QR and polar retractions on Stiefel and Grassmann

By [F5], tangency gives $X^{\mathsf{T}}\eta +{\eta }^{\mathsf{T}}X=0$. Therefore for $v\mathrm{\ne }0$, $\mathrm{\parallel }(X+\eta )v{\mathrm{\parallel }}_2^2=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+v^{\mathsf{T}}(X^{\mathsf{T}}\eta +{\eta }^{\mathsf{T}}X)v+\mathrm{\parallel }\eta v{\mathrm{\parallel }}_2^2=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\eta v{\mathrm{\parallel }}_2^2>0.$ Thus $X+\eta$ has full column rank. More generally, for any full-rank $Y$, $v^{\mathsf{T}}{Y}^{\mathsf{T}}Yv=\mathrm{\parallel }Yv{\mathrm{\parallel }}_2^2>0$ for $v\mathrm{\ne }0$, so its Gram matrix is positive definite; [F1] supplies the QR factor, while [F2] diagonalizes the Gram matrix and [F3] supplies its positive square root and inverse.

For full-column-rank $Y=[y_1\cdots y_r]$, its columns are linearly independent. Apply Gram–Schmidt [F1], writing $u_j=y_j-{\sum }_{i<j}(q_i^{\mathsf{T}}{y}_j)q_i$ and $q_j=u_j\mathrm{/}\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_2$. The residuals are nonzero, so $Q=[q_1\cdots q_r]$ has orthonormal columns. Set $R=Q^{\mathsf{T}}Y$. For $i>j$, $q_i$ is orthogonal to $\mathrm{span}(y_1,\dots ,y_j)$, so $R_{ij}=0$; and $R_{jj}=q_j^{\mathsf{T}}{y}_j=\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_2>0$. Since each $y_j$ lies in $\mathrm{span}(q_1,\dots ,q_j)$, $QR=Y$. This constructs a reduced QR factorization with positive diagonal. To check uniqueness, if $Y=Q_1{R}_1=Q_2{R}_2$ are two such factorizations, then $Q_2^{\mathsf{T}}{Q}_1=R_2{R}_1^{-1}$ is orthogonal and upper triangular with positive diagonal. Its first column has only its first entry, which must be $1$ by unit norm and positivity; orthogonality zeros the rest of its first row. Induction on the remaining block makes it $I_r$, so $Q_1=Q_2$ and $R_1=R_2$. For the polar factor $U=Y(Y^{\mathsf{T}}Y{)}^{-1\mathrm{/}2}$, symmetry of the root and its defining square give $U^{\mathsf{T}}U=(Y^{\mathsf{T}}Y{)}^{-1\mathrm{/}2}(Y^{\mathsf{T}}Y)(Y^{\mathsf{T}}Y{)}^{-1\mathrm{/}2}=I_r$. Hence both outputs have orthonormal columns. For a Grassmann input, [F6] says a frame change $X\mapsto X{Q}_0$ changes $\eta$ to $\eta Q_0$, so the input $Y=X+\eta$ becomes $Y{Q}_0$ with the same column space; uniqueness of the positive root gives $(Q_0^{\mathsf{T}}{Y}^{\mathsf{T}}Y{Q}_0{)}^{-1\mathrm{/}2}=Q_0^{\mathsf{T}}(Y^{\mathsf{T}}Y{)}^{-1\mathrm{/}2}{Q}_0$, so the polar output is multiplied by $Q_0$. Thus the Grassmann outputs are well-defined.