samples/texts/227520/page_16.md

implies

$$\begin{array}{cccc}& \xi (A_i)=l+r=\xi (B_i)=r+\xi (R_i)\mathrm{.}& & \text{(4.20)}\end{array}\backslash xi(A\_i)\; =\; l\; +\; r\; =\; \backslash xi(B\_i)\; =\; r\; +\; \backslash xi(R\_i).\; \backslash tag\{4.20\}$$

This implies that $\xi(R_i) = l$, and, since $l \ge 1$, that $R_i$ is singular, and that $l \le b+r$. Consequently, there exists a matrix $(X_i | Y_i)$ of rank $l$ with $X_i \in \mathbb{R}^{b \times l}$ and $Y_i \in \mathbb{R}^{r \times l}$, such that

$$R_i\left(\begin{array}{c}{X}_i\\ Y_i\end{array}\right)=0,(4.21)R\_i\; \backslash begin\{pmatrix\}\; X\_i\; \backslash \backslash \; Y\_i\; \backslash end\{pmatrix\}\; =\; 0,\; \backslash quad\; (4.21)$$

or, equivalently,

$$(I_b+U(\mathrm{\Sigma }-{\sigma }_i^{\ast }{I}_n)U)X_i+U{W}_i{Y}_i=0(4.22)(I\_b\; +\; U\; (\backslash Sigma\; -\; \backslash sigma\_i^*\; I\_n)\; U)\; X\_i\; +\; U\; W\_i\; Y\_i\; =\; 0\; \backslash quad\; (4.22)$$

$$\begin{array}{cccc}& W_{i}U{X}_i=0.& & \text{(4.23)}\end{array}W\_i\; U\; X\_i\; =\; 0.\; \backslash tag\{4.23\}$$

Computation of the Eigenvectors. Given the null space $(X_i | Y_i)$ of $R_i$ (a discussion of how it can be computed will be deferred for the moment), define

$${\widehat{Z}}_i:=(\mathrm{\Sigma }-{\sigma }_i^{\ast }{I}_n)U{X}_i+W_i{Y}_i\in {\mathbb{R}}^{n\times t}\mathrm{.}(4.24)\backslash hat\{Z\}\_i\; :=\; (\backslash Sigma\; -\; \backslash sigma\_i^*\; I\_n)\; U\; X\_i\; +\; W\_i\; Y\_i\; \backslash in\; \backslash mathbb\{R\}^\{n\; \backslash times\; t\}.\; \backslash quad\; (4.24)$$

The columns of $\hat{Z}_i$ form a set of $l$ linearly independent eigenvectors of $S$ corre- sponding to $\sigma_i^*$. Because of (4.23) and (4.17)

Because of

$$\begin{array}{cccc}& \mathrm{rank}({\widehat{Z}}_i)=l,& & \text{(4.25)}\end{array}\backslash operatorname\{rank\}(\backslash hat\{Z\}\_i)\; =\; l,\; \backslash tag\{4.25\}$$

the columns of $\hat{Z}_i$ span the whole eigenspace corresponding to $\sigma_i^*$. (4.25) can be proved indirectly. Assume, that rank($\hat{Z}_i$) < $l$. Then there must be another eigenvector $z_i \in \mathbb{R}^n$ with

$$\mathrm{\parallel }{z}_i{\mathrm{\parallel }}_2=1,z_i{\widehat{Z}}_i=0,\backslash |z\_i\backslash |\_2\; =\; 1,\; \backslash quad\; z\_i\; \backslash hat\{Z\}\_i\; =\; 0,$$

and

$$\begin{array}{cccc}& (\mathrm{\Sigma }+UU-{\sigma }_i^{\ast }{I}_n)z_i=0.& & \text{(4.26)}\end{array}(\backslash Sigma\; +\; UU\; -\; \backslash sigma\_i^*\; I\_n)\; z\_i\; =\; 0.\; \backslash tag\{4.26\}$$

Premultiplication with $U (\Sigma - \sigma_i^* I_n)$ implies

$$(I_b+U(\mathrm{\Sigma }-{\sigma }_i^{\ast }{I}_n)U)U{z}_i-U{W}_i{W}_i{z}_i=0.(4.27)(I\_b\; +\; U\; (\backslash Sigma\; -\; \backslash sigma\_i^*\; I\_n)\; U)\; U\; z\_i\; -\; U\; W\_i\; W\_i\; z\_i\; =\; 0.\; \backslash quad\; (4.27)$$

At the same time, (4.26) implies

$$\begin{array}{cccc}& UU{z}_i=-(\mathrm{\Sigma }-{\sigma }_i^{\ast }{I}_n)z_i,& & \text{(4.28)}\end{array}UU\; z\_i\; =\; -(\Sigma \; -\; \sigma \_i^*\; I\_n)\; z\_i,\; \backslash tag\{4.28\}$$

and premultiplication by $W_i$ (using(4.17)) yields

$$W_{i}UU{z}_i=-W_i(\mathrm{\Sigma }-{\sigma }_i^{\ast }{I}_n)z_i=0.(4.29)W\_i\; U\; U\; z\_i\; =\; -W\_i\; (\backslash Sigma\; -\; \backslash sigma\_i^*\; I\_n)\; z\_i\; =\; 0.\; \backslash qquad\; (4.29)$$