## 每週問題 August 4, 2014

Let

$\displaystyle A=\frac{1}{2}\left[\!\!\begin{array}{crrr} 1&1&\sqrt{2}&0\\ 1&1&-\sqrt{2}&0\\ 1&-1&0&\sqrt{2}\\ 1&-1&0&-\sqrt{2} \end{array}\!\!\right]\left[\!\!\begin{array}{ccr} 1&1&0\\ 0&2&0\\ 0&0&-2\\ 0&0&0 \end{array}\!\!\right]\left[\!\!\begin{array}{rrc} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\[0.5em] -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\[0.5em] 0&-\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}}\end{array}\!\!\right]$.

Find $\text{rank}\begin{bmatrix} AA^T&A\\ A^T&A^TA \end{bmatrix}$.

$\displaystyle \begin{bmatrix} AA^T&A\\ A^T&A^TA \end{bmatrix}=\begin{bmatrix} A&0\\ 0&A^T \end{bmatrix}\begin{bmatrix} A^T&I_3\\ I_4&A \end{bmatrix}$

$\displaystyle \text{rank}\begin{bmatrix} AA^T&A\\ A^T&A^TA \end{bmatrix}=\text{rank}\begin{bmatrix} A&0\\ 0&A^T \end{bmatrix}=\text{rank}A+\text{rank}A^T=2\text{rank}A$

$\displaystyle B=\left[\!\!\begin{array}{ccr} 1&1&0\\ 0&2&0\\ 0&0&-2\\ 0&0&0 \end{array}\!\!\right]$

$\displaystyle Q=\frac{1}{2}\left[\!\!\begin{array}{crrr} 1&1&\sqrt{2}&0\\ 1&1&-\sqrt{2}&0\\ 1&-1&0&\sqrt{2}\\ 1&-1&0&-\sqrt{2} \end{array}\!\!\right],~~P=\left[\!\!\begin{array}{rrc} \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\[0.5em] -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\[0.5em] 0&-\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}}\end{array}\!\!\right]$

$\displaystyle \text{rank}A=\text{rank}B=\text{rank}\left[\!\!\begin{array}{ccr} 1&1&0\\ 0&2&0\\ 0&0&-2\\ 0&0&0 \end{array}\!\!\right]=3$

$\displaystyle \begin{bmatrix} A^T&I_3\\ I_4&A \end{bmatrix}\to\begin{bmatrix} I_4&A\\ A^T&I_3 \end{bmatrix}\to\begin{bmatrix} I_4&A\\ 0&I_3-A^TA \end{bmatrix}$

\displaystyle \begin{aligned} \text{rank}\begin{bmatrix} A^T&I_3\\ I_4&A \end{bmatrix}&=\text{rank}\begin{bmatrix} I_4&A\\ 0&I_3-A^TA \end{bmatrix}\\ &=\text{rank}I_4+\text{rank}(I_3-A^TA)\\ &=4+\text{rank}(I_3-A^TA),\end{aligned}

$\displaystyle B^TB=\begin{bmatrix} 1&1&0\\ 1&5&0\\ 0&0&4 \end{bmatrix}$

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