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The characteristic polynomials, in the convention $\det (\lambda I-T)$, areSolving the eigenvalue equations gives the complete eigenspacesEvery nonzero vector in a listed eigenspace is an eigenvector, and these are all the eigenvectors. For $A$, the repeated eigenvalue $2$ has a two-dimensional eigenspace; for $B$, all three eigenvalues are simple. The three displayed generating vectors form a basis because their column matrixhas determinant $2$. Thus both matrices are diagonalizable, with diagonal entries $(1,2,2)$ for $A$ and $(2,4,6)$ for $B$ in this basis.

Direct multiplication givesTake $D=B$ and $C=A$ in the preceding result: $B$ has the three distinct eigenvalues $2,4,6$ and the matrices commute. The common eigenbasis from part (i) givesIndeed $AS=S\mathrm{diag}(1,2,2)$ and $BS=S\mathrm{diag}(2,4,6)$, and $\det S=2\ne 0$. These equalities directly verify the simultaneous diagonalization.

If $Ax=\lambda x$, then $M=A-\lambda I$ is a normal matrix by part (ii), and $Mx=0$. Part (i), applied to $M$, yields $\Vert M^{†}x\Vert =\Vert Mx\Vert =0$. ThusThe nonzero vector $x$ is consequently an eigenvector of the conjugate transpose with eigenvalue $\overline{\lambda }$. This adjoint eigenvector identity for a normal matrix will also determine the special spectral restrictions below.

Start with the assumed basis of eigenvectors, grouping them by eigenvalue. In each eigenspace, apply the Gram-Schmidt process, which replaces a linearly independent list by an orthonormal list with the same span. Every new vector stays within that eigenspace, since an eigenspace is a vector subspace.

By part (iv), different eigenspaces are orthogonal. The combined lists therefore form an orthonormal basis of the entire space, still consisting of eigenvectors. Let $U$ have these vectors as columns. Then $U$ is a unitary matrix, soThis gives the required unitary diagonalization of a normal matrix, including repeated eigenvalues.

For a skew-Hermitian matrix, $A^{†}=-A$, so both $A^{†}A$ and $A{A}^{†}$ equal $-A^2$. It is therefore a normal matrix. On an eigenvector with eigenvalue $\lambda$, part (a)(iii) givesThus $\overline{\lambda }=-\lambda$, and every eigenvalue lies in $i\mathbb{R}$, including the possible eigenvalue zero.

For a unitary matrix, $A^{†}=A^{-1}$, so both defining products for a normal matrix are $I$. The matrix is invertible, hence no eigenvalue is zero. If $Ax=\lambda x$, then $A^{-1}x={\lambda }^{-1}x$, while part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Therefore $\overline{\lambda }={\lambda }^{-1}$, andThus all eigenvalues have unit modulus.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

The three linearly independent eigenvectors still form a basis, so the same coefficient calculation applies to this diagonalizable matrix matrix. The general solution isA repeated eigenvalue alone does not create a polynomial factor in time; that factor is needed only when a nontrivial Jordan block prevents a full eigenvector basis.

The independence of eigenvectors for distinct eigenvalues ensures that such eigenvectors form a linearly independent set. For completeness, apply ${\prod }_{j\ne k}(M-{\lambda }_{j}I)$ to a proposed relation ${\sum }_i{c}_i{\mathbf{e}}_i=0$. It leaves $c_k{\prod }_{j\ne k}({\lambda }_k-{\lambda }_j){\mathbf{e}}_k=0$, so every $c_k=0$. Thus the three eigenvectors form a basis over the relevant field and $\mathbf{x}={\sum }_i{a}_i(t){\mathbf{e}}_i$. Substitution into the linear system of ordinary differential equations gives ${\sum }_i({\dot{a}}_i-{\lambda }_i{a}_i){\mathbf{e}}_i=0$; linear independence implies ${\dot{a}}_i={\lambda }_i{a}_i$. HenceA real matrix may have a complex conjugate pair of eigenvalues. In that case the calculation is over $\mathbb{C}$ and conjugate coefficients yield real solutions; equivalently use real and imaginary parts of the complex modes. No unstated assumption that all three eigenvalues are real is needed.

Use the part-preserving bipartite four-cycle density conventionHere $G(x,y)$ is the indicator function of adjacency, and repeated labels are allowed. This is the homomorphism density convention appropriate to the four-cycle norm. We derive the needed counting and operator norm facts directly.

Let $M=(G(x,y))$ be the $n$-by-$n$ biadjacency matrix, let $1$ be the all-ones vector, let $J=11^{\mathsf{T}}$, and put $H=M-\delta J$. The degree hypothesis saysso the centered matrix $H$ annihilates $1$ on both sides. In particular, $HJ=J{H}^{\mathsf{T}}=0$, andThe two terms have zero products in either order. Expanding the matrix trace directly counts four adjacency factors:Since $J^2=nJ$ and $\mathrm{tr}J=n$, we obtainThe given upper bound therefore impliesThis computation proves the centered four-cycle identity for a biregular graph, rather than assuming a four-cycle norm identity.

The matrix $Q=H{H}^{\mathsf{T}}$ is a positive semidefinite matrix. By the spectral theorem for real symmetric matrices it has nonnegative eigenvalues ${\lambda }_j$, andMoreover, for the Euclidean norm,Indeed $\Vert H^{\mathsf{T}}v{\Vert }_2^2=v^{\mathsf{T}}Qv$, whose maximum on unit vectors is ${\lambda }_{\max }$; the Cauchy-Schwarz inequality characterization $\Vert H{\Vert }_{\mathrm{op}}={\sup }_{\Vert u{\Vert }_2=\Vert v{\Vert }_2=1}|u^{\mathsf{T}}Hv|$ also shows that $H$ and $H^{\mathsf{T}}$ have the same operator norm. Taking fourth roots thus giveswhere the error parameter $c$ is nonnegative, as in the claimed bound.

Finally use the indicator vectors $a=1_A$, $b=1_B$. Their Euclidean norms are $\sqrt{|A|}$ and $\sqrt{|B|}$, so the Cauchy-Schwarz inequality and the operator norm estimate yieldThis is the required discrepancy bound. It also holds when either subset is empty or $\delta =0$. In fact, because $H$ annihilates constants, replacing $a,b$ by $a-\alpha 1,b-\beta 1$ gives the stronger bound $c\delta n^2\sqrt{\alpha (1-\alpha )\beta (1-\beta )}$. This is the spectral discrepancy bound for a biregular graph.

A nilpotent endomorphism $x$ satisfies $x^N=0$ for some positive integer $N$. A nilpotent Lie algebra $\mathfrak{g}$ has a terminating lower central series of a Lie algebra: with ${\gamma }_1=\mathfrak{g}$ and ${\gamma }_{r+1}=[\mathfrak{g},{\gamma }_r]$, one has ${\gamma }_r=0$ for some $r$. These are different conditions: the first concerns a particular linear map, whereas the second concerns repeated Lie brackets.

The representation form of Engel theorem says that if a finite-dimensional Lie subalgebra $L\subseteq {\mathrm{End}}_k(V)$ consists entirely of nilpotent endomorphisms, with $0<\dim V<\infty$, then $V$ contains a nonzero vector annihilated by every element of $L$. Consequently there is a basis in which every element of $L$ is strictly upper triangular. No assumption on the characteristic of a field or algebraic closure is required.

We prove the common-vector assertion by induction on $\dim L$, simultaneously for every nonzero finite-dimensional representation space. The zero algebra is immediate. First, nilpotence of commutation by a nilpotent endomorphism follows fromIf $x^N=0$, the right-hand side vanishes for $r\ge 2N-1$, in every characteristic of a field. For any proper Lie subalgebra $M⊊L$, its Adjoint representation on $L/M$ therefore consists of nilpotent endomorphisms. Its image has dimension at most $\dim M<\dim L$, so the inductive assertion gives a nonzero class $y+M$ annihilated by $M$. Equivalently $y\notin M$ and $[M,y]\subseteq M$. Thus the Engel normalizer lemma gives $M⊊N_L(M)$.

Choose a maximal proper Lie subalgebra $M$. Its normalizer of a Lie subalgebra must be all of $L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has dimension one: otherwise a one-dimensional Lie subalgebra of this quotient would have a proper inverse image strictly between $M$ and $L$. By induction the space $W=\{v\in V:Mv=0\}$ is nonzero. It is invariant under $L$, since $m(xw)=x(mw)+[m,x]w=0$. Write $L=M+kx$. The restriction of the nilpotent endomorphism $x$ to $W$ has a nonzero kernel, and any vector in that kernel is annihilated by all of $L$. This completes the induction. Apply the assertion repeatedly to $V/kv$ to obtain a complete invariant flag $0=V_0\subset V_1\subset \cdots \subset V_d=V$ with $L{V}_j\subseteq V_{j-1}$. A basis adapted to that flag makes every matrix strictly upper triangular.

A nilpotent abstract Lie algebra need not act by nilpotent matrices. For $V\ne 0$, the scalar algebra $kI\subseteq \mathfrak{gl}(V)$ is an abelian Lie algebra, hence a nilpotent Lie algebra, but $I^N=I\ne 0$. No change of basis makes $I$ strictly upper triangular. The failed implication is precisely the distinction described in nilpotent Lie algebras need not act nilpotently.

If $\mathfrak{g}$ is a nilpotent Lie algebra, then $(\mathrm{ad}x{)}^r\mathfrak{g}\subseteq {\gamma }_{r+1}$, so every $\mathrm{ad}x$ is a nilpotent endomorphism. Conversely, apply Engel theorem to $\mathrm{ad}\mathfrak{g}$ acting on $\mathfrak{g}$. Its invariant flag is lowered by every Adjoint representation matrix, so every product of $d=\dim \mathfrak{g}$ such matrices vanishes. Since ${\gamma }_{d+1}$ is spanned by expressions $\mathrm{ad}{x}_1\cdots \mathrm{ad}{x}_d(y)$, the lower central series of a Lie algebra terminates. ThereforeFinally suppose $k$ is an algebraically closed field. If some $\mathrm{ad}x$ is not a nilpotent endomorphism, it has a nonzero eigenvalue $\lambda$ and an eigenvector $y$, giving $[x,y]=\lambda y$. The vectors $x,y$ are linearly independent and span a nonabelian two-dimensional Lie subalgebra. Conversely, a Lie subalgebra of a nilpotent Lie algebra is nilpotent because its lower central series of a Lie algebra lies termwise in the ambient series. A nonabelian two-dimensional Lie algebra has a basis $u,v$ with $[u,v]=v$: choose $v$ spanning the nonzero derived algebra and rescale a complementary vector. Its Adjoint representation has the nonzero eigenvalue $1$, so it is not nilpotent. This proves the two-dimensional subalgebra criterion for Lie algebra nilpotence:

Work over $\mathbb{C}$. Decompose $V$ as the direct sum of the generalized eigenspaces $V_{\lambda }$ of $X$. Define $X_s$ to be $\lambda I$ on $V_{\lambda }$ and set $X_n=X-X_s$. Then $X_s$ is a diagonalisable endomorphism, $X_n$ is a nilpotent endomorphism, and they commute. For uniqueness, any commuting decomposition $X=S+N$ has $S$ and $N$ commuting with $X$, hence preserving each $V_{\lambda }$. Decompose $V_{\lambda }$ further into eigenspaces of $S$. On a nonzero such space with eigenvalue $\sigma$, the operator $X=\sigma I+N$ has only the eigenvalue $\sigma$, so $\sigma =\lambda$. Since $S$ is diagonalizable, $S=\lambda I$ throughout $V_{\lambda }$. Thus $S=X_s$ and $N=X_n$. This is the Additive Jordan decomposition.

The Chinese remainder theorem for the pairwise coprime polynomials $(t-\lambda {)}^{r_{\lambda }}$ also gives a polynomial $p$ with $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^{r_{\lambda }})$, where $r_{\lambda }$ is the largest Jordan block size. Consequently $X_s=p(X)$ and $X_n=X-p(X)$. This polynomial description shows that both parts preserve every $X$-invariant subspace.

On $\mathrm{Hom}(V_{\lambda },V_{\mu })$ the operator $\mathrm{ad}{X}_s$ is the scalar $\mu -\lambda$, so it is diagonalizable. The nilpotence of commutation by a nilpotent endomorphism makes $\mathrm{ad}{X}_n$ nilpotent. They commute, since $[\mathrm{ad}{X}_s,\mathrm{ad}{X}_n]=\mathrm{ad}[X_s,X_n]=0$. Uniqueness of the Additive Jordan decomposition therefore gives the adjoint compatibility of additive Jordan decomposition:Now let $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ be a complex semisimple Lie algebra and $X\in \mathfrak{g}$. Since the semisimple part of $\mathrm{ad}X$ is a polynomial in $\mathrm{ad}X$, it preserves $\mathfrak{g}$. Thus $[X_s,\mathfrak{g}]\subseteq \mathfrak{g}$. By the Weyl complete reducibility theorem, the Adjoint representation of $\mathfrak{g}$ on $\mathrm{End}(V)$ has a decomposition $\mathrm{End}(V)=\mathfrak{g}\oplus M$ into invariant subspaces. Write $X_s=y+z$ with $y\in \mathfrak{g}$ and $z\in M$. For $a\in \mathfrak{g}$, the vector $[z,a]=[X_s,a]-[y,a]$ lies in $\mathfrak{g}$, and invariance of $M$ puts it in $M$ as well. Hence $[z,a]=0$.

Decompose $V={⨁}_j{U}_j$ into Irreducible Lie algebra representations. Each $U_j$ is preserved by $X_s=p(X)$ and by $y$, hence by $z$. The Schur lemma makes $z{|}_{U_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every representing element of $\mathfrak{g}$ has zero trace on every $U_j$. Also $\mathrm{tr}(X_s{|}_{U_j})=\mathrm{tr}(X{|}_{U_j})$, because $X_n$ is nilpotent there. It follows that $c_j\dim U_j=\mathrm{tr}(z{|}_{U_j})=0$. In characteristic zero, $c_j=0$, so $z=0$. Therefore $X_s\in \mathfrak{g}$ and $X_n=X-X_s\in \mathfrak{g}$. This proves that semisimple matrix Lie algebras are closed under additive Jordan decomposition, including representations with repeated isomorphic irreducible summands.

Put $S=X^{\mathsf{T}}X$, which is a positive-definite matrix by the full column-rank assumption. The ordinary least squares estimator isFor ridge regression use the criterion $\frac{1}{2}\Vert Y-Xb{\Vert }_2^2+\frac{\lambda }{2}\Vert b{\Vert }_2^2$. Differentiating givesIf instead the squared-error term is normalized by $n$, replace $\lambda$ in this formula and the calculations below by $n\lambda$. Only the scale of the tuning parameter changes.

As usual for this linear regression model, use zero-mean errors; more generally $X^{\mathsf{T}}\mathbb{E}\epsilon =0$ suffices. Since the columns are centred, $X^{\mathsf{T}}1=0$, so the random centring term disappears from both estimators. ThusLet $A=(S+\lambda I{)}^{-1}$. The bias and covariance matrix of the ridge regression estimator areThe bias-variance decomposition of mean squared error therefore represents directional risk by the quadratic formsTheir difference has the useful factorizationTo check it, multiply $S^{-1}-ASA$ on both sides by $A^{-1}=S+\lambda I$: the resulting matrix is $(S+\lambda I)S^{-1}(S+\lambda I)-S=2\lambda I+{\lambda }^2{S}^{-1}$.

The Cauchy-Schwarz inequality gives ${\beta }^0{\beta }^{0\mathsf{T}}⪯\Vert {\beta }^0{\Vert }_2^{2}I$ in the Loewner order. Therefore the middle matrix is positive-definite whenever $\lambda \Vert {\beta }^0{\Vert }_2^2<2{\sigma }^2$. For example,works, independently of the eigenvalues of $S$. Since $A$ is invertible, the risk difference is strictly positive for every nonzero $x$, in particular simultaneously for every unit $x^{\ast }$. This proves uniform directional risk improvement by ridge regression. For the normalized-loss convention, take the displayed choice divided by $n$.

For the final assertion, fix any $\lambda >0$ and choose a unit eigenvector $v$ of $S$ with eigenvalue $d>0$. Put $x^{\ast }=v$ and ${\beta }^0=Bv$. Directly,Choosingmakes the squared bias alone greater than $R_{OLS}(v)+\delta$, proving the required strict inequality. Thus unbounded directional risk of fixed ridge shrinkage is compatible with the existence of a beneficial signal-dependent small penalty: no fixed positive penalty uniformly improves risk over all possible signals.

The mean assumption matters if the printed covariance condition is read in isolation. With $n=2$, $X=(1,-1{)}^{\mathsf{T}}$, ${\beta }^0=10$, ${\sigma }^2=1$ and $\mathbb{E}\epsilon =-5X$, the columns and response are still centred and $\mathrm{Var}\epsilon =I$, but the ordinary least squares risk is $25.5$. Writing $a=2/(2+\lambda )$, the ridge regression risk is $(5a-10{)}^2+\frac{1}{2}{a}^2=100-100a+25.5a^2>25.5$ for every $0<a<1$. Hence the standard zero-mean regression-error convention, or its stated orthogonality version, is necessary for the first risk comparison; covariance alone is insufficient.