Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 326 / 1

For a bounded operator $A:X\to Y$ between Hilbert spaces, the Moore--Penrose inverse of a Hilbert-space operator is defined onand has range $\mathcal{R}(A^{†})=\mathcal{N}(A{)}^{\perp }$. Its Penrose equations areEquivalently, $A^{†}A=P_{\mathcal{N}(A{)}^{\perp }}$ and $A{A}^{†}$ is the restriction of $P_{\overline{\mathcal{R}(A)}}$ to $\mathcal{D}(A^{†})$.

The statement is true without an additional rank assumption. If $A=U\Sigma V^H$ is a singular value decomposition, then

This statement is also always true. The same singular value decomposition gives

This statement is true. The matrix $A^{†}A$ is the orthogonal projection onto the row space $\mathcal{R}(A^H)$, so its eigenvalues are zero and one. Consequently

The Reverse-order law for the Moore--Penrose inverse is false in general. TakeThenA sufficient condition isso that $A$ has full column rank and $B$ has full row rank. Indeed $A^{†}A=I_n$ and $B{B}^{†}=I_n$; these identities make $B^{†}{A}^{†}$ satisfy all four Penrose equations for $AB$.

If $B={\sup }_j|d_j|<\infty$, then for every $f\in {\ell }^2$,so the diagonal operator on sequence space is bounded. Conversely, for the standard unit vector $e_j$,so boundedness of $D$ implies ${\sup }_j|d_j|\le \Vert D\Vert <\infty$.

Part i gives $\Vert D\Vert \le B$. Conversely, choose indices $j_k$ with $|d_{j_k}|\to B$. Since $\Vert e_{j_k}{\Vert }_2=1$,Taking the limit proves $\boxed{\Vert D\Vert =B}$, whether or not the supremum is attained.

The Moore-Penrose inverse acts coordinatewise ason the domainThe coordinates supported where $d_j=0$ form $\mathcal{R}(D{)}^{\perp }$ and are sent to zero. The inverse is continuous exactly when the nonzero diagonal entries are bounded away from zero,apart from the trivial all-zero operator, whose inverse is zero. Otherwise unit vectors along entries tending to zero show that $D^{†}$ is unbounded.

For $d_j=1/j$, the equation $Du=f$ forces $u_j=j{f}_j$. It has a solution in ${\ell }^2$ exactly whenFor example, $f_j=1/j$ defines an element of ${\ell }^2$, but its forced preimage is the constant sequence $u_j=1$, which is not in ${\ell }^2$. Thus existence can fail. Since every $d_j$ is nonzero, $D$ has trivial kernel and any solution is unique.

The proposed stability property is false. Set $u^{\ast }=0$, $f^{\ast }=0$, and take $u_n=e_n$. Thenwhile $\Vert u_n-u^{\ast }{\Vert }_2=1$ for every $n$. Hence the inverse is not continuous on its range and the recovery problem is not stable with respect to perturbations. This is the standard unbounded inverse on a nonclosed operator range.