in\mathbb{B}_{+-}, $$ with the normalisation $$ \vec\Upsilon_+\vec\Upsilon_-=\tilde\Pi_-,\qquad \vec\Upsilon_-\vec\Upsilon_+=\tilde\Pi_+,\qquad \vec\Upsilon_\pm^2=0. $$ The model $\mathsf{M}_2^{Q}$ is defined by the assignment $$ \tilde\Pi_+\longmapsto\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad \vec\Upsilon_-\longmapsto\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad \vec\Upsilon_+\longmapsto\begin{pmatrix}0&0\\1&0\end{pmatrix},\qquad \tilde\Pi_-\longmapsto\begin{pmatrix}0&0\\0&1\end{pmatrix}. $$ It is the composition of $\mathsf{M}_2$ with the change of basis that carries the eigenbasis to the matrix units. Every element has a unique expansion $$ \tilde Q_{\mathrm{sp}}=Q_{\mathrm{sp},11}\tilde\Pi_++Q_{\mathrm{sp},12}\vec\Upsilon_-+Q_{\mathrm{sp},21}\vec\Upsilon_++Q_{\mathrm{sp},22}\tilde\Pi_-, \qquad Q_{\mathrm{sp},ij}\in\mathbb{C}, $$ and is represented by $$ \mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})= \begin{pmatrix} Q_{\mathrm{sp},11}&Q_{\mathrm{sp},12}\\ Q_{\mathrm{sp},21}&Q_{\mathrm{sp},22} \end{pmatrix}. $$ For $\tilde Q$ itself the coordinates are $Q_{\mathrm{sp},11}=\lambda_+$, $Q_{\mathrm{sp},22}=\lambda_-$, $Q_{\mathrm{sp},12}=Q_{\mathrm{sp},21}=0$, so $$ \boxed{\; \mathsf{M}_2^{Q}(\tilde Q)= \begin{pmatrix} \lambda_+&0\\ 0&\lambda_- \end{pmatrix}\;}. $$ The four basis elements and the general element are collected in the table.

The algebra in matrix form. The Peirce identities of the eigenbasis read as the matrix-unit identities: $$ \begin{pmatrix}1&0\\0&0\end{pmatrix}^2=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad \begin{pmatrix}0&0\\0&1\end{pmatrix}^2=\begin{pmatrix}0&0\\0&1\end{pmatrix},\qquad \begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=0, $$ $$ \begin{pmatrix}0&1\\0&0\end{pmatrix}^2=0,\qquad \begin{pmatrix}0&0\\1&0\end{pmatrix}^2=0,\qquad \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad \begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{pmatrix}, $$ together with $$ \begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}, $$ $$ \begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}. $$ These are the corner relations of the Peirce decomposition, read as matrix multiplication. The two diagonal units are the identity elements of the two minimal left ideals, and the two off-diagonal units are the nilpotents.

General objects. For two elements with matrices $A=\mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})$ and $N=\mathsf{M}_2^{Q}(\tilde Q_3)$, the product, inverse and invariants are $$ \mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}}\tilde Q_3)=AN,\qquad \mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}}^{-1})=A^{-1}\quad(\det A\neq0), $$ $$ \det\mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})=Q_{\mathrm{sp},11}Q_{\mathrm{sp},22}-Q_{\mathrm{sp},12}Q_{\mathrm{sp},21}=\langle\tilde Q_{\mathrm{sp}}\rangle_{\natural}, \qquad \operatorname{tr}\mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})=Q_{\mathrm{sp},11}+Q_{\mathrm{sp},22}=2Q_{\mathrm{sp},t}. $$ Quaternion conjugation is the adjugate, $$ \mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}}^{\natural})= \begin{pmatrix}Q_{\mathrm{sp},22}&-Q_{\mathrm{sp},12}\\-Q_{\mathrm{sp},21}&Q_{\mathrm{sp},11}\end{pmatrix}, \qquad \mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}}\tilde Q_{\mathrm{sp}}^{\natural})=\langle\tilde Q_{\mathrm{sp}}\rangle_{\natural}\,\mathrm I_2. $$ The identity is $\mathrm I_2$, and a power is the power of the matrix. The model is an isomorphism, so no information is lost and no relation is added: the eigenbasis is a choice of coordinates on the same algebra.

Base Changes

The fixed model, the standard eigenbasis and the eigenbasis of $\tilde Q$ are three bases of the same matrix space, and the passage between them is a conjugation by an invertible matrix.

The Change of Basis of a Matrix Model

Let $P\in GL_2(\mathbb{C})$ be the matrix whose columns are the coordinates of a new basis in the old one. The matrix of the same endomorphism in the new basis is $$ M'=P^{-1}MP. $$ The trace and the determinant are unchanged, $$ \operatorname{tr}M'=\operatorname{tr}M,\qquad \det M'=\det M, $$ and so are the characteristic polynomial, the spectrum, the rank and the Jordan form. The idempotents and the nilpotents are not unchanged as matrices: they are conjugated, $$ \rho\longmapsto P^{-1}\rho P, \qquad N\longmapsto P^{-1}NP, $$ and they remain idempotent and square-zero respectively. The two columns of $P$ are the new basis vectors; a column of $P$ that is an eigenvector of $M$ produces a column of $P^{-1}MP$ with the eigenvalue on the diagonal.

The two normal forms. The fixed model and the eigenbasis model are the two ends of one change. The fixed model is $\mathsf{M}_2$; the eigenbasis model is $$ \mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})=\mathsf{C}_Q^{-1}\mathsf{M}_2(\tilde Q_{\mathrm{sp}})\mathsf{C}_Q, $$ where $\mathsf{C}_Q$ is the matrix whose columns are the coordinates of a $\lambda_+$-eigenvector and a $\lambda_-$-eigenvector in the fixed basis. The first column is a nonzero column of $\rho_+$ and the second a nonzero column of $\rho_-$. The matrix $\mathsf{C}_Q$ is invertible because the two columns are independent: $\rho_+\rho_-=0$ and both are nonzero.

The Explicit Diagonalising Matrix

For $B-Q_z\neq0$ put $$ \mathsf{C}_Q= \begin{pmatrix} B-Q_z&Q_x-iQ_y\\ -Q_x-iQ_y&B-Q_z \end{pmatrix}, \qquad \det\mathsf{C}_Q=-2Q_z(B-Q_z)\neq0. $$ The first column is the first column of $2B\rho_+$, the second is the second column of $2B\rho_-$, and the two are eigenvectors of $\mathsf{M}_2(\tilde Q)$ with eigenvalues $\lambda_+$ and $\lambda_-$ respectively. Therefore $$ \mathsf{C}_Q^{-1}\mathsf{M}_2(\tilde Q)\mathsf{C}_Q=\operatorname{diag}(\lambda_+,\lambda_-). $$ When $Q_z=0$ or $B=Q_z$ the first formula is singular: at $B=Q_z$ one of the two chosen columns of the spectral projectors vanishes, and at $Q_z=0$ the two chosen columns are proportional. The other pair of nonzero columns gives $$ \mathsf{C}_Q'= \begin{pmatrix} -Q_x+iQ_y&B+Q_z\\ B+Q_z&Q_x+iQ_y \end{pmatrix}, \qquad \det\mathsf{C}_Q'=-2B(B+Q_z)\neq0, $$ with columns the second column of $2B\rho_+$ and the first column of $2B\rho_-$. It satisfies the same diagonalisation. Since $B\neq0$, the determinants $-2Q_z(B-Q_z)$ and $-2B(B+Q_z)$ cannot both vanish, so one of the two formulas applies. When $Q_x=Q_y=0$, the sign $B=-Q_z$ is the first case, $\mathsf{C}_Q=-2Q_z\,\mathrm I_2$, and the sign $B=Q_z$ is the second, $\mathsf{C}_Q'=2Q_z\begin{pmatrix}0&1\\1&0\end{pmatrix}$; both are invertible because $Q_z\neq0$ when $B\neq0$. The scalar case $B=0$ is excluded, as in the eigenbasis article. The matrix $\mathsf{C}_Q$ is not the most general diagonaliser: it is the one whose columns are the two columns of the spectral projectors.

The inverse and the conjugation. For the first matrix, with $D=-2Q_z(B-Q_z)$, $$ \mathsf{C}_Q^{-1}=\frac{1}{D} \begin{pmatrix} B-Q_z&-(Q_x-iQ_y)\\ Q_x+iQ_y&B-Q_z \end{pmatrix}. $$ The second matrix has the analogous inverse. In either case the conjugation $\Phi(M)=\mathsf{C}_Q^{-1}M\mathsf{C}_Q$, with $\mathsf{C}_Q$ the chosen matrix, sends $$ \rho_+\longmapsto\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad \rho_-\longmapsto\begin{pmatrix}0&0\\0&1\end{pmatrix},\qquad \mathsf{M}_2(\tilde Q)\longmapsto\operatorname{diag}(\lambda_+,\lambda_-). $$ It sends each mixed corner onto the mixed corner with the same indices, $$ \mathsf{M}_2(\mathbb{B}_{+-})\longmapsto\mathbb{C}\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad \mathsf{M}_2(\mathbb{B}_{-+})\longmapsto\mathbb{C}\begin{pmatrix}0&0\\1&0\end{pmatrix}, $$ and it is the algebra isomorphism $\mathsf{M}_2^{Q}$.

The Nilpotents in the Change

The two nilpotents are recovered from the matrix units by the inverse conjugation: $$ \mathsf{M}_2(\vec\Upsilon_-)=\mathsf{C}_Q\begin{pmatrix}0&1\\0&0\end{pmatrix}\mathsf{C}_Q^{-1}, \qquad \mathsf{M}_2(\vec\Upsilon_+)=\mathsf{C}_Q\begin{pmatrix}0&0\\1&0\end{pmatrix}\mathsf{C}_Q^{-1}. $$ They are square-zero and, in the eigenbasis model, they are the two off-diagonal matrix units: $$ \mathsf{M}_2^{Q}(\vec\Upsilon_+)\mathsf{M}_2^{Q}(\vec\Upsilon_-) =\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix} =\begin{pmatrix}0&0\\0&1\end{pmatrix}, \qquad \mathsf{M}_2^{Q}(\vec\Upsilon_-)\mathsf{M}_2^{Q}(\vec\Upsilon_+) =\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix} =\begin{pmatrix}1&0\\0&0\end{pmatrix}, $$ that is $\vec\Upsilon_+\vec\Upsilon_-=\tilde\Pi_-$ and $\vec\Upsilon_-\vec\Upsilon_+=\tilde\Pi_+$. In the fixed model the same identities read $$ \mathsf{M}_2(\vec\Upsilon_+)\mathsf{M}_2(\vec\Upsilon_-)=\rho_-, \qquad \mathsf{M}_2(\vec\Upsilon_-)\mathsf{M}_2(\vec\Upsilon_+)=\rho_+. $$ The construction fixes the two matrices, not merely the corner lines: the normalisation of the pair is exactly the condition that the off-diagonal entries of the eigenbasis matrix be $1$ rather than an arbitrary nonzero scalar.

The Freedom of the Change

The diagonalising matrix is not unique. If $\mathsf{C}_Q$ diagonalises $\mathsf{M}_2(\tilde Q)$, then so does $\mathsf{C}_Q D$ for every invertible diagonal $D=\operatorname{diag}(\alpha,\beta)$: $$ (\mathsf{C}_Q D)^{-1}\mathsf{M}_2(\tilde Q)(\mathsf{C}_Q D) =D^{-1}\operatorname{diag}(\lambda_+,\lambda_-)D =\operatorname{diag}(\lambda_+,\lambda_-). $$ The two scales are the two independent rescalings of the eigenvectors. A common scale $\alpha=\beta$ leaves the conjugation unchanged, because $(\alpha \mathsf{C}_Q)^{-1}M(\alpha \mathsf{C}_Q)=\mathsf{C}_Q^{-1}M\mathsf{C}_Q$. A relative scale multiplies the two off-diagonal entries by $\alpha^{-1}\beta$ and $\alpha\beta^{-1}$, which is the reciprocal rescaling $$ \vec\Upsilon_+\longmapsto\mu\vec\Upsilon_+,\qquad \vec\Upsilon_-\longmapsto\mu^{-1}\vec\Upsilon_-,\qquad \mu\in\mathbb{C}^{\times}, $$ of the nilpotents. The normalised pair is the choice $\alpha=\beta=1$ in the chosen matrix $\mathsf{C}_Q$ above; the general diagonaliser is $\mathsf{C}_Q\operatorname{diag}(\alpha,\beta)$.

What depends on $\tilde Q$. The matrix $\mathsf{C}_Q$, the projectors $\rho_\pm$, the nilpotents $\mathsf{M}_2(\vec\Upsilon_\pm)$ and the diagonal matrix $\operatorname{diag}(\lambda_+,\lambda_-)$ depend on $\tilde Q$. The fixed model $\mathsf{M}_2$, the four elementary matrices and the assignment $\mathsf{M}_2^{Q}$ as a map from the eigenbasis do not: they are the same for every $\tilde Q$ for which the construction exists. The dependence is carried by the basis, not by the algebra.

Changing from One Element to Another

For two elements $\tilde Q$ and $\tilde Q'$ with distinct eigenvalues, the two eigenbases are related by a further change of basis, $$ \mathsf{C}_{Q'}=\mathsf{C}_Q\,\mathsf{D}_{Q,Q'}, $$ where $\mathsf{D}_{Q,Q'}$ is the matrix of the eigenbasis of $\tilde Q'$ in that of $\tilde Q$. The matrix $\mathsf{D}_{Q,Q'}$ is generally full, not diagonal, because the two bases have different idempotents and different nilpotents. The composite change satisfies the cocycle relation $$ \mathsf{D}_{Q,Q''}=\mathsf{D}_{Q,Q'}\mathsf{D}_{Q',Q''}, $$ and the eigenvalues are unchanged by it. This is the matrix form of the fact that the eigenbasis is attached to the element: there is no single eigenbasis of the whole algebra.

The standard eigenbasis as a special case. For $\tilde Q'$ commuting with $\vec e_z$, the eigenbasis of $\tilde Q'$ is the standard one up to the exchange of the two labels and the reciprocal rescaling, and $\mathsf{C}_{Q'}$ is the fixed matrix of §The Fixed Change in the $4\times4$ Model, composed with the exchange when the labels are exchanged. The general $\mathsf{C}_Q$ reduces to it when $Q_x=Q_y=0$ with the sign of $B$ chosen so that the root is $\pm i\vec e_z$. The reduction is the statement that the standard idempotents are the eigenprojections of the elements on the $e_z$ axis, up to the exchange.

Spinors and the Columns

The two columns of the $2\times2$ model are the two minimal left ideals, and a spinor is a column.

The matrices with zero second column are the image of $S_+=\mathbb{B}\tilde\Pi_+$, and the matrices with zero first column are the image of $S_-=\mathbb{B}\tilde\Pi_-$: $$ S_+\longleftrightarrow\left\{\begin{pmatrix}Q_{\mathrm{sp},11}&0\\Q_{\mathrm{sp},21}&0\end{pmatrix}:Q_{\mathrm{sp},11},Q_{\mathrm{sp},21}\in\mathbb{C}\right\}, \qquad S_-\longleftrightarrow\left\{\begin{pmatrix}0&Q_{\mathrm{sp},12}\\0&Q_{\mathrm{sp},22}\end{pmatrix}:Q_{\mathrm{sp},12},Q_{\mathrm{sp},22}\in\mathbb{C}\right\}. $$ A spinor of first chirality is an element of $S_+$, $$ \xi_+=Q_{\mathrm{sp},11}\tilde\Pi_++Q_{\mathrm{sp},21}\vec\Upsilon_+, $$ and is represented by the first column, that is, by the matrix $$ \xi_+\longleftrightarrow \begin{pmatrix}Q_{\mathrm{sp},11}&0\\ Q_{\mathrm{sp},21}&0\end{pmatrix}, \qquad \xi_+\longleftrightarrow \begin{pmatrix}Q_{\mathrm{sp},11}\\ Q_{\mathrm{sp},21}\end{pmatrix} \text{ in column }1. $$ A spinor of second chirality is an element of $S_-$, $$ \xi_-=Q_{\mathrm{sp},12}\vec\Upsilon_-+Q_{\mathrm{sp},22}\tilde\Pi_-, $$ and is represented by the second column, $$ \xi_-\longleftrightarrow \begin{pmatrix}0&Q_{\mathrm{sp},12}\\ 0&Q_{\mathrm{sp},22}\end{pmatrix}, \qquad \xi_-\longleftrightarrow \begin{pmatrix}Q_{\mathrm{sp},12}\\ Q_{\mathrm{sp},22}\end{pmatrix} \text{ in column }2. $$ The two spinor spaces and the column correspondence are collected in the table.

Under left multiplication by $\tilde Q$ the matrix is multiplied on the left by $\operatorname{diag}(\lambda_+,\lambda_-)$: $$ \tilde Q\xi_+\longleftrightarrow \begin{pmatrix}\lambda_+&0\\0&\lambda_-\end{pmatrix} \begin{pmatrix}Q_{\mathrm{sp},11}\\ Q_{\mathrm{sp},21}\end{pmatrix} \text{ in column }1, \qquad \tilde Q\xi_-\longleftrightarrow \begin{pmatrix}\lambda_+&0\\0&\lambda_-\end{pmatrix} \begin{pmatrix}Q_{\mathrm{sp},12}\\ Q_{\mathrm{sp},22}\end{pmatrix} \text{ in column }2. $$ The same matrix acts on both columns. This is the precise sense in which the two spinor spaces are two copies of the same module.

Right multiplication and the rows. Right multiplication by $\tilde Q$ sends $\mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})$ to $\mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})\operatorname{diag}(\lambda_+,\lambda_-)$; it scales the first column by $\lambda_+$ and the second by $\lambda_-$. The two rows are the two minimal right ideals: the first row is $\tilde\Pi_+\mathbb{B}$, the second is $\tilde\Pi_-\mathbb{B}$. Left multiplication scales the first row by $\lambda_+$ and the second by $\lambda_-$. The rows and the columns are the two-sided Peirce structure read in the matrix.

The $4\times4$ Matrix Element Representation

The left and right regular matrices carry the multiplication on the four-vector space, and in the eigenbasis order they are Kronecker products of the $2\times2$ Peirce matrix with an identity.

The Regular Matrices

In the fixed basis $e_t,\vec e_x,\vec e_y,\vec e_z$, the left regular matrix $\mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}})$ has as $\mu$-th column the coordinate column of $\tilde Q_{\mathrm{sp}}e_\mu$: $$ \mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}})= \begin{pmatrix} Q_{\mathrm{sp},t}&-Q_{\mathrm{sp},x}&-Q_{\mathrm{sp},y}&-Q_{\mathrm{sp},z}\\ Q_{\mathrm{sp},x}&Q_{\mathrm{sp},t}&-Q_{\mathrm{sp},z}&Q_{\mathrm{sp},y}\\ Q_{\mathrm{sp},y}&Q_{\mathrm{sp},z}&Q_{\mathrm{sp},t}&-Q_{\mathrm{sp},x}\\ Q_{\mathrm{sp},z}&-Q_{\mathrm{sp},y}&Q_{\mathrm{sp},x}&Q_{\mathrm{sp},t} \end{pmatrix}. $$ The right regular matrix $\mathsf{M}_4^{R}(\tilde Q_{\mathrm{sp}})$ has as $\mu$-th column the coordinate column of $e_\mu\tilde Q_{\mathrm{sp}}$: $$ \mathsf{M}_4^{R}(\tilde Q_{\mathrm{sp}})= \begin{pmatrix} Q_{\mathrm{sp},t}&-Q_{\mathrm{sp},x}&-Q_{\mathrm{sp},y}&-Q_{\mathrm{sp},z}\\ Q_{\mathrm{sp},x}&Q_{\mathrm{sp},t}&Q_{\mathrm{sp},z}&-Q_{\mathrm{sp},y}\\ Q_{\mathrm{sp},y}&-Q_{\mathrm{sp},z}&Q_{\mathrm{sp},t}&Q_{\mathrm{sp},x}\\ Q_{\mathrm{sp},z}&Q_{\mathrm{sp},y}&-Q_{\mathrm{sp},x}&Q_{\mathrm{sp},t} \end{pmatrix}. $$ They satisfy $$ \mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}})\mathsf{M}_4^{L}(\tilde Q_3)=\mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}}\tilde Q_3), \qquad \mathsf{M}_4^{R}(\tilde Q_{\mathrm{sp}})\mathsf{M}_4^{R}(\tilde Q_3)=\mathsf{M}_4^{R}(\tilde Q_3\tilde Q_{\mathrm{sp}}), $$ and $$ \operatorname{Tr}\mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}})=4Q_{\mathrm{sp},t}, \qquad \det\mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}})=\langle\tilde Q_{\mathrm{sp}}\rangle_{\natural}^2. $$ The left matrix is multiplicative and the right matrix is anti-multiplicative; both are injective, and the left one alone is the regular representation of the algebra.

The Eigenbasis Order

Write $M=\mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})$. In the eigenbasis order $$ (\tilde\Pi_+,\vec\Upsilon_-,\vec\Upsilon_+,\tilde\Pi_-) $$ the two regular matrices are block matrices: $$ [L_{\tilde Q_{\mathrm{sp}}}]_{\mathrm Q}= \begin{pmatrix} Q_{\mathrm{sp},11}&0&Q_{\mathrm{sp},12}&0\\ 0&Q_{\mathrm{sp},11}&0&Q_{\mathrm{sp},12}\\ Q_{\mathrm{sp},21}&0&Q_{\mathrm{sp},22}&0\\ 0&Q_{\mathrm{sp},21}&0&Q_{\mathrm{sp},22} \end{pmatrix} =M\otimes\mathrm I_2, $$ $$ [R_{\tilde Q_{\mathrm{sp}}}]_{\mathrm Q}= \begin{pmatrix} Q_{\mathrm{sp},11}&Q_{\mathrm{sp},21}&0&0\\ Q_{\mathrm{sp},12}&Q_{\mathrm{sp},22}&0&0\\ 0&0&Q_{\mathrm{sp},11}&Q_{\mathrm{sp},21}\\ 0&0&Q_{\mathrm{sp},12}&Q_{\mathrm{sp},22} \end{pmatrix} =\mathrm I_2\otimes M^{\mathsf T}. $$ In this order the two blocks of two coordinates are the two minimal right ideals $\tilde\Pi_+\mathbb{B}$ and $\tilde\Pi_-\mathbb{B}$. Left multiplication acts on each block by $M$, which is the block form $M\otimes\mathrm I_2$; right multiplication preserves each block and acts on it by $M^{\mathsf T}$, which is the block form $\mathrm I_2\otimes M^{\mathsf T}$. The two blocks are the rows of the $2\times2$ model, and the correspondence between the rows and the right ideals is the one stated in §The Eigenbasis of Left Multiplication.

In the column order $$ (\tilde\Pi_+,\vec\Upsilon_+\,;\,\vec\Upsilon_-,\tilde\Pi_-), $$ which groups the two vectors of $S_+$ and then the two of $S_-$, the left matrix is block diagonal and the right matrix is not: $$ [L_{\tilde Q_{\mathrm{sp}}}]_{\mathrm{col}}= \begin{pmatrix}M&0\\0&M\end{pmatrix} =\mathrm I_2\otimes M, \qquad [R_{\tilde Q_{\mathrm{sp}}}]_{\mathrm{col}}= \begin{pmatrix} Q_{\mathrm{sp},11}\mathrm I_2&Q_{\mathrm{sp},21}\mathrm I_2\\ Q_{\mathrm{sp},12}\mathrm I_2&Q_{\mathrm{sp},22}\mathrm I_2 \end{pmatrix} =M^{\mathsf T}\otimes\mathrm I_2. $$ Left multiplication preserves each left ideal $S_\pm$ and acts on each by $M$, so $[L]_{\mathrm{col}}$ is block diagonal. Right multiplication does not preserve the left ideals; it mixes $S_+$ and $S_-$ according to the entries of $M$, so $[R]_{\mathrm{col}}$ is not block diagonal. The two orders differ by the permutation of the second and third basis vectors. In the eigenbasis order the blocks are the right ideals and the right matrix is block diagonal; in the column order the blocks are the left ideals and the left matrix is block diagonal. The two block forms record the right-ideal decomposition $\mathbb{B}=\tilde\Pi_+\mathbb{B}\oplus\tilde\Pi_-\mathbb{B}$ and the left-ideal decomposition $\mathbb{B}=S_+\oplus S_-$, which are different groupings of the same four basis vectors.

The Four Basis Elements in the Eigenbasis

The four elements of the eigenbasis order have the four elementary matrices for left multiplication, and the fixed basis has four constant matrices in the standard eigenbasis.

In the order $(\tilde\Pi_+,\vec\Upsilon_-,\vec\Upsilon_+,\tilde\Pi_-)$ the left regular matrices are the Kronecker products $\mathsf M_2^{Q}(P)\otimes\mathrm I_2$ of the four matrices of the $2\times2$ table with the identity: $$ [L_{\tilde\Pi_+}]_{\mathrm Q}=\operatorname{diag}(1,1,0,0), \qquad [L_{\tilde\Pi_-}]_{\mathrm Q}=\operatorname{diag}(0,0,1,1), $$ $$ [L_{\vec\Upsilon_-}]_{\mathrm Q}= \begin{pmatrix} 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}, \qquad [L_{\vec\Upsilon_+}]_{\mathrm Q}= \begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 1&0&0&0\\ 0&1&0&0 \end{pmatrix}. $$ The right regular matrices are the same units on the other factor: $$ [R_{\tilde\Pi_+}]_{\mathrm Q}=\operatorname{diag}(1,0,1,0), \qquad [R_{\tilde\Pi_-}]_{\mathrm Q}=\operatorname{diag}(0,1,0,1), $$ $$ [R_{\vec\Upsilon_-}]_{\mathrm Q}= \begin{pmatrix} 0&0&0&0\\ 1&0&0&0\\ 0&0&0&0\\ 0&0&1&0 \end{pmatrix}, \qquad [R_{\vec\Upsilon_+}]_{\mathrm Q}= \begin{pmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0 \end{pmatrix}. $$ These eight matrices are the multiplication table of the four basis elements. The four left matrices reproduce the Peirce relations $$ [L_{\vec\Upsilon_+}][L_{\vec\Upsilon_-}]=[L_{\tilde\Pi_-}], \qquad [L_{\vec\Upsilon_-}][L_{\vec\Upsilon_+}]=[L_{\tilde\Pi_+}], \qquad [L_{\vec\Upsilon_\pm}]^2=0, \qquad [L_{\tilde\Pi_+}]+[L_{\tilde\Pi_-}]=\mathrm I_4, $$ and the four right matrices satisfy the same relations with the two factors in the reverse order, because $[R_P][R_{P'}]$ is the matrix of $P'P$.

The fixed basis in the standard eigenbasis. At $Q_x=Q_y=0$ with the standard root $i\vec e_z$, that is $B=-Q_z$, the change matrix of §The Fixed Change in the $4\times4$ Model is $$ \mathsf C_4^{(0)}= \frac12 \begin{pmatrix} 1&0&0&1\\ 0&i&i&0\\ 0&-1&1&0\\ i&0&0&-i \end{pmatrix}, \qquad (\mathsf C_4^{(0)})^{-1}= \begin{pmatrix} 1&0&0&-i\\ 0&-i&-1&0\\ 0&-i&1&0\\ 1&0&0&i \end{pmatrix}. $$ The columns of $(\mathsf C_4^{(0)})^{-1}$ are the coordinate columns of the fixed basis in the eigenbasis: $$ [e_t]=(1,0,0,1),\quad [\vec e_x]=(0,-i,-i,0),\quad [\vec e_y]=(0,-1,1,0),\quad [\vec e_z]=(-i,0,0,i). $$ Conjugating the fixed regular matrices gives the four matrices of the fixed basis in the eigenbasis: $$ (\mathsf C_4^{(0)})^{-1}\mathsf M_4^{L}(e_t)\mathsf C_4^{(0)}=\mathrm I_4, $$ $$ (\mathsf C_4^{(0)})^{-1}\mathsf M_4^{L}(\vec e_x)\mathsf C_4^{(0)}= \begin{pmatrix} 0&0&-i&0\\ 0&0&0&-i\\ -i&0&0&0\\ 0&-i&0&0 \end{pmatrix}, \qquad (\mathsf C_4^{(0)})^{-1}\mathsf M_4^{L}(\vec e_y)\mathsf C_4^{(0)}= \begin{pmatrix} 0&0&-1&0\\ 0&0&0&-1\\ 1&0&0&0\\ 0&1&0&0 \end{pmatrix}, $$ $$ (\mathsf C_4^{(0)})^{-1}\mathsf M_4^{L}(\vec e_z)\mathsf C_4^{(0)}= \begin{pmatrix} -i&0&0&0\\ 0&-i&0&0\\ 0&0&i&0\\ 0&0&0&i \end{pmatrix}. $$ In the $2\times2$ model the same four elements are $$ \mathsf M_2^{Q}(e_t)=\mathrm I_2,\qquad \mathsf M_2^{Q}(\vec e_x)= \begin{pmatrix}0&-i\\-i&0\end{pmatrix}, \qquad \mathsf M_2^{Q}(\vec e_y)= \begin{pmatrix}0&-1\\1&0\end{pmatrix}, \qquad \mathsf M_2^{Q}(\vec e_z)= \begin{pmatrix}-i&0\\0&i\end{pmatrix}, $$ and $[L_{P}]_{\mathrm Q}=\mathsf M_2^{Q}(P)\otimes\mathrm I_2$ returns the four $4\times4$ matrices above. In the standard case the eigenbasis model is the fixed model itself: the assignment of §The $2\times2$ Matrix Element Representation already sends $\tilde\Pi_{\pm,0}$ to $\operatorname{diag}(1,0)$ and $\operatorname{diag}(0,1)$, and the standard nilpotents to the two off-diagonal matrices with a single $1$, so the four matrices above are the four matrices displayed there. They satisfy $$ \mathsf M_2^{Q}(\vec e_x)^2=\mathsf M_2^{Q}(\vec e_y)^2=\mathsf M_2^{Q}(\vec e_z)^2=-\mathrm I_2, \qquad \mathsf M_2^{Q}(\vec e_x)\mathsf M_2^{Q}(\vec e_y)=\mathsf M_2^{Q}(\vec e_z), $$ which are the relations $\vec e_x^2=\vec e_y^2=\vec e_z^2=-e_t$ and $\vec e_x\vec e_y=\vec e_z$ of the fixed basis. Off the axis the identity $[L_P]_{\mathrm Q}=\mathsf M_2^{Q}(P)\otimes\mathrm I_2$ holds with $\mathsf M_2^{Q}(\vec e_\mu)$ the Peirce matrix of $\vec e_\mu$ in the general eigenbasis model, which is full.

The Diagonal Form

For $\tilde Q_{\mathrm{sp}}=\tilde Q$ the Peirce matrix is $M=\operatorname{diag}(\lambda_+,\lambda_-)$, and the two regular matrices become diagonal: $$ \boxed{\; [L_{\tilde Q}]_{\mathrm Q}=\operatorname{diag}(\lambda_+,\lambda_+,\lambda_-,\lambda_-), \qquad [R_{\tilde Q}]_{\mathrm Q}=\operatorname{diag}(\lambda_+,\lambda_-,\lambda_+,\lambda_-) \;} $$ These are the matrices of the eigenbasis of left multiplication and of right multiplication of Spectral Regularity and Degeneracy in the Biquaternion Algebra, §The Eigenbasis of Left Multiplication, read in the $4\times4$ representation. The eigenvalues of $L_{\tilde Q}$ are $\lambda_+,\lambda_+,\lambda_-,\lambda_-$; the eigenvalues of $R_{\tilde Q}$ are $\lambda_+,\lambda_-,\lambda_+,\lambda_-$. The two multiplicities are the dimensions of the two spinor spaces.

The trace and the determinant in the diagonal form. The trace of $L_{\tilde Q}$ is $2(\lambda_++\lambda_-)=4Q_t$, the trace of $R_{\tilde Q}$ is the same, the determinant of $L_{\tilde Q}$ is $\lambda_+^2\lambda_-^2=\langle\tilde Q\rangle_{\natural}^2$, and the determinant of $R_{\tilde Q}$ is the same product in the other order. The two matrices have the same characteristic polynomial, $$ (X-\lambda_+)^2(X-\lambda_-)^2, $$ which is the square of the characteristic polynomial of $\mathsf{M}_2(\tilde Q)$. The doubling is the doubling of the module: the regular representation is two copies of the $2\times2$ one.

The Fixed Change in the $4\times4$ Model

The eigenbasis order is reached from the fixed basis by the change-of-basis matrix whose columns are the coordinates of $\tilde\Pi_+,\vec\Upsilon_-,\vec\Upsilon_+,\tilde\Pi_-$ in the fixed basis. For the standard root $i\vec e_z$ and the standard nilpotents of the fixed model, this matrix is $$ \mathsf{C}_4^{(0)}= \begin{pmatrix} \tfrac12&0&0&\tfrac12\\ 0&\tfrac{i}{2}&\tfrac{i}{2}&0\\ 0&-\tfrac12&\tfrac12&0\\ \tfrac{i}{2}&0&0&-\tfrac{i}{2} \end{pmatrix}, \qquad (\mathsf{C}_4^{(0)})^{-1}\mathsf{M}_4^{L}(\tilde Q)\mathsf{C}_4^{(0)} =\operatorname{diag}(\lambda_+,\lambda_+,\lambda_-,\lambda_-) $$ when $\tilde Q$ commutes with $\vec e_z$ and the sign of $B$ is chosen so that the root of the eigenbasis is the standard root $i\vec e_z$; for $Q_x=Q_y=0$ this means $B=-Q_z$, so that $\lambda_\pm=Q_t\mp iQ_z$. The four columns are the coordinate columns of the four matrices $$ \begin{pmatrix}1&0\\0&0\end{pmatrix},\quad \begin{pmatrix}0&1\\0&0\end{pmatrix},\quad \begin{pmatrix}0&0\\1&0\end{pmatrix},\quad \begin{pmatrix}0&0\\0&1\end{pmatrix} $$ read back through the inverse of $\mathsf{M}_2^{Q}$. The matrix is fixed because the standard root and the standard nilpotents are fixed.

For a general $\tilde Q$ the change matrix $\mathsf{C}_4(\tilde Q)$ is not fixed: its columns are the coordinates of the eigenbasis, which depends on $\tilde Q$. It satisfies $$ \mathsf{C}_4(\tilde Q)^{-1}\mathsf{M}_4^{L}(\tilde Q)\mathsf{C}_4(\tilde Q) =\operatorname{diag}(\lambda_+,\lambda_+,\lambda_-,\lambda_-), $$ $$ \mathsf{C}_4(\tilde Q)^{-1}\mathsf{M}_4^{R}(\tilde Q)\mathsf{C}_4(\tilde Q) =\operatorname{diag}(\lambda_+,\lambda_-,\lambda_+,\lambda_-). $$ The two diagonalisations use the same change of basis, because the eigenbasis order diagonalises both the left and the right multiplication. This is the $4\times4$ statement that the eigenbasis is a basis of eigenvectors for both operators.

The change as a Kronecker product. In the matrix-unit order the change from the fixed model to the eigenbasis model is the conjugation $X\mapsto\mathsf{C}_Q^{-1}X\mathsf{C}_Q$ on $M_2(\mathbb{C})$. On the four coordinates, with the column-major order $$ \begin{pmatrix}1&0\\0&0\end{pmatrix},\quad \begin{pmatrix}0&0\\1&0\end{pmatrix},\quad \begin{pmatrix}0&1\\0&0\end{pmatrix},\quad \begin{pmatrix}0&0\\0&1\end{pmatrix}, $$ this conjugation is the Kronecker product $$ \mathsf{C}_Q^{\mathsf T}\otimes\mathsf{C}_Q^{-1}. $$ The fixed change $\mathsf{C}_4^{(0)}$ is the same construction for the standard root, composed with the fixed passage from the basis $e_t,\vec e_x,\vec e_y,\vec e_z$ to the matrix units. The composition is the reason the $4\times4$ change is not a single Kronecker product in the order of the fixed basis: the fixed basis is not the matrix-unit basis, and the permutation between them is the one displayed in $\mathsf{C}_4^{(0)}$.

Base Changes in the $4\times4$ Model

A change of basis $P\in GL_4(\mathbb{C})$ acts on the regular matrices by $$ \mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}})\longmapsto P^{-1}\mathsf{M}_4^{L}(\tilde Q_{\mathrm{sp}})P, \qquad \mathsf{M}_4^{R}(\tilde Q_{\mathrm{sp}})\longmapsto P^{-1}\mathsf{M}_4^{R}(\tilde Q_{\mathrm{sp}})P. $$ The trace, the determinant, the characteristic polynomial and the spectrum are unchanged. The eigenbasis change is the particular $P=\mathsf{C}_4(\tilde Q)$ that diagonalises both matrices. The basis is fixed by the Peirce conditions: the two idempotents $\tilde\Pi_\pm$ must remain idempotent, so their scales are $1$, and the two nilpotents satisfy $\vec\Upsilon_+\vec\Upsilon_-=\tilde\Pi_-$ and $\vec\Upsilon_-\vec\Upsilon_+=\tilde\Pi_+$, so their scales are reciprocal. The only remaining freedom is $$ \vec\Upsilon_+\longmapsto\mu\vec\Upsilon_+,\qquad \vec\Upsilon_-\longmapsto\mu^{-1}\vec\Upsilon_-,\qquad \mu\in\mathbb{C}^{\times}, $$ which leaves the diagonal form and the Peirce multiplication table unchanged. This is the $4\times4$ form of the relative scale of §The Freedom of the Change; the common scale of the $2\times2$ diagonaliser is the part that leaves the conjugation unchanged.

The two models compared. The $2\times2$ model and the $4\times4$ model carry the same element but different linear objects. The $2\times2$ matrix is the element; the $4\times4$ matrix is the operator of left or right multiplication by the element. The $2\times2$ model is irreducible, the $4\times4$ model is the direct sum of two copies of it, and the eigenvalues of the $4\times4$ matrix are the eigenvalues of the $2\times2$ matrix, each repeated twice. The $2\times2$ model has the two columns for the two spinor spaces; the $4\times4$ model has the two blocks for the two copies, and the eigenbasis order interleaves them.

Summary

The eigenbasis in the $2\times2$ model. The eigenprojections are the spectral projectors $\rho_\pm=\mathsf{M}_2(\tilde\Pi_\pm)$, the nilpotents are the two square-zero matrices $\mathsf{M}_2(\vec\Upsilon_\pm)$ in the two mixed corners, and the eigenbasis model $\mathsf{M}_2^{Q}$ sends them to the four matrix units. The element is diagonal in that model, $\mathsf{M}_2^{Q}(\tilde Q)=\operatorname{diag}(\lambda_+,\lambda_-)$, and a general element has a full $2\times2$ matrix.

The base change. The fixed model and the eigenbasis model are related by $$ \mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})=\mathsf{C}_Q^{-1}\mathsf{M}_2(\tilde Q_{\mathrm{sp}})\mathsf{C}_Q, $$ with $\mathsf{C}_Q$ the eigenvector matrix of §Base Changes. The change is unique up to the rescaling of its two columns, and the relative scale is the reciprocal rescaling of the nilpotents. The change is attached to the element, not to the algebra: two elements with different roots have different eigenbases, related by a further change of basis.

The $4\times4$ model. In the eigenbasis order the left and right regular matrices are $$ [L_{\tilde Q_{\mathrm{sp}}}]_{\mathrm Q}=M\otimes\mathrm I_2, \qquad [R_{\tilde Q_{\mathrm{sp}}}]_{\mathrm Q}=\mathrm I_2\otimes M^{\mathsf T}, $$ with $M=\mathsf{M}_2^{Q}(\tilde Q_{\mathrm{sp}})$; for $\tilde Q_{\mathrm{sp}}=\tilde Q$ they are diagonal, with eigenvalues $(\lambda_+,\lambda_+,\lambda_-,\lambda_-)$ and $(\lambda_+,\lambda_-,\lambda_+,\lambda_-)$. The $4\times4$ change of basis is the same eigenbasis change, read on the four-vector space.

The four basis elements. In the eigenbasis order the four basis elements of the eigenbasis have the four elementary regular matrices, $$ [L_{\tilde\Pi_+}]_{\mathrm Q}=\operatorname{diag}(1,1,0,0),\quad [L_{\tilde\Pi_-}]_{\mathrm Q}=\operatorname{diag}(0,0,1,1),\quad [L_{\vec\Upsilon_-}]_{\mathrm Q}=\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix},\quad [L_{\vec\Upsilon_+}]_{\mathrm Q}=\begin{pmatrix}0&0&0&0\\0&0&0&0\\1&0&0&0\\0&1&0&0\end{pmatrix}, $$ and the right matrices are the same units on the other factor. In the standard eigenbasis the fixed basis has the constant matrices $$ [L_{e_t}]_{\mathrm Q}=\mathrm I_4,\quad [L_{\vec e_x}]_{\mathrm Q}=\begin{pmatrix}0&0&-i&0\\0&0&0&-i\\-i&0&0&0\\0&-i&0&0\end{pmatrix},\quad [L_{\vec e_y}]_{\mathrm Q}=\begin{pmatrix}0&0&-1&0\\0&0&0&-1\\1&0&0&0\\0&1&0&0\end{pmatrix},\quad [L_{\vec e_z}]_{\mathrm Q}=\operatorname{diag}(-i,-i,i,i), $$ which are the conjugates of the fixed regular matrices by $\mathsf C_4^{(0)}$.

The limits. For $Q_x^2+Q_y^2+Q_z^2=0$ the construction fails: there is no spectral projector of rank one attached to $\tilde Q$, no eigenbasis attached to $\tilde Q$, no nilpotent normalisation and no diagonal form. The fixed model still exists, and the matrix of $\tilde Q$ is a Jordan block when $Q_t\neq0$ and a nonzero nilpotent when $Q_t=0$. The fixed $4\times4$ matrices still exist and still satisfy their multiplicativity; they are not diagonalisable. The scalar case $\vec Q=0$ has $\mathsf{M}_2(\tilde Q)=Q_t\mathrm I_2$, every basis is an eigenbasis, and the construction from the root is absent although an eigenbasis exists.

Summary of Notation

Further Reading

• Richard S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88 (Springer, 1982), for idempotents, Peirce decompositions and the matrix units of a Peirce decomposition.
• Fuzhen Zhang, Matrix Theory: Basic Results and Techniques, 2nd edition (Springer, 2011), for the change of basis, the spectral projectors and the Kronecker product.