Past exam of the numerical linear algebra and continuous optimisation course of the University of Oxford / 2007 / A / 1

The outer product has entries $\boxed{(x{y}^T{)}_{ij}=x_i{y}_j}$. Every column is a multiple of $x$, and at least one column is nonzero because $y\ne 0$. Its column space is therefore exactly $\mathrm{span}\{x\}$, giving $\boxed{\mathrm{rank}(x{y}^T)=1}$.

A Householder matrix is $H=I-2v{v}^T/(v^{T}v)$, $v\ne 0$. With $P=v{v}^T/(v^{T}v)$ one has $P^T=P$ and $P^2=P$, so $H^{T}H=(I-2P{)}^2=I$. Thus it is an orthogonal matrix and is its own inverse.

To express an arbitrary orthogonal $Q$ as a product of these reflections, argue by induction on its dimension. Its first column $q$ is a unit vector. If $q\ne e_1$, take $v=q-e_1$. Then $v^{T}q=1-q_1$ and $v^{T}v=2(1-q_1)$, so the corresponding reflection satisfies $Hq=q-v=e_1$. If $q=e_1$, no reflection is required. The resulting orthogonal matrix has first column $e_1$; orthogonality of its columns then makes its first row $e_1^T$ as well. It is consequently $\mathrm{diag}(1,Q^{\prime })$ with $Q^{\prime }$ orthogonal. Apply the induction hypothesis to $Q^{\prime }$ and embed its reflections in the full space. In dimension one the only possibilities are $1$, the empty product, and $-1$, itself a reflection. Reversing the performed transformations proves the Householder factorization of an orthogonal matrix, with at most the dimension many reflections.

A full real singular value decomposition iswhere $\Sigma \in {\mathbb{R}}^{m\times n}$ has only diagonal entries, with ${\sigma }_1\ge \cdots \ge {\sigma }_{\min (m,n)}\ge 0$. These are the singular values. The number of positive singular values is the matrix rank. The columns of $V$ are right singular vectors and those of $U$ are left singular vectors, giving orthonormal input and output coordinate systems.

For a rectangular diagonal matrix $D$, define $D^{+}\in {\mathbb{R}}^{n\times m}$ by transposing its shape and takingEntry by entry, $D{D}^{+}D=D$ and $D^{+}D{D}^{+}=D^{+}$. For $A=U\Sigma V^T$, orthogonality therefore givessince $A{A}^{+}A=U\Sigma {\Sigma }^{+}\Sigma V^T=A$, and similarly $A^{+}A{A}^{+}=A^{+}$. This candidate is the Moore-Penrose inverse. It also makes $A{A}^{+}$ and $A^{+}A$ symmetric, although those two extra conditions are not required by the question's stated definition. The two displayed conditions alone need not specify a unique generalized inverse.