Numerical Linear Algebra with Applications

$\begin{array}{l}{H}_{u_i}H=\left[\begin{array}{ccc}{I}^{(i–1)\times (i–1)}& 0^{(i–1)\times 3}& 0^{(i–1)\times (n–i–2)}\\ 0^{3\times (i–1)}& {\overline{H}}_{u_i}^{3\times 3}& 0^{3\times (n–i–2)}\\ 0^{(n–i–2)\times (i–1)}& 0^{(n–i–2)\times 3}& I^{(n–i–2)\times (n–i–2)}\end{array}\right]\text{?}\left[\begin{array}{ccc}{H}_1^{(i–1)\times (i–1)}& H_2^{(i–1)\times 3}& H_3^{(i–1)\times (n–i–2)}\\ H_4^{3\times (i–1)}& H_5^{3\times 3}& H_6^{3\times (n–i–2)}\\ H_7^{(n–i–2)\times (i–1)}& H_8^{(n–i–2)\times 3}& H_9^{(n–i–2)\times (n–i–2)}\end{array}\right]\\ \text{?}=\left[\begin{array}{ccc}{H}_1^{(i–1)\times (i–1)}& H_2^{(i–1)\times 3}& H_3^{(i–1)\times (n–i–2)}\\ {\overline{H}}_{u_i}^{3\times 3}{H}_4^{3\times (i–1)}& {\overline{H}}_{u_i}^{3\times 3}{H}_5^{3\times 3}& {\overline{H}}_{u_i}^{3\times 3}{H}_6^{3\times (n–i–2)}\\ H_7^{(n–i–2)\times (i–1)}& H_8^{(n–i–2)\times 3}& H_9^{(n–i–2)\times (n–i–2)}\end{array}\right]\end{array}$

The only computation that need be done involves rows $i:i+\text{2}$, and columns 1:n. The same type of analysis shows that the product $\left(H_{u_i}H\right)H_{u_i}$ is done using rows 1:n and columns $i:i+\text{2}$. The final preproduct product ...