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Interpret the norms as Euclidean norms and set $B=A^{\mathsf{T}}A$. This is a real symmetric matrix with ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}=\Vert A\mathbf{x}{\Vert }^2\ge 0$, hence a positive semidefinite matrix. Squaring the nonnegative objective does not change its argmin. Let ${\lambda }_{\min }$ be the smallest eigenvalue of $B$ and $E_{\min }$ its eigenspace.

First minimize on the unit sphere. The Lagrange multiplier function $L(\mathbf{x},\lambda )={\mathbf{x}}^{\mathsf{T}}B\mathbf{x}-\lambda ({\mathbf{x}}^{\mathsf{T}}\mathbf{x}-1)$ has stationarity condition $B\mathbf{x}=\lambda \mathbf{x}$. Multiplication by ${\mathbf{x}}^{\mathsf{T}}$ shows that the squared objective at such a point is $\lambda$. The spectral theorem for real symmetric matrices also gives, for every unit vector, ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}\ge {\lambda }_{\min }$, with equality exactly in $E_{\min }$. This proves global minimality, rather than merely finding stationary candidates.

If ${\lambda }_{\min }>0$, every point with $\Vert \mathbf{x}\Vert >1$ has a strictly smaller objective after rescaling to unit length. ThereforeIf ${\lambda }_{\min }=0$, the kernel of a linear map satisfies $\ker B=\ker A$, since $B\mathbf{x}=0$ implies $\Vert A\mathbf{x}{\Vert }^2=0$, and the converse is immediate. The minimum is then zero, attained precisely atIn this case minimizers need not have unit length. This distinction is essential for a rank-deficient $A$, including any real $m\times n$ matrix with $m<n$. The result is the minimum image norm outside a Euclidean unit ball, governed by the least singular value.

For $\dot{A}=D^{2}f(t)B$ and $\dot{B}=iA$, with $f$ alternating between $+1$ and $-1$ for intervals of length $T$, the monodromy matrix of a periodic linear system has determinant one and trace $\cosh (\sqrt{2}DT)+\cos (\sqrt{2}DT)>2$. Its reciprocal Floquet multipliers give positive dominant Floquet growth rate for $D,T>0$, despite the zero mean of $f$. An exceptional initial state in the stable eigenspace decays.

For a nontrivial finite-dimensional anti-Hermitian Lie algebra representation of a compact Lie algebra that is simple, its Trace form of a Lie algebra representation is a positive multiple of the Killing form. Define $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance makes $T$ commute with the Adjoint representation of a Lie algebra. Symmetry makes $T$ self-adjoint for $-\kappa$, so its eigenspaces are ideals. Simplicity implies $T=\mu I$. The kernel of the representation is an ideal and hence vanishes; therefore $H(X,X)=-\mathrm{Tr}(d(X{)}^{†}d(X))<0$ for nonzero $X$, proving $\mu >0$.

// Target: lie-theory.bigb

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

For a unit vector $u$ spanning the target eigenspace and a normalized iterate $x_k$, defineThis is the norm of the component in the orthogonal complement of $u$, equivalently the sine of the acute angle to its line. It ignores the arbitrary sign or complex unit factor of an eigenvector. Under power iteration convergence for a diagonalizable matrix, ${\epsilon }_k=O({\gamma }^k)$ in Big O notation. The orthogonal projection defines how we measure direction; it does not assume that the eigenbasis of $A$ is orthogonal. In computation, the projection form avoids subtracting nearly equal numbers inside the square root.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. For a self-adjoint operator with a unique dominant eigenvalue separated in modulus from the others, a starting vector that is nonorthogonal to its dominant eigenvector gives convergence of the eigendirection at the spectral-ratio rate. For a general diagonalizable matrix, the condition is a nonzero coefficient in the dominant component of its eigenbasis, detected by the corresponding left eigenvector; an ordinary dot product with the right eigenvector does not generally detect that coefficient.

The power method is power iteration; its convergence depends on the dominant spectral component of the starting vector.

For example, $B=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has dominant right eigenvector $(1,0{)}^T$ and a dominant left eigenvector $(1,1{)}^T$. The starting vector $q^{(0)}=(-1,1{)}^T/\sqrt{2}$ has nonzero dot product with the right eigenvector, but $B{q}^{(0)}=q^{(0)}$: it stays forever in the eigenspace for the smaller eigenvalue $1$. Its coefficient along the dominant direction is zero, as the left eigenvector detects. An orthonormal eigenbasis makes the two tests agree in the self-adjoint case.

A principal component is the projection of centered data onto a unit eigenvector of its covariance matrix. Ordering the components by their eigenvalues orders their variances. Repeated eigenvalues allow any orthonormal basis of the corresponding eigenspace.

The principal moments of inertia about a specified point $O$ are the eigenvalues $I_1,I_2,I_3$ of the inertia tensor ${\mathbf{I}}_O$. Their associated unit vectors that are eigenvectors specify principal axes through $O$. The spectral theorem for real symmetric matrices guarantees an orthonormal basis of these eigenvectors, even for an irregular rigid body. Choose their signs to make a right-handed basis and collect them as the columns of an orthogonal matrix $Q$:Thus diagonalization removes the off-diagonal coupling in the rotational kinetic energy. The eigenvalues belong to the mass distribution and reference point, independently of which orthonormal basis initially describes them. ^[ref]

In this principal body frame, write angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{\mathbf{e}}_i$. ThenUse these as total angular momentum and kinetic energy only under the reference-point conditions stated under inertia tensor. Along a single principal axis with $I_i>0$, angular momentum is parallel to angular velocity. For a mixture of directions with different principal moments of inertia, it generally is not. A principal axis need not be a geometric symmetry axis: it is selected by the complete mass distribution. A principal axis also need not give stable free rotation; for three distinct principal moments of inertia, the intermediate axis theorem supplies the stability distinction.

For any unit vector $\mathbf{n}={\sum }_i{n}_i{\mathbf{e}}_i$,The axial moment of inertia is a weighted average of the principal moments of inertia. At fixed angular speed, the least and greatest rotational kinetic energies occur in the corresponding eigenspaces. At fixed nonzero angular momentum magnitude $L$, instead, $K=\frac{1}{2}{\sum }_i{L}_i^2/I_i$, so the largest principal moment of inertia gives the smallest rotational kinetic energy. These comparisons hold the reference point fixed.

There is an additional physical restriction beyond positive semidefiniteness. In principal axes coordinates,with two analogous cyclic inequalities. Therefore three arbitrary positive numbers do not always represent a possible inertia tensor. For a distribution in the $xy$ plane through $O$, $I_3=I_1+I_2$, as in the planar inertia tensor of paired point masses. A uniform solid ellipsoid of mass $M$ and semiaxes $a,b,c$ supplies a continuous three-dimensional example: using its mass second-moment tensor, its central principal moments of inertia are $M(b^2+c^2)/5$, $M(a^2+c^2)/5$, $M(a^2+b^2)/5$. Setting $a=b=c=R$ recovers the moment of inertia of a uniform solid sphere.

For closed manifolds with the product Riemannian metric, the positive Laplace-Beltrami operator is the sum of the two factor operators. Products of their orthonormal eigenbases form a complete basis, and eigenvalues add. Hence the product eigenspace at $\lambda$ is the displayed finite orthogonal sum of tensor products. The discrete assertion requires a discrete spectral setting.

A projective measurement has self-adjoint orthogonal projections $P_i$ satisfying $P_i{P}_j={\delta }_{ij}{P}_i$ and ${\sum }_i{P}_i=I$. Measuring an observable with discrete spectral decomposition $A={\sum }_i{a}_i{P}_i$ reports eigenvalue $a_i$ with Born rule probability $p_i=\mathrm{Tr}(P_i\rho )$. A projector can have rank greater than one: the outcome then identifies an eigenspace, rather than a particular eigenvector inside it. Projectors specify the outcome statistics; the usual ideal conditional update is the additional Lüders rule.

In the qubit example from measurement in quantum mechanics, $Z=|0⟩⟨0|-|1⟩⟨1|$. At $\theta =\pi /3$, the zero and one probabilities are $3/4$ and $1/4$, regardless of the relative quantum phase $\varphi$. Under the Lüders rule, obtaining zero prepares $|0⟩$; obtaining one prepares $|1⟩$, up to global phase. Repeating the same measurement returns the same label with certainty, provided no intervening evolution or operation changes the measured sector. Repeatability concerns the conditional output, and does not mean the first outcome was predetermined.

Change to $X$ and its Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. Adding the two probability amplitudes before squaring givesAt $\theta =\pi /3$ and $\varphi =0$, $p_{+}=(2+\sqrt{3})/4\simeq 0.9330$ and $p_{-}\simeq 0.0670$. At $\varphi =\pi /2$, both are $1/2$; at $\varphi =\pi$, the first pair is exchanged. The $Z$ statistics remain $3/4,1/4$ throughout. This is the same addition-and-cancellation principle as wave interference: relative quantum phase changes how amplitudes reinforce or cancel in another basis.

The beauty of this basis dependence is that information invisible to one measurement becomes accessible to another. A $Z$ measurement of an ensemble estimates its quantum state populations, while $X$ probes one component of its quantum coherence. These are experiments on separate identically prepared copies. Sequentially measuring $Z$ and then $X$ on one copy instead probes the disturbed post-measurement state. Quantum state tomography combines suitable ensemble statistics to reconstruct the original state.

For real three-dimensional vectors, Lagrange's identity gives $|a\times b{|}^2=a^{T}Ma$ with symmetric matrix $M=|b{|}^{2}I-b{b}^T$. For $b\ne 0$, the normal line spanned by $b$ is its kernel and the perpendicular plane is its $|b{|}^2$ eigenspace. Consequently $M$ is positive semidefinite, has rank two and has zero determinant. For $b=0$, it is the zero matrix.

Suppose a unitary matrix $U$ is symmetric under ordinary transpose, and write its entrywise real and imaginary parts as $A$ and $B$. They are real symmetric matrices. Expanding $U^{†}U=I$ gives $A^2+B^2+i(AB-BA)=I$, proving the displayed identities. Commuting real symmetric matrices admit simultaneous diagonalization: diagonalize $A$, then diagonalize the restriction of $B$ to each invariant eigenspace of $A$. A common real orthonormal eigenbasis givesIf $A$ has simple eigenvalues, every eigenvector of $A$ is already an eigenvector of $B$, because its one-dimensional eigenspace is invariant. The eigenvalues of $A$ and $B$ are paired as the real and imaginary parts of the same eigenvalue of $U$.

If a real linear operator $A$ on an $n$-dimensional vector space has $n$ distinct real eigenvalues, any real square root $B$ commutes with $A$ and preserves each one-dimensional eigenspace. There $B$ is multiplication by a real $b_i$, hence ${\lambda }_i=b_i^2\ge 0$. Conversely, if all the eigenvalues are nonnegative, choosing $b_i=\sqrt{{\lambda }_i}$ on an eigenbasis supplies a real square root. Without simple spectrum the necessity fails: a real quarter-turn squares to $-I$.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.

Suppose $D$ has $n$ distinct eigenvalues over a field and $C$ commutes with $D$. Each eigenspace of $D$ is one-dimensional. Since commuting maps preserve eigenspaces, $C$ acts as a scalar on each of them. The eigenvectors of $D$ give a common eigenbasis and hence simultaneous diagonalization. For real matrices, the result gives a real eigenbasis only if the eigenvalues of $D$ are real; otherwise it applies over $\mathbb{C}$.

The displayed parts of a unitary matrix are skew-Hermitian matrices. Expanding $U{U}^{†}=U^{†}U=I$ gives $AB=BA$ and $A^2+B^2=-I$. They admit a common orthonormal eigenbasis. For a unitary eigenvalue $u$, the corresponding part eigenvalues are $i\mathrm{Im}u$ and $-i\mathrm{Re}u$. If $U^T=-U$, then $U^{†}=-\overline{U}$, so these are exactly its real and imaginary parts and both are real skew-symmetric matrices. Such an invertible antisymmetric matrix has even dimension, by its determinant. If one part has distinct eigenvalues, commutation alone shows its one-dimensional eigenspaces are preserved by the other part.

A spectral decomposition expresses a linear operator through its eigenvalues and projections onto its eigenspaces.

For a diagonalizable matrix with distinct eigenvalues $\lambda ,\mu ,\dots$, this polynomial is the projection operator onto the eigenspace for $\lambda$ along the direct sum of the other eigenspaces. It is one on its own eigenspace and zero on the others. Consequently $P_{\lambda }{P}_{\mu }=0$ for different eigenvalues, $\sum P_{\lambda }=I$ and $A=\sum \lambda P_{\lambda }$. These projections need not be orthogonal projections; that stronger property holds for a normal matrix. This is the polynomial interpolation form of the spectral decomposition.

For a finite-dimensional Hermitian matrix $A$, its spectral projection $P_{\lambda }$ is the orthogonal projection onto the eigenspace for $\lambda$. In an orthonormal eigenbasis, the displayed sum is over every basis vector with that eigenvalue. The projection is independent of the choice of basis within a repeated eigenspace. For distinct eigenvalues $\lambda ,\mu$,This is the spectral decomposition grouped by distinct eigenvalues: decompose a vector into orthogonal pieces, then scale each piece. For the matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$,Thus $P_3(1,0{)}^T=(1/2,1/2{)}^T$ and $P_1(1,0{)}^T=(1/2,-1/2{)}^T$: the two pieces add back to $(1,0{)}^T$. Squaring the matrix simply squares each scalar, $A^2=9P_3+P_1$, and similarly the matrix exponential is $e^{tA}=e^{3t}{P}_3+e^t{P}_1$. Spectral projections of a self-adjoint matrix turn repeated matrix calculations into scalar calculations on fixed vector subspaces.

For the positive Laplace-Beltrami operator on the unit sphere $S^d$, with $d\ge 1$, the eigenvalues are $n(n+d-1)$ for integers $n\ge 0$. Their eigenspaces consist of restrictions of degree-$n$ harmonic polynomials and have dimensions $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)-\left(\genfrac{}{}{0ex}{}{n+d-2}{d}\right)$, with the second term zero for $n<2$. The Laplacian in polar coordinates gives the eigenvalue; Fischer decomposition and the Stone-Weierstrass theorem prove completeness.