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$\begin{array}{l}{H}_{{u}_{i}}H=\left[\begin{array}{ccc}{I}^{\left(i–1\right)×\left(i–1\right)}& {0}^{\left(i–1\right)×3}& {0}^{\left(i–1\right)×\left(n–i–2\right)}\\ {0}^{3×\left(i–1\right)}& {\overline{H}}_{{u}_{i}}^{3×3}& {0}^{3×\left(n–i–2\right)}\\ {0}^{\left(n–i–2\right)×\left(i–1\right)}& {0}^{\left(n–i–2\right)×3}& {I}^{\left(n–i–2\right)×\left(n–i–2\right)}\end{array}\right]\text{?}\left[\begin{array}{ccc}{H}_{1}^{\left(i–1\right)×\left(i–1\right)}& {H}_{2}^{\left(i–1\right)×3}& {H}_{3}^{\left(i–1\right)×\left(n–i–2\right)}\\ {H}_{4}^{3×\left(i–1\right)}& {H}_{5}^{3×3}& {H}_{6}^{3×\left(n–i–2\right)}\\ {H}_{7}^{\left(n–i–2\right)×\left(i–1\right)}& {H}_{8}^{\left(n–i–2\right)×3}& {H}_{9}^{\left(n–i–2\right)×\left(n–i–2\right)}\end{array}\right]\\ \text{?}=\left[\begin{array}{ccc}{H}_{1}^{\left(i–1\right)×\left(i–1\right)}& {H}_{2}^{\left(i–1\right)×3}& {H}_{3}^{\left(i–1\right)×\left(n–i–2\right)}\\ {\overline{H}}_{{u}_{i}}^{3×3}{H}_{4}^{3×\left(i–1\right)}& {\overline{H}}_{{u}_{i}}^{3×3}{H}_{5}^{3×3}& {\overline{H}}_{{u}_{i}}^{3×3}{H}_{6}^{3×\left(n–i–2\right)}\\ {H}_{7}^{\left(n–i–2\right)×\left(i–1\right)}& {H}_{8}^{\left(n–i–2\right)×3}& {H}_{9}^{\left(n–i–2\right)×\left(n–i–2\right)}\end{array}\right]\end{array}$

The only computation that need be done involves rows $i:i+2$, and columns 1:n. The same type of analysis shows that the product $(HuiH)Hui$ is done using rows 1:n and columns $i:i+2$. The final preproduct product ...

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