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First divide each of the first two equations by two. ThenThe equal-component and opposite-component directions diagonalize the upper block, while the third coordinate is independent. Direct multiplication gives the eigenvalues and unit eigenvectorsThey form an orthonormal eigenbasis because $K$ is a real symmetric matrix.

A left eigenvector satisfies $(M-\lambda I{)}^T\mathbf{x}=0$. Here $M$ is a symmetric matrix, so its left and right eigenspaces have the same components. The characteristic polynomial factors asThe eigenvalues and all corresponding nonzero row eigenvectors areFor example, in the $xy$ block, equal components give the eigenvalue $3$ and opposite components give $-1$, while the $z$ coordinate is unaffected.

For any real square matrix $N$, the symmetric and antisymmetric parts of a matrix give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a symmetric matrix and an antisymmetric matrix. Here it givesTransposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original matrix. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.

The two orthogonal directions $(1,1{)}^T$ and $(1,-1{)}^T$ are preserved by this symmetric matrix. Direct multiplication, using the hyperbolic function identities $\cosh u\pm \sinh u=e^{\pm u}$, gives the eigenvalue–eigenvector pairsEach eigenvector can be multiplied by any nonzero scalar. For $u\ne 0$ these give the two one-dimensional eigenspaces. When $u=0$, the matrix is the identity: the only eigenvalue is $1$, with multiplicity two, and every nonzero vector is an eigenvector.

The first coordinate direction is already an eigenvector of $A$ with eigenvalue $4$. On the last two coordinates, the matrix is $\left(\begin{array}{cc}2& 4\\ 4& 2\end{array}\right)$, with characteristic polynomialThus the three eigenvalues are $4,6,-2$. For $\lambda =6$, the last two components are equal; for $\lambda =-2$, they are opposite. Normalising by the Euclidean norm gives the complete listEach sign-reversed vector is an equally valid choice. The three displayed eigenvectors are mutually orthogonal, as expected for a real symmetric matrix with distinct eigenvalues.

The symmetric and antisymmetric parts of a matrix areIndeed $S+A=M$, while the transpose identities $(M^T{)}^T=M$ give $S^T=S$ and $A^T=-A$. Thus $S$ is a symmetric matrix and $A$ is an antisymmetric matrix. The decomposition is also unique: a matrix that is both symmetric and antisymmetric must equal its own negative and hence vanish.

Tensor transformation. Under an orthogonal transformation $R$, the magnetic moment and magnetic field intensity have transformed components $M^{\prime }=RM$, $H^{\prime }=RH$ for a proper rotation. The physical relation $M=\chi H$ must retain its form, giving $RM={\chi }^{\prime }RH$, henceThis is precisely the law for a Cartesian second-rank tensor. Both magnetic quantities are axial vectors; under reflections they each acquire the same determinant sign, which cancels from the transformation of the magnetic susceptibility tensor.

Parallel directions. The condition $M=\lambda H$ is the eigenvalue equation $\chi H=\lambda H$. For the matrix in the paper, expansion of $\det (\chi -\lambda I)$ givesSolving the three resulting homogeneous systems givesThe three directions are mutually orthogonal. For the zero eigenvalue, the induced magnetic moment is zero: the proportionality equation still holds with factor zero, though there is then no nonzero moment direction to assign.

Energy extrema. The stipulated energy is $\mathcal{E}=H^T\chi H$. At fixed $H^{T}H=H_0^2>0$, a Lagrange multiplier calculation givesbecause $\chi$ is a real symmetric matrix. Thus all stationary directions are the eigenvectors, for which $M=\lambda H$. Equivalently write $H=q_0{e}_0+q_1{e}_1+q_6{e}_6$; thenThe Rayleigh quotient immediately givesThe intermediate direction $e_1$ is stationary with $\mathcal{E}=H_0^2$ but is not an extremum on the sphere. At the minimum $M=0$, so the precise extremum condition is proportionality, including a vanishing proportionality factor.

Constant-energy surface. In the original field components,For a fixed positive value $E$, $q_0$ is unrestricted and the cross-section satisfies $q_1^2/E+q_6^2/(E/6)=1$. The surface is therefore an elliptic cylinder, with axis along $(1,-1,2)$ and transverse semiaxes $\sqrt{E}$ along $e_1$ and $\sqrt{E/6}$ along $e_6$. This is an example of level sets of a positive semidefinite quadratic form. At $E=0$ the surface degenerates to the axis line; no negative-energy level exists.

Nilpotence and eigenvalues. If $Av=\lambda v$ with $v\ne 0$, then $A^{n}v={\lambda }^{n}v$. Thus $A^n=0$ forces $\lambda =0$ for every eigenvalue. Conversely, if all $n$ complex eigenvalues, counted with algebraic multiplicity, vanish, the monic characteristic polynomial is ${\chi }_A(t)=t^n$. The Cayley-Hamilton theorem then gives $A^n=0$. HenceThis is the criterion for a nilpotent matrix.

Symmetric case. Here a symmetric matrix is understood to be real. The spectral theorem gives $A=QD{Q}^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\ge n$, so $A^{2^r}=0$. For every real vector $x$,It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the quadratic form $x^T{M}^{T}Mx$. The restriction to real symmetric, or complex Hermitian matrices, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.

The parameters and powers. The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha ,\beta$. The sum of its principal two-by-two determinants, each a minor of a matrix, isTogether with $\mathrm{tr}T=2+\beta$, this givesFor every eigenvalue to vanish both remaining coefficients must be zero. ThereforeDirect multiplication at these values givesIndeed the first row of $T^2$ multiplied by $T$ is $3{\mathrm{row}}_1(T)+3{\mathrm{row}}_2(T)-6{\mathrm{row}}_3(T)=0$, and the second row is its negative.

Geometry of repeated application. For any parameters the image of a linear map lies in the span of $a=(1,1,1{)}^T$ and $b=(1,\alpha ,\beta {)}^T$, since $Tx=(x_1+x_2)a+x_{3}b$. Thus it lies in a fixed plane through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta -\alpha ,1-\beta ,\alpha -1)$; if they coincide, the image is a line and is still contained in a plane.

For $\alpha =-5,\beta =-2$, put $s=x_1+x_2$ and $t=x_3$. ThenThe columns span this entire plane, so $T$ has rank two. The next two steps areThus the image of a linear map contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the kernel of $T$.

The spectral theorem gives a real orthonormal eigenbasis $v_1,\dots ,v_n$ for a real symmetric matrix $A$. Distinct eigenspaces are orthogonal; within a repeated eigenspace, choose an orthonormal basis, for example by Gram-Schmidt orthogonalization. Put these eigenvectors in the columns of $T$. Then $T^{-1}=T^T$ andFor a change of variables $y=Sx$, choose $\boxed{S=T^T}$. Since $x=Ty$, the quadratic form becomes $x^{T}Ax={\sum }_i{\lambda }_i{y}_i^2$; the ${\lambda }_i$ are precisely the eigenvalues of $A$.

In the numerical case, the off-diagonal entries are half the cross-term coefficients, soA direct eigenvector calculation gives the unit eigenvectors $(1,1,0)/\sqrt{2}$, $(1,-1,1)/\sqrt{3}$, $(1,-1,-2)/\sqrt{6}$, with eigenvalues two, three and minus six. They are pairwise orthogonal. Therefore the required orthogonal diagonalization isand the transformed quadratic form is $\boxed{2y_1^2+3y_2^2-6y_3^2}$.

Since $A$ and $B$ are real symmetric matrices, $U^{†}=A-iB$. Expanding the unitarity condition givesEquality of the entrywise real and imaginary parts yieldsThese are the identities for the real and imaginary parts of a symmetric unitary matrix.

Since $M$ is real, $(M^2{)}^T=(M^T{)}^2=(-M{)}^2=M^2$. For every real vector $x$,Thus $M^2$ is a real symmetric matrix and is negative semidefinite. When $n$ is odd, part (i) supplies a nonzero $x$ in the kernel of $M$, and then $M^{2}x=0$. Hence zero is an eigenvalue of $M^2$. In fact $\ker M^2=\ker M$, as the displayed norm identity shows.

The spectral theorem for a real symmetric matrix gives real eigenvalues and an orthonormal eigenbasis. Compute the roots of $\det (A-\lambda I)=0$, find a basis for each eigenspace, and orthonormalize within any repeated-eigenvalue eigenspace. Eigenvectors belonging to distinct eigenvalues are already orthogonal: if $Av=\lambda v$ and $Aw=\mu w$, symmetry gives $(\lambda -\mu )v^{T}w=0$.

Putting the normalized basis vectors in the columns of an orthogonal matrix $Q$ yieldsFor repeated eigenvalues, the choice of axes inside their eigenspace is not unique.

Since $a^2=I$, a representation satisfies $D(a{)}^2=I_2$. Orthogonality gives $D(a{)}^{-1}=D(a{)}^T$, while the involution relation gives $D(a{)}^{-1}=D(a)$. Hence $D(a)$ is a symmetric matrix, so $\gamma =\delta$.

Its entries obeyIf $\gamma \ne 0$, then $\beta =-\alpha$; putting $\alpha =\cos \theta$ and $\gamma =\sin \theta$ gives a reflection. If $\gamma =0$, the diagonal entries are independently $\pm 1$: the opposite-sign cases are already included among these reflections, while equal signs give $I_2$ or $-I_2$. Faithfulness excludes $I_2$, since $a\ne I$. Therefore the orthogonal involutions of the plane giveThe first is rotation by $\pi$. The second is reflection in the line making angle $\theta /2$ with the first axis: the unit vector along that line is fixed, and its perpendicular is negated. These are all the nonidentity orthogonal transformations of order two in the plane.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.

Every row sums to six, so $\mathbf{e}=(1,1,1{)}^T$ is an eigenvector with eigenvalue six. Since $M$ is a real symmetric matrix, its other two eigenvalues, say ${\lambda }_2,{\lambda }_3$, are real. The trace givesAlso, for a real symmetric matrix, $\mathrm{tr}(M^2)$ is the sum of the squares of all its entries. HereThus ${\lambda }_2^2+{\lambda }_3^2=42-36=6$. Together with their zero sum, this implies

The entries of $M^{T}M$ are real, and the transpose rule gives $(M^{T}M{)}^T=M^{T}M$. Thus it is a real symmetric matrix, hence also a Hermitian matrix.

For any complex vector $v$,In particular, if $M^{T}Mv=\lambda v$ with $v\ne 0$,Thus $M^{T}M$ is positive semidefinite. Its zero eigenspace is precisely the kernel of $M$, since $\Vert Mv{\Vert }^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative eigenvalues.

A real symmetric matrix has a real orthonormal eigenbasis. Put those eigenvectors into an orthogonal matrix $Q$, so $Q^{T}AQ=\mathrm{diag}(a_1,\dots ,a_n)$. With the change of coordinates $x=Q{x}^{\prime }$, or $x^{\prime }=Q^{T}x$,Thus the coefficients are the eigenvalues and the new coordinate directions are the eigenvectors. This orthogonal diagonalization preserves Euclidean lengths and angles while removing cross terms from the quadratic form.

A real matrix is a Hermitian matrix exactly when it is a symmetric matrix. Since $R_1$ is orthogonal, this is equivalent to $R_1=R_1^{-1}$, or $R_1^2=I$. A generic rotation does not satisfy this; for example $\theta =0$, $\varphi =\pi /2$ gives a nonsymmetric quarter-turn. Thus $\boxed{R_1\text{ is not Hermitian in general.}}$

The exceptions are the identity and half-turns, as characterized by real rotation matrices that are Hermitian. For this product, a half-turn occurs when either $\theta$ or $\varphi$ is an odd multiple of $\pi$; the identity occurs when both are even multiples. For example, the half-turn condition follows from $\mathrm{tr}{R}_1=(1+\cos \theta )(1+\cos \varphi )-1=-1$.

For the two sine modes, orthogonality givesThe diagonal potential integrals vanish, while the mixed one isThus ${\int }_0^1\cos (\pi x){\tilde{\psi }}^{2}dx=\alpha /2$, giving the two-sine variational bound for a Dirichlet Mathieu equation:Its derivative vanishes when $1+6{\pi }^2\alpha -{\alpha }^2=0$. Hence the two stationary parameters are $3{\pi }^2\pm \sqrt{9{\pi }^4+1}$. To identify the minimum, use the normalized sine basis: the restricted self-adjoint operator has real symmetric matrixIts smaller eigenvalue is the minimum of the Rayleigh quotient, attained at the negative stationary parameter. ThereforeThe estimate improves on the single-sine value ${\pi }^2$. The Rayleigh-Ritz variational principle remains valid here even though $q(x)=\cos (\pi x)$ changes sign: on this finite interval the Sturm-Liouville operator is self-adjoint and bounded below, which is sufficient for the upper bound.

Since $M^T=-M$, $(M^2{)}^T=(M^T{)}^2=M^2$, so $M^2$ is a real symmetric matrix. Moreover,proving that it is a negative semidefinite matrix.

For odd $n$, determinant invariance under transpose givesso $\det M=0$. Hence $M$ and $M^2$ have a zero eigenvalue. Finally, if $M^{2}w=0$, then $0=w^T{M}^{2}w=-\Vert Mw{\Vert }^2$, so $Mw=0$. Thuswhere equality of these kernels includes the reverse implication, which is immediate.