Past exam of the numerical linear algebra course of the University of Oxford / 2025 / 1 / b

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.

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.

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.

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.