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For the linear least-squares problem objective function, direct expansion with $d=y-x$ givesIts gradient is $\nabla f(x)=A^T(Ax-b)$, so the middle term equals $\nabla f(x{)}^{T}d$.

Use the supplied thin singular value decomposition $A=US{V}^T$. Since $U^{T}U=I_n$, the linear map $w\mapsto Uw$ preserves the Euclidean norm. The square matrix $V$ is an orthogonal matrix, so $\Vert V^{T}d\Vert =\Vert d\Vert$. ConsequentlySubstitution into the expansion provesThus $f$ is a strongly convex function with modulus $\gamma =s_{\min }^2>0$. This modulus is sharp: choose $d=V{e}_i$ for an index attaining the smallest singular value. The Hessian matrix is $A^{T}A=V{S}^2{V}^T$, so the same conclusion says that its smallest eigenvalue is $s_{\min }^2$.

Write ${\Delta }_0=f(x^0)-f(x^{\ast })$. Dividing the corrected estimate from part (c) by $k/(2L)$ gives the averaged stationarity bound for stochastic gradient descentThe Cauchy-Schwarz inequality, first within each expectation and then over the $k$ indices, givesThusEquivalently, if $J$ is chosen independently of the run with the discrete uniform distribution on $\{0,\dots ,k-1\}$, then $\mathbb{E}\Vert \nabla f(X^J){\Vert }^2\le M+2L{\Delta }_0/k$. The same bound for the expected Euclidean norm follows by the Cauchy-Schwarz inequality.

The guarantee is averaged or best-iterate stationarity. It does not assert that every last iterate converges, or that the limiting Euclidean norm equals $\sqrt{M}$. The squared gradient bound has excess $O(k^{-1})$ above $M$; the displayed expected Euclidean norm bound is within $O(k^{-1/2})$ of $\sqrt{M}$, including the zero-noise case. For fixed $M>0$, it is sharper:For example, to ensure ${\min }_{j<k}\mathbb{E}\Vert \nabla f(X^j)\Vert \le \sqrt{M}+\epsilon$ with $\epsilon >0$, it suffices that $k\ge 2L{\Delta }_0/(2\sqrt{M}\epsilon +{\epsilon }^2)$. The big O notation here concerns these upper bounds, rather than an asserted asymptotic equality for the stochastic gradient noise floor.

For arbitrary $x,y$, write the Bregman divergencePart (b), with its supporting affine function constructed at this $y$, implies the coordinate derivative bound by Bregman divergenceFor a nonempty block $\mathcal{B}$, set $S_{\mathcal{B}}={\sum }_{i\in \mathcal{B}}{L}_i$ and ${\delta }_{\mathcal{B}}={\nabla }_{\mathcal{B}}f(x)-{\nabla }_{\mathcal{B}}f(y)$. Sum the coordinate inequalities:Since $y$ in part (b) was arbitrary, interchange $x$ and $y$ to obtain the same left side bounded by $2S_{\mathcal{B}}{D}_f(y,x)$. Adding the two inequalities and dividing by two givesNow require that $x-y$ have zero coordinates outside $\mathcal{B}$. The last dot product is then ${\delta }_{\mathcal{B}}^T(x-y)$. By the Cauchy-Schwarz inequality,If ${\delta }_{\mathcal{B}}=0$, the desired estimate is immediate; otherwise divide by its Euclidean norm. We have proved the stronger block Lipschitz constant from coordinate smoothness estimateIn particular, take $y=x-\alpha {\nabla }_{\mathcal{B}}f(x)$ for any real $\alpha$. This proves (6), with a valid choiceThe least admissible Lipschitz constant is therefore no larger than this choice. For the empty block, the block coordinate gradient vanishes identically and its least nonnegative Lipschitz constant is zero. The proof uses only $C^1$ differentiability, convexity and coordinate Lipschitz continuity, with no assumed Hessian matrix.

The supporting inequality in part 2(a) bounds $f$ below by an affine function plus a positive squared Euclidean norm, making it a coercive function. Its continuity gives a global minimizer $x^{\ast }$, which is unique by strict convexity. Let $L_{\max }={\max }_i{L}_i$, $\ell =n/m$ and ${\Delta }_0=f(x^0)-f(x^{\ast })$. Here $m$ is the block size, so its role differs from the number of component functions in Question 2. Use randomized block coordinate descent with uniformly sampled blocks of size $m$, independently at each update. Sampling uniformly from a fixed partition into $\ell$ equal-sized blocks gives the same calculation, since each coordinate has inclusion probability $1/\ell =m/n$.

Let $\beta$ be a common upper bound on the block Lipschitz constants. Part (c) permits $\beta =m{L}_{\max }$. With step size $1/\beta$, part (a) givesSumming squared partial derivatives according to their inclusion probabilities,For a strongly convex function, part 2(a) also yields the strong convexity gradient inequalityIndeed, substitute $y=x^{\ast }$ into the supporting inequality and complete the square in $x^{\ast }-x$: the expression $\nabla f(x{)}^{T}d+\gamma \Vert d{\Vert }^2/2$ is at least $-\Vert \nabla f(x){\Vert }^2/(2\gamma )$ for every $d$.

Combining the last three estimates gives a contraction in conditional expectation:By the tower property and repeated application,An optimization epoch counts $n$ coordinate partial derivative evaluations: it comprises $\ell$ block updates or one full gradient update. With sampling with replacement, an optimization epoch is this work measure; it need not visit every coordinate. Consequently the block method needsto bring the expected objective function gap below $\epsilon$, for $0<\epsilon <{\Delta }_0$, up to integer rounding.

For the steepest descent method, use a global Lipschitz gradient constant $L_f$. Part (c) with the full block permits $L_f={\sum }_i{L}_i\le n{L}_{\max }$. The descent lemma with step size $1/L_f$, together with the strong convexity gradient inequality, givesAn exact line search along the negative gradient does at least as well in objective function decrease per update. Ignoring its additional work, the resulting epoch complexity isThe common coordinate-based guarantees favour blocks by a factor $n/m=\ell$ in epochs. In fact, with the conservative constants $\beta =m{L}_{\max }$ and $L_f=n{L}_{\max }$, both methods have the same guaranteed contraction per update, while a block update uses only $m$ coordinate partial derivatives.

This compares the bounds justified by the given constants. For a particular objective function, sharper constants give the comparison $\beta /\gamma$ against $L_f/\gamma$; an actual speedup of $\ell$ is not universal. Computing a block coordinate gradient also need not cost exactly $m/n$ of a full gradient evaluation. The epoch complexity comparison assumes that coordinate partial derivative counts are a useful measure of work, and does not itself establish a wall-clock comparison.

Put $v=P_2-P_1$ and $w=Q_2-Q_1$. They are nonzero vectors with the same Euclidean norm. Choose the rotation matrix $R$ taking the direction $v/\Vert v\Vert$ to $w/\Vert w\Vert$; equal lengths then give $Rv=w$. DefineThis is a translation by $Q_1-P_1$ followed by a rotation about $Q_1$, so it is an isometry. Direct substitution gives $t(P_1)=Q_1$ and $t(P_2)=Q_1+w=Q_2$.

Put $v=P_2-P_1$ and $w=Q_2-Q_1$. They are nonzero vectors with the same Euclidean norm. Choose the rotation matrix $R$ taking the direction $v/\Vert v\Vert$ to $w/\Vert w\Vert$; equal lengths then give $Rv=w$. DefineThis is a translation by $Q_1-P_1$ followed by a rotation about $Q_1$, so it is an isometry. Direct substitution gives $t(P_1)=Q_1$ and $t(P_2)=Q_1+w=Q_2$.

Multiply $Ax=b$ by $Q^T$ to get the equivalent triangular systemFor nonsingular $A$, compute $c=Q^{T}b$ and use back substitution:Orthogonal transformations preserve the Euclidean norm, making this a numerically stable reduction. The factors can be applied as stored Householder transformations without explicitly forming $Q$.

Since $Q$ is invertible, $A$ is singular exactly when $R$ is singular. For an upper triangular matrix,Thus a singular $A$ forces at least one zero diagonal entry in $R$. In that case, ordinary division-based back substitution must be replaced by a consistency/rank analysis: the system may have no solution or infinitely many solutions, and a rank-revealing factorization can identify the free parameters.

Expand the vector field about the reference trajectory:Over a sufficiently short interval, the tangent-linear propagator isConsequentlyIf $S=J+J^T$ has $S\mathbf{e}=\lambda \mathbf{e}$ and the perturbation is initially parallel to $\mathbf{e}$, its squared Euclidean norm obeysPositive $\lambda$ means instantaneous norm-squared growth, negative $\lambda$ decay, and zero $\lambda$ no first-order change. The corresponding norm growth rate is $\lambda /2$, not $\lambda$. These are eigenvalues of the symmetric part governing the chosen error metric, not generally the eigenvalues of the nonsymmetric Jacobian matrix or the long-time Lyapunov exponents. The direction and the reference Jacobian matrix evolve during a finite forecast.

A Givens rotation acts as the identity except on coordinates $i,j$, where its block can be writtenThe block has orthonormal columns because ${\cos }^2\theta +{\sin }^2\theta =1$ and their inner product is zero, so $J^{T}J=I$. For $r=\sqrt{v_i^2+v_j^2}>0$, choose $\cos \theta =v_i/r$, $\sin \theta =v_j/r$. The new coordinates are $(r,0)$, eliminating the $j$th entry. If both entries are zero, the identity already works.

For the specified vector, one suitable matrix isIts rows are orthonormal, so it is orthogonal; in fact its determinant is one. The positive nonzero output component must be five because an orthogonal matrix preserves Euclidean norm.

The matrix is not unique. Left multiplication by any orthogonal matrix fixing the second coordinate vector preserves this output. For example arbitrary rotations of the first and third coordinates provide infinitely many alternatives.

A QR decomposition is $A=QR$, with $Q$ orthogonal and $R$ upper triangular. Orthogonality gives $Q^{\mathsf{T}}Q=I$, and therefore $\Vert Qx{\Vert }^2=x^{\mathsf{T}}{Q}^{\mathsf{T}}Qx=x^{\mathsf{T}}x=\Vert x{\Vert }^2$, proving invariance of the Euclidean norm.

The unshifted QR algorithm starts with $A_0=A$. At step $k$, factor $A_k=Q_k{R}_k$ and set $A_{k+1}=R_k{Q}_k$. ThenThus consecutive matrices are related by orthogonal similarity, so they have the same characteristic polynomial and eigenvalues. Induction proves every iterate has the original eigenvalues. Symmetry is also preserved. When the iteration converges to diagonal form, its diagonal entries are the eigenvalues; practical implementations use shifts and deflation to improve convergence.

Use Haar sampling by Gaussian QR decomposition:

The matrix $G$ is invertible almost surely. Indeed, successively sampled Gaussian columns have probability zero of lying in the proper linear subspace spanned by the previous columns. Thus the diagonal signs used above exist with probability one. Equivalently, apply the Gram-Schmidt process to the columns, always dividing each residual by its positive Euclidean norm; this produces $M$ directly.

To prove uniformity, fix $H\in O(n)$. Each column of $G$ has density proportional to $e^{-\Vert x{\Vert }^2/2}$, so orthogonal transformations preserve its probability distribution. Column independence therefore gives $HG\stackrel{\mathrm{d}}{=}G$.

The QR decomposition with positive diagonal is unique: if $G=Q_1{R}_1=Q_2{R}_2$, then $Q_2^T{Q}_1=R_2{R}_1^{-1}$ is both an orthogonal matrix and an upper triangular matrix, hence diagonal with entries $\pm 1$. Its diagonal entries are positive, so it is the identity. It follows that the orthogonal factor of $HG=HMR$ is $HM$. ThereforeBy uniqueness of normalized Haar measure on the compact group $O(n)$, the returned matrix has exactly the required probability distribution. Reflection invariance gives each determinant sign probability $1/2$, so the algorithm samples both components of the orthogonal group. The positive-diagonal correction is essential when a numerical QR decomposition chooses its own signs.

The inertial Lagrangian is $L=m|\dot{\mathbf{r}}{|}^2/2-U(\mathbf{r},t)$. Put $\mathbf{r}=R(t){\mathbf{r}}^{\prime }$ and define $U^{\prime }({\mathbf{r}}^{\prime },t)=U(R(t){\mathbf{r}}^{\prime },t)$. The transport identity and invariance of the Euclidean norm under a rotation giveIts expansion contains the ordinary kinetic energy, $m{\dot{\mathbf{r}}}^{\prime }\cdot ({\mathbf{\Omega }}^{\prime }\times {\mathbf{r}}^{\prime })$, and $m|{\mathbf{\Omega }}^{\prime }\times {\mathbf{r}}^{\prime }{|}^2/2$. These velocity and position terms generate the apparent forces in the rotating frame.

Write $s={\sigma }^2\ge 1$. The two multivariate normal distribution probability density functions have ratioHere $\Vert x\Vert$ is the Euclidean norm. Consequently the smallest rejection sampling envelope constant, and its acceptance probability, areThe supremum is attained at $x=0$; hence no smaller constant can dominate the ratio. Draw $Y\sim N(0,s{I}_d)$ and an independent $U$ from the uniform distribution on $(0,1)$. Accept $Y$ whenotherwise repeat with fresh independent random variables. One can generate $Y=\sqrt{s}G$ using a random vector $G$ with independent standard normal distribution coordinates.

For any measurable set $B$, the joint acceptance probability is ${\int }_{B}q(y)p(y)/(M_{\ast }q(y))dy=M_{\ast }^{-1}{\int }_{B}p(y)dy$. Dividing by $1/M_{\ast }$ proves that the returned random vector has probability density function $p$. The number of proposals has a geometric distribution on the positive integers, with expected value $M_{\ast }$. When $s=1$, every proposal is accepted.

Write $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$ and let $|\Lambda |={\max }_i|{\lambda }_i|$. The stated Euclidean norm givesTaking nonnegative square roots proves $\Vert f_n\Vert \le |\Lambda {|}^n\Vert f_0\Vert$. At $n=0$ this is equality. This is a contraction bound in the transformed coordinates; an arbitrary nonorthogonal $S$ does not give the same unit-constant bound for the original error norm.

The characteristic polynomial isThus the eigenvalues are 1 and 11. Corresponding eigenvectors are $(3,-1{)}^T$ and $(1,3{)}^T$. Their dot product is zero and each has Euclidean norm $\sqrt{10}$. Normalizing givesThe columns are orthonormal, so $P^{T}P=I$. The eigenvector equations give $AP=PD$ and hence $P^{T}AP=D$, proving the claimed orthogonal diagonalization of a real symmetric matrix.

For the vector equation, the Jacobian matrix isWrite the updated coordinates as $(X,Y)$. The second equation of the multivariable Newton iteration gives $X-Y=0$. Substituting $X=Y$ into its first linearized equation givesHence $h(x,y)=6x^5-20y^3$, and, whenever $h\ne 0$,

Starting from $(0,1)$ gives $(x_1,y_1)=(21/20,21/20)$. Every subsequent iterate remains on the diagonal, where this is exactly the scalar Newton root-finding iteration from (b). The value $f(21/20)>0$, so $21/20<r$.

For $1\le x<r$, concavity again puts the tangent's zero at or above $r$. Alsoso $r\le N(21/20)\le N(1)=8/7<6/5$. After one scalar step the sequence therefore enters the invariant interval proved in (b). It converges to $(r,r)$ at least quadratically. If the vector error is measured in the Euclidean norm, then $\Vert (x_k,y_k)-(r,r)\Vert =\sqrt{2}|x_k-r|$, so the scalar quadratic estimate also gives a quadratic vector estimate. The initial off-diagonal step causes no difficulty because its Jacobian matrix is invertible and lands exactly on the diagonal.

Write $r=x-a$. The squared Euclidean norm is $r^{T}r$, whose gradient is $2r$. The chain rule for the exponential function therefore gives the column gradientThe row derivative, if that convention is used for $df/dx$, is its transpose.

By the diagonal reduction in a two-variable Newton iteration, $y_n=x_n$ for $n\ge 1$. Put $e_n=x_n-1$. Direct factorization gives the exact Newton map in optimization error identityThe assumed convergence $x_n\to 1$ makes the final factor tend to $3/2$, so it is bounded for all sufficiently large $n$. Hence $|e_{n+1}|\le C|e_n{|}^2$. For the Euclidean norm error $E_n=\Vert (x_n,y_n)-(1,1)\Vert =\sqrt{2}|e_n|$, this becomes $E_{n+1}\le (C/\sqrt{2})E_n^2$, proving quadratic convergence. Unless the exact limit is reached in finitely many steps, $E_{n+1}/E_n^2\to 3/(2\sqrt{2})$.

The origin is feasible and has objective zero, while a Euclidean norm is nonnegative. Thus the minimum is $0$, and the complete argmin isIf $A$ is injective, this set contains only the origin. If it has a nontrivial kernel, every feasible kernel vector is also a minimizer. No Lagrange multipliers are needed.

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.