By the Eckart-Young-Mirsky theorem, truncate the singular value decomposition:The same $A_r$ minimizes the spectral norm and Frobenius norm errors over matrices of matrix rank at most $r$. This conventional meaning of rank-$r$ low-rank approximation includes the case $\mathrm{rank}A<r$, when $A_r=A$.

No uniform absolute error bound of order $ϵ$ follows. Take the Jordan blockHere $\Vert B{\Vert }_2=1$ and $\Vert E{\Vert }_2=ϵ$. The exact eigenvalues of $B$ are both zero, whereasThese are admissible outputs under a backward error guarantee of size $ϵ$, yet their forward errors are $\sqrt{ϵ}$. This is the square-root splitting of a defective double eigenvalue. More generally, eigenvalue sensitivity depends on the matrix; backward stability alone does not control it. The singular value perturbation bound used in part (ii) has no equally uniform counterpart for the eigenvalues of arbitrary nonsymmetric matrices.

An error of order $ϵ$ in a squared singular value can become an error of order $\sqrt{ϵ}$ after taking its square root. To keep $A$ nonsingular as required, take $0<ϵ<1/2$ andThen $\Vert A{\Vert }_2=1$, and the computed eigendecomposition has exactly the permitted backward error:Nevertheless the second computed singular value has errorBy the singular value perturbation bound, every product with these computed singular values and exactly orthonormal outer factors has spectral norm error at least this large. In particular, even the favorable choice $\hat{U}=I_2$ giveswhich has no uniform $O(ϵ)$ bound in big O notation. Therefore backward stability for the eigendecomposition of the Gram matrix does not imply backward stability for the singular value decomposition.

There is a related loss of orthonormality in the raw formula: evaluating $A\hat{V}{\hat{\Sigma }}^{-1}$ exactly for these perturbed factors gives $\mathrm{diag}(1,ϵ/\sqrt{{ϵ}^2+ϵ})$, whose second column has length tending to zero. Its product with $\hat{\Sigma }{\hat{V}}^T$ equals $A$, but its diagonal factors are consequently not a singular value decomposition. The paper's idealized orthonormality assumption lets us isolate the unavoidable singular value error even if this additional problem is repaired.

Let $\tilde{A}=\hat{U}\hat{\Sigma }{\hat{V}}^T$. Interpret the diagonal of $\hat{\Sigma }$ as nonnegative computed singular values and sort them consistently. If signed diagonal entries are used instead, take their absolute values and absorb signs into the orthonormal columns. Because the two outer factors have the stated orthonormality, these sorted numbers are exactly ${\sigma }_j(\tilde{A})$.

For completeness, the singular value perturbation bound follows from the Eckart-Young-Mirsky theorem. If $B_{j-1}$ is the truncated singular value decomposition of a matrix $B$, the triangle inequality givesExchange $A$ and $B$ to get $|{\sigma }_j(A)-{\sigma }_j(B)|\le \Vert A-B{\Vert }_2$. Applying this to $B=\tilde{A}$ and the given backward error provesThus backward stability of the singular value decomposition gives an absolute forward error bound for every singular value, even a very small one.

Choose the nonnegative square roots of the eigenvalues of $A^{T}A$ for the diagonal entries of $\Sigma$, in nonincreasing order. Its assumed matrix inverse exists, so these entries are positive. The eigendecomposition givesHence $U$ has orthonormal columns. AlsoTogether with the orthogonal matrix $V$ and the nonnegative diagonal matrix $\Sigma$, these are precisely the properties of a thin singular value decomposition.

Using the orthonormal columns of $U_1,U_2$ in the factorization givesMultiplying by the orthogonal matrices $V^T$ and $V$ gives

Put $B=A_1^T{A}_1$. It is a symmetric matrix and a positive semidefinite matrix, while $A_2^T{A}_2=I_n-B$ is also a positive semidefinite matrix. By the spectral theorem, choose an orthogonal matrix $V=[v_1,\dots ,v_n]$ such thatThe bounds follow directly from $z^{T}Bz=\Vert A_{1}z{\Vert }_2^2\in [0,\Vert z{\Vert }_2^2]$. Define the diagonal matricesThe columns of $A_{1}V$ are mutually orthogonal, since $(A_1{v}_i{)}^T(A_1{v}_j)={\lambda }_j{\delta }_{ij}$. For each ${\lambda }_j>0$, set $u_{1j}=A_1{v}_j/\sqrt{{\lambda }_j}$. These columns are orthonormal. At a zero ${\lambda }_j$, one has $A_1{v}_j=0$; choose $u_{1j}$ by completing the already prescribed columns to an orthonormal set of $n$ columns in ${\mathbb{R}}^k$. Such a completion exists because $k\ge n$. This gives $A_{1}V=U_1{\Sigma }_1$, including the zero columns.

Likewise, the Gram matrix of $A_{2}V$ is $\mathrm{diag}(1-{\lambda }_j)$. Normalize its nonzero columns and complete at its zero columns in ${\mathbb{R}}^{m-k}$, using $m-k\ge n$. The resulting $U_2$ has orthonormal columns and $A_{2}V=U_2{\Sigma }_2$. Combining the identities givesThis is a form of the cosine-sine decomposition. Completing the zero columns is essential: neither block was assumed to have full column rank.

For every vector $z\in {\mathbb{R}}^n$, splitting the rows givesBoth summands are nonnegative. Taking the supremum over unit vectors in the definition of the spectral norm proves

The orthonormal columns give $A^{T}A=I_n$. Thus $P=A{A}^T$ is the orthogonal projection onto the column space of $A$, andThe two terms are orthogonal, so the Pythagorean identity givesOnly the first term depends on $x$. Consequently the linear least-squares problem has the unique answer

Yes: for $k=2$ the stated inequality guarantees zero-stability. Part (ii) puts every polynomial root in the closed unit disc. The remaining root condition for a multistep method is that every root on the unit circle has multiplicity of a root one.

If this failed for a quadratic polynomial, it would have a repeated root $\xi$ with $|\xi |=1$, andConsequently $|{\alpha }_1|=2|{\alpha }_2|$ and $|{\alpha }_0|=|{\alpha }_2|$, contradictingbecause ${\alpha }_2\ne 0$. Thus all roots on the unit circle are simple andThis conclusion does not assert consistency of a numerical method: part (i) rules out consistency of a numerical method whenever the coefficient inequality is strict. Zero-stability and consistency of a numerical method are separate requirements.

As appropriate for a genuine $k$-step linear multistep method, take ${\alpha }_k\ne 0$. Suppose the characteristic polynomial has a polynomial root $\xi$ with modulus $|\xi |>1$. From $\rho (\xi )=0$ and the triangle inequality,Dividing by the positive number $|{\alpha }_k||\xi {|}^{k-1}$ gives $|\xi |\le 1$, a contradiction. HenceThe nonzero leading coefficient matters: if ${\alpha }_k=0$, the assumed inequality forces all the ${\alpha }_j$ to vanish, so $\rho$ is the zero polynomial rather than a characteristic polynomial with $k$ roots. This degenerate case does not define the method in the question.

No. A necessary condition for consistency of a numerical method is reproduction of the constant solution $y(x)=1$ of the ordinary differential equation $y^{\prime }=0$. In a linear multistep method this givesBut the reverse triangle inequality gives, under the strict coefficient inequality,Thus the necessary order conditions for a linear multistep method already fail at order zero, regardless of the values of the ${\beta }_j$.

With $f(x,y)=-x{y}^2-y^5$, write $t=x_{n+1}=x_n+h$ and $z=y_{n+1}$. The Backward Euler method requiresor the monic quintic polynomial equationConstruct its $5\times 5$ companion matrixIts characteristic polynomial is $\det (zI-C_n)=p(z)$. Thus the polynomial roots needed for this update are exactly the eigenvalues of $C_n$.

At each step, assemble $C_n$, compute its eigenvalues, for example with the QR algorithm, and choose its unique positive real eigenvalue as $y_{n+1}$. To justify this selection, letIf $y_n>0$, then $F(0)=-y_n<0$, $F(y_n)=h{y}_n^5+ht{y}_n^2>0$, andThe intermediate value theorem and strict monotonicity prove there is exactly one positive polynomial root, located in $(0,y_n)$. Starting at $y_0=1$ therefore makes every update well defined, positive and decreasing. In numerical arithmetic, select the nearly real eigenvalue in this interval and check the polynomial residual $|F(z)|$; a final Newton root-finding iteration can refine it. This reduces each update to the requested eigenvalue problem.

The Backward Euler method uses the new point:Writing $a=h(x_n+h)>0$ gives the quadratic equation $a{y}_{n+1}^2+y_{n+1}-y_n=0$. Starting from $y_n>0$, its two polynomial roots have opposite signs. The positive root, which continues the solution as $h\to 0$, isThe other root is negative and does not approach $y_n$ as $h\to 0$. Rationalizing the numerator gives an equivalent form that avoids cancellation error when the step size is small:The defining quadratic equation gives $0<y_{n+1}<y_n$ for this root, so positivity is maintained inductively from $y_0=1$. The branch choice is part of specifying the Backward Euler method here: the algebraic update alone has two solutions.

For $f(x,y)=-x{y}^2$, the explicit Euler method evaluates the right-hand side at the known point $(x_n,y_n)$. ThereforeThis update requires no root finding.

For $0\le x\le 1$ and every real $z$,where $2|z|\le 1+z^2$ was used. The mean value theorem therefore proves Lipschitz continuity in $y$ with constant $1$.

Let $e_n=y(x_n)-y_n$. The explicit Euler method and the exact-solution identity from part (i) giveUsing the requested local truncation error bound $|{\tau }_n|\le h/2$, including at $n=0$, yieldsIteration and the finite geometric series giveSince $1+h\le e^h$ and $Nh=1$,Keeping the stronger bound $|{\tau }_n|\le h/4$ from part (i) improves the same proof to $(e-1)h/4$. The requested inequality follows already from the weaker bound.

Here $y^{\prime }=x/(1+y^2)$ and $y(0)=1$. Since $x\ge 0$, the solution is nondecreasing, so $y(x)\ge 1$. In fact $0\le y^{\prime }\le 1/2$ on $[0,1]$, which also bounds the solution and ensures it extends across this entire interval. The smooth right-hand side makes $y$ twice continuously differentiable, with second derivativeFor $y\ge 1$ and $0\le x\le 1$,so $y^{\prime \prime }\ge 0$. The subtracted term in the first expression is nonnegative, giving the useful boundThe integral form of the Taylor remainder now givesTherefore the normalized local truncation error satisfies the stronger estimateThis proves the requested bound for $n=1,\dots ,N-1$, and also for $n=0$, which will be needed to start the global error estimate. For $N=1$ the stated range is empty, but the estimate at $n=0$ still applies.

No. Distinct singular values do not force uniqueness in the spectral norm. For an explicit example takeThe singular values of $A$ are $2>1>0$. Every $B_t$ has matrix rank one, and its error in the spectral norm isThus there are infinitely many best rank-one approximations even with a strict gap between the singular values.

The same mechanism works for every $1\le r<n$ under the question's positive singular values. Starting with $A_r$, defineWhen the singular values are distinct, ${\sigma }_1+t\ge {\sigma }_1-{\sigma }_{r+1}>0$, so every $B_t$ has matrix rank $r$, whileThe spectral norm measures only the largest remaining singular value, leaving room to change retained components without increasing that maximum. This is the nonuniqueness of spectral-norm low-rank approximation. The exceptional case $r=0$ is unique for the elementary reason given above.

Take the identity matrixBoth $B_1$ and $B_2$ have matrix rank one, andThey are distinct and both attain the minimum spectral norm error from the preceding part. Hence a best low-rank approximation of rank one need not be unique. More generally, every orthogonal projection $u{u}^T$ with $\Vert u{\Vert }_2=1$ gives another such minimizer.

Retain the first $r$ terms in the singular value decomposition:The Eckart-Young-Mirsky theorem says that this truncated singular value decomposition is a best low-rank approximation in the spectral norm. The remaining singular values of the difference are ${\sigma }_{r+1},\dots ,{\sigma }_n$, together with zeros, soFor $r\ge 1$ the displayed matrix has matrix rank exactly $r$, since every retained singular value is positive. For $r=0$, the only matrix of matrix rank zero is the zero matrix, and the same formula gives error ${\sigma }_1$.