## 每週問題 January 5, 2015

Let $\mathbf{u}=(1,-1,-1,1,1)^T$. Determine $(I+2\mathbf{u}\mathbf{u}^T)(I+\mathbf{u}\mathbf{u}^T)^{-1}\mathbf{u}$.

$\displaystyle \left(I+2\mathbf{u}\mathbf{u}^T\right)\mathbf{x}=\left(I+2\mathbf{u}\mathbf{u}^T\right)\frac{1}{6}\mathbf{u}=\frac{1}{6}\left(\mathbf{u}+2\mathbf{u}(\mathbf{u}^T\mathbf{u})\right)=\frac{11}{6}\mathbf{u}=\frac{11}{6}\left[\!\!\begin{array}{r} 1\\ -1\\ -1\\ 1\\ 1 \end{array}\!\!\right]$

\displaystyle\begin{aligned} I&=(I+\mathbf{u}\mathbf{u}^T)^{-1}(I+\mathbf{u}\mathbf{u}^T)\\ &=(I+c\mathbf{u}\mathbf{u}^T)(I+\mathbf{u}\mathbf{u}^T)\\ &=I+\mathbf{u}\mathbf{u}^T+c\mathbf{u}\mathbf{u}^T+c\mathbf{u}\mathbf{u}^T\mathbf{u}\mathbf{u}^T\\ &=I+(1+6c)\mathbf{u}\mathbf{u}^T. \end{aligned}

\displaystyle\begin{aligned} (I+2\mathbf{u}\mathbf{u}^T)(I+\mathbf{u}\mathbf{u}^T)^{-1}\mathbf{u} &=(I+2\mathbf{u}\mathbf{u}^T)\left(I-\frac{1}{6}\mathbf{u}\mathbf{u}^T\right)\mathbf{u}\\ &=(I+2\mathbf{u}\mathbf{u}^T)\left(\mathbf{u}-\frac{1}{6}\mathbf{u}\mathbf{u}^T\mathbf{u}\right)\\ &=(I+2\mathbf{u}\mathbf{u}^T)\frac{1}{6}\mathbf{u}\\ &=\frac{11}{6}\mathbf{u}. \end{aligned}

### One Response to 每週問題 January 5, 2015

1. SmallSun Lai 說道：

good