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The two-dimensional Brouwer fixed-point theorem states that every continuous map from a closed filled triangle to itself has a fixed point. Use the triangle $\Delta =\{x\in {\mathbb{R}}^3:x_i\ge 0,{\sum }_i{x}_i=1\}$, which lies in a two-dimensional affine plane. Strict positivity of the matrix entries ensures $(Ax{)}_i>0$ for every $x\in \Delta$. Thereforeis a continuous self-map of $\Delta$, in fact into its interior. Its fixed point $x$ satisfies $Ax=\lambda x$, where $\lambda ={\sum }_i(Ax{)}_i>0$. Since $x=T(x)$ lies in the interior, $A$ has an eigenvector all of whose entries are strictly positive. This is positive matrix eigenvector from simplex normalization; it uses strict positivity, not merely nonnegative entries.

If $e_1,\dots ,e_n$ are orthonormal, any relation ${\sum }_i{c}_i{e}_i=0$ gives $c_j=0$ on taking the dot product with $e_j$. Thus the vectors are linearly independent. A linearly independent set of $n$ vectors in the $n$-dimensional vector space ${\mathbb{R}}^n$ is a basis, proving the claim.

Expand both $x$ and $f$ in this orthonormal basis, writing $f_i=e_i\cdot f$. Since $A{e}_i={\lambda }_i{e}_i$, the linear equation becomesComparing the unique basis coefficients gives, when $\mu$ is not an eigenvalue,If $\mu ={\lambda }_1$, the equation is solvable exactly when $e_i\cdot f=0$ for every $i$ with ${\lambda }_i=\mu$. For those indices $a_i$ is arbitrary; for all other indices the displayed formula still applies. Thus repeated eigenvalues require orthogonality to the entire corresponding eigenspace, not merely to one chosen eigenvector. This is the finite-dimensional Fredholm solvability condition for a self-adjoint operator.

For the specific real symmetric matrix, chooseTheir dot products with $f$ are $-1/\sqrt{2}$, $3/\sqrt{2}$ and $3$. At $\mu =2$ this givesOne can check $(A-2I)x=f$ directly. At $\mu =1$, the first coefficient equation would require $0\cdot a_1=-1/\sqrt{2}$, which is impossible. Hence there is no solution when $\mu =1$; the nonzero projection onto its eigenspace is the obstruction.

Let $H$ be a Hermitian matrix and $He=\lambda e$ with $e\ne 0$. The inner product $e^{†}He$ is real because its conjugate equals $e^{†}{H}^{†}e=e^{†}He$. But it also equals $\lambda e^{†}e$, with $e^{†}e>0$. Thus every eigenvalue is real. For two eigenvectors,When ${\lambda }_i\ne {\lambda }_j$, this proves $e_i^{†}{e}_j=0$, so the eigenvectors are orthogonal.

For a real antisymmetric matrix, $(iA{)}^{†}=-i{A}^T=iA$, so $iA$ is a Hermitian matrix. Its real eigenvalues imply that the eigenvalues of $A$ are purely imaginary. To ensure a nonzero one without assuming it, writeExpansion of the characteristic polynomial gives $\det (tI-A)=t(t^2+|w{|}^2)$. Therefore the nonzero eigenvalue $-i\theta$ exists with $\theta =|w|>0$, and a nonzero complex eigenvector $e_1=u+iv$ satisfiesSeparating real and imaginary parts givesNeither real vector can vanish, since either vanishing would force the other to vanish. Antisymmetry yields $0=u\cdot Au=\theta u\cdot v$ andThus $u,v$ are orthogonal and have equal length.

In odd dimension $\det A=\det A^T=\det (-A)=-\det A$, so $\det A=0$. Its real homogeneous system therefore has a nonzero real eigenvector $e_3$ with $A{e}_3=0$. It is perpendicular to $u,v$, because $e_3\cdot Au=-(A{e}_3)\cdot u=0$ and similarly for $Av$. Normalize it and choose its sign so $(u/|u|,v/|v|,e_3)$ is positively oriented; this is an orthonormal basis.

The matrix exponential series converges absolutely, allowing even and odd terms to be grouped. Since $A^{2}u=-{\theta }^{2}u$ and $A^{2}v=-{\theta }^{2}v$, it acts asConsequently the skew-symmetric exponential as an axial rotation has, in the displayed orthonormal basis, the matrixIt is orthogonal with determinant one, fixes its axis and rotates the perpendicular plane. Hence $R=e^A$ is a rotation matrix, with angle $\theta$ about the chosen oriented axis.

The real orthogonal group and special orthogonal group are, under matrix multiplication,An orthogonal matrix preserves Euclidean inner products; taking determinants of $Q^{T}Q=I$ shows that its determinant is $1$ or $-1$. Its inverse is $Q^T$, so $Q{Q}^T=I$ as well.

For $Q\in SO(3)$, use $Q-I=Q(I-Q^T)$. The determinant then satisfiesIt follows that $\det (Q-I)=0$. Thus the real matrix $Q-I$ has a nonzero vector in its kernel of a linear map, yieldingThis is the three-dimensional case of odd-dimensional special orthogonal transformation has a fixed vector; geometrically it supplies the fixed rotation axis, unless the transformation is the identity and every direction is fixed.

The claim is false for $O(3)$. The matrix $Q=-I_3$ is orthogonal with determinant $-1$ and has only the eigenvalue $-1$. It therefore has no nonzero eigenvector with eigenvalue $1$.

In the position representation, the orbital angular momentum operator isOn a common smooth domain use $[x_a,p_b]=i\hslash {\delta }_{ab}$, $[x_a,x_b]=[p_a,p_b]=0$. Expanding the product commutator givesContracting with the two alternating tensors gives the orbital angular momentum commutation relations:For $L_{\pm }=L_1\pm i{L}_2$, these imply $[L_3,L_{\pm }]=\pm \hslash L_{\pm }$. Therefore $L_3\psi =0$ yieldsThusThe nonzero qualification is necessary: a rotationally invariant wavefunction is annihilated by all three components, and the zero vector is not an eigenvector.

A Hermitian matrix satisfies $A=A^{†}$, where $†$ means conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda \Vert v{\Vert }^2$ is real, since it equals its complex conjugate $v^{†}{A}^{†}v$. Thus every eigenvalue is real. To prove the finite-dimensional spectral theorem, choose a unit eigenvector $v$; one exists because the complex characteristic polynomial has a root. Its orthogonal complement is invariant, since for $w\perp v$,The restriction to that complement is again Hermitian. Induction on dimension supplies an orthonormal basis of eigenvectors, so $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)U^{†}$ for a unitary matrix $U$.

In this basis, $x^{†}Ax={\sum }_i{\lambda }_i|(U^{†}x{)}_i{|}^2$. Therefore $\boxed{A>0⟺{\lambda }_i>0\text{ for every }i}$. In that case the square root of a matrixis Hermitian, positive definite and satisfies $C^2=A$. For uniqueness, any positive-definite Hermitian $D$ satisfying $D^2=A$ commutes with $A$ and hence preserves each eigenspace of $A$. Its restriction to the $\lambda$-eigenspace is Hermitian, and its positive eigenvalues all solve ${\mu }^2=\lambda$. Thus that restriction is $\sqrt{\lambda }I$, forcing $D=C$. This is the principal square root of a positive semidefinite matrix, here strictly positive definite.

Now let $D=\sqrt{B}$ and $X=C-D$. Expanding without assuming that $C$ and $D$ commute givesThe Hermitian matrix $X$ has $X^2\ge 0$ because $v^{†}{X}^{2}v=\Vert Xv{\Vert }^2$, whereas $A-B>0$. Hence $CX+XC>0$. If $Xv=\mu v$, $v\ne 0$, thenSince $C>0$, every real eigenvalue $\mu$ of $X$ is positive. The positive-eigenvalue criterion now gives the strict finite-dimensional operator monotonicity of the square root:

A Jordan normal form of a complex $n\times n$ matrix $A$ is a similarity transformation $P^{-1}AP=J$ in which $J$ is a direct sum of Jordan blocksFor each eigenvalue, the multiset of block sizes is uniquely determined by $A$; only the order of the blocks is arbitrary. The change-of-basis matrix $P$ is generally not unique.

For the given matrix, expansion of the characteristic polynomial givesThe upper-left $2\times 2$ minor of $A-I$ is $-5$, so its rank is $2$ and the eigenspace for $1$ has dimension $1$. Thus the two occurrences of the eigenvalue $1$ belong to a single size-$2$ Jordan block. More explicitly, the Jordan chain and the remaining eigenvector can be chosen asConsequently, with $P=(v_1{v}_2{v}_3)$, for which $\det P=1$, the answer is

Use complex inner products linear in the first argument throughout. The spectral theorem for compact self-adjoint operators states that, for a compact operator $T=T^{\ast }$ on a Hilbert space, its nonzero spectral values are real eigenvalues, each of finite multiplicity, with no possible accumulation point except zero. Moreover,where $P_{\lambda }$ are the mutually orthogonal eigenspace projections. The operator series converges in operator norm when the nonzero eigenvalues are listed in decreasing absolute value. The sum is at most countable, even if the kernel is nonseparable. Zero need not be an eigenvalue.

First prove that a nonzero compact self-adjoint operator has an eigenvalue of magnitude its norm, without assuming a spectral decomposition. Put $r=\Vert T\Vert >0$ and choose unit vectors ${\xi }_n$ with $\Vert T{\xi }_n\Vert \to r$. Since $\Vert T^2\xi \Vert \le r\Vert T\xi \Vert$, expansion givesCompactness of $T^2$ supplies a subsequence for which $T^2{\xi }_n$ converges; the displayed estimate then forces ${\xi }_n$ itself to converge to a unit vector $\xi$ satisfying $T^2\xi =r^2\xi$. The vectors $\xi +T\xi /r$ and $\xi -T\xi /r$ are eigenvectors for $r$ and $-r$, respectively, whenever nonzero, and they cannot both vanish. This proves the claim.

Self-adjointness gives reality of every eigenvalue and orthogonality of eigenspaces for distinct eigenvalues. An eigenspace for $\lambda \ne 0$ cannot contain an infinite orthonormal sequence: its images under $T$ would have pairwise distance $\sqrt{2}|\lambda |$, contradicting compactness. The same argument proves that there are only finitely many mutually orthogonal eigenvectors with eigenvalues of magnitude at least any given $\epsilon >0$. Hence all nonzero eigenspaces together have a countable orthonormal basis, and their eigenvalues tend to zero if there are infinitely many.

Let $K$ be the orthogonal complement of the closed span of all these eigenspaces. It is invariant under $T$. If $T{|}_K$ were nonzero, the norm-attaining eigenvalue argument applied to that compact self-adjoint restriction would produce another nonzero eigenvector, a contradiction. Thus $K\subseteq \ker T$, and the converse inclusion follows from orthogonality. This establishes the asserted decomposition. On its orthogonal summands the tail of the operator series has norm equal to the supremum of the omitted eigenvalue magnitudes, proving norm convergence.

Finally, a nonzero complex number not among these eigenvalues has positive distance from their set and from zero. Inverting $T-zI$ separately on the eigenspaces and kernel therefore gives a bounded inverse. This proves that no other nonzero spectral values occur and completes the theorem.

The Schmidt decomposition theorem says that a normalized vector in the tensor product of two finite-dimensional Hilbert spaces can be writtenwhere both families are orthonormal. The numbers $\sqrt{{\lambda }_j}$ are its Schmidt coefficients. The Schmidt rank $r$ and coefficients, including multiplicities, are uniquely determined by the state; the local vectors may be changed within degenerate eigenspaces and by compensating phases. Zero coefficients may be appended to make $n_1$ terms, since $n_2\ge n_1$.

For a proof, form the reduced density matrix ${\rho }_1={\mathrm{Tr}}_2|\psi ⟩⟨\psi |$. It is a Hermitian matrix that is positive semidefinite, with trace one. The spectral theorem supplies an orthonormal basis $u_1,\dots ,u_{n_1}$ of eigenvectors with nonnegative eigenvalues ${\lambda }_j$. Define the unnormalized relative vectors in the second Hilbert space byCompleteness of the first orthonormal basis gives $|\psi ⟩={\sum }_j|u_j⟩|w_j⟩$. In any orthonormal basis $e_b$ of the second factor, the definition of the partial trace givesThus a zero eigenvalue gives $w_j=0$, and for each positive eigenvalue the vector $v_j=w_j/\sqrt{{\lambda }_j}$ has unit norm. Distinct such vectors are orthogonal by the same identity. Substituting them gives the required Schmidt decomposition, with at most $n_1$ nonzero terms.

Taking the other partial trace now givesTherefore the two reduced density matrices have identical nonzero eigenvalues, namely the squared Schmidt coefficients. This also proves uniqueness of the coefficient multiset and shows that the pure state is a product state exactly when $r=1$.

For real nonnegative envelopes, put $\Omega =\sqrt{{\Omega }_1^2+{\Omega }_2^2}$ and choose a continuous $\theta =atan2({\Omega }_1,{\Omega }_2)$. The state $|D⟩=\cos \theta |1⟩-\sin \theta |3⟩$ is dark because its two amplitudes feeding $|2⟩$ cancel: ${\Omega }_1\cos \theta -{\Omega }_2\sin \theta =0$. The orthogonal bright combination $|B⟩=\sin \theta |1⟩+\cos \theta |3⟩$ couples to $|2⟩$ with strength $\Omega$. Thus the other eigenstates are $(|B⟩\pm |2⟩)/\sqrt{2}$ with energies $\pm \Omega$.

Start with ${\Omega }_1=0$ and ${\Omega }_2\ne 0$, so $\theta =0$ and $|D⟩=|1⟩$. Slowly increase ${\Omega }_1/{\Omega }_2$ to infinity while maintaining a nonzero gap; end with ${\Omega }_2=0$ and ${\Omega }_1\ne 0$, so $|D⟩=-|3⟩$. This target-to-intermediate pulse first is the counterintuitive ordering of stimulated Raman adiabatic passage. Its quantitative slowness condition isIt follows from $|⟨\pm |\dot{D}⟩|=|\dot{\theta }|/\sqrt{2}$ and the gap $\Omega$. Avoid switching both envelopes off while $\theta$ is changing; switch off together only after the transfer angle has become stationary. The dark eigenvalue is zero, so there is no dynamical phase; with this real eigenvector $⟨D|\dot{D}⟩=0$ as well. The ideal adiabatic quantum control therefore transfers $|1⟩$ to $-|3⟩$, physically the same target ray.

Write $u=u_x$, $w=u_y$ and $B=B_y$. The isothermal equation of state replaces the adiabatic thermal equation by $p=c_s^2\rho$. Since the fields depend only on $x,t$, the solenoidal constraint gives ${\partial }_x{B}_x=0$ and the $x$ component of the ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_x=0$. Hence $B_x$ is constant in both space and time.

The continuity, two momentum and transverse induction equations areHere $B$ is the transverse magnetic component and the constant $B_x$ is the longitudinal component. The longitudinal magnetic force is the gradient of transverse magnetic pressure; the transverse force is magnetic tension. In primitive variables the isothermal coplanar magnetohydrodynamic characteristic matrix isThe imposed $z$ components stay zero because their corresponding equations have zero right-hand sides under this reduction.

Let $c=v-u$ be a characteristic speed relative to the fluid and define the Alfvén velocity ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$. To derive the characteristic polynomial, an eigenvector with components $(\delta \rho ,\delta u,\delta w,\delta B)$ obeysEliminating $\delta u$ and $\delta w$ on a nondegenerate branch givesTheir consistency yields $(c^2-c_s^2)(c^2-v_{ax}^2)=v_{ay}^2{c}^2$. Equivalently, direct expansion of $\det (\mathbf{A}-vI)$ gives the same polynomial without dividing by $c$ or a possibly vanishing field component:ThusThe discriminant is $(c_s^2-v_a^2{)}^2+4c_s^2{v}_{ay}^2\ge 0$, and both squared speeds are nonnegative. These are the fast magnetosonic wave and slow magnetosonic wave families, transported with the mean flow. The equations are hyperbolic; on a generic state the four distinct real eigenvalues give four characteristic directions. Special field alignments can make speeds coincide or the slow speed vanish. There is no separate entropy characteristic because the thermal model is isothermal, and the independent out-of-plane Alfvén wave polarization is excluded by $u_z=B_z=0$.

To construct a simple wave in magnetohydrodynamics, take $\mathbf{U}=\mathbf{U}(a)$ along a state-space curve whose tangent is a right eigenvector of one selected branch:Substitution into the partial differential equations then gives the scalar transport equationSuch curves exist locally by integrating the eigenvector field away from degeneracies. For example, parameterizing by $a=\rho$ where its eigenvector density component is nonzero gives the explicit state-curve equationswith signed $c$ equal to one of the four relative wave speeds. These formulas are used only where their denominators are nonzero.

A particularly transparent nonlinear example is an isothermal perpendicular magnetosonic simple wave. Set the allowed constant $B_x=0$, choose $B=K\rho$ and keep $w$ constant. For the fast branch,These relations satisfy the eigenvector equations and therefore all four original equations wheneverThe amplitude-dependent characteristic speed has derivativeThis already proves the existence of genuinely nonlinear waves in the stated system.

More generally, the method of characteristics applied to initial data $a(x,0)=a_0(x)$ givesIf the initial speed decreases towards increasing $x$ somewhere, the denominator reaches zero in finite time. When the negative minimum is attained, the first steepening time isFaster trailing states catch slower leading states, giving shock formation by characteristic intersection. For the right-moving perpendicular fast example, a negative density gradient provides such a compression. Beyond $t_{\ast }$ a smooth simple wave is no longer sufficient and an appropriate shock solution is required. Rarefactive profiles instead spread. A linearly degenerate characteristic field has $v^{\prime }=0$ along its state curve and avoids this mechanism; the conclusion is that compressive nonlinear magnetosonic simple waves steepen, not that every possible MHD waveform must do so.