每週問題 December 29, 2014

Let $\mathbb{R}^{2\times 2}$ be the vector space formed by $2\times 2$ real matrices. Let $A=\begin{bmatrix} 1&2\\ 0&1 \end{bmatrix}$ and $B=\begin{bmatrix} 0&1\\ 1&1 \end{bmatrix}$. Consider the following transformation from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$:

$\displaystyle T(X)=XA-AX$.

Determine a matrix $\hat{X}$ that minimizes $\Vert T(\hat{X})-B\Vert_F$, where $\Vert\cdot\Vert_F$ denotes Frobenius norm.

$X=\begin{bmatrix} a&b\\ c&d \end{bmatrix}$，則

$\displaystyle T(X)=\begin{bmatrix} a&b\\ c&d \end{bmatrix}\begin{bmatrix} 1&2\\ 0&1 \end{bmatrix}-\begin{bmatrix} 1&2\\ 0&1 \end{bmatrix}\begin{bmatrix} a&b\\ c&d \end{bmatrix}=\begin{bmatrix} -2c&2a-2d\\ 0&2c \end{bmatrix}$

$\displaystyle \boldsymbol{\beta}=\{V_1,V_2,V_3,V_4\}=\left\{\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix},\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix},\begin{bmatrix} 0&0\\ 1&0 \end{bmatrix},\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}\right\}$

$\displaystyle T(V_1)=\begin{bmatrix} 0&2\\ 0&0 \end{bmatrix},~T(V_2)=\begin{bmatrix} 0&0\\ 0&0 \end{bmatrix},~T(V_3)=\left[\!\!\begin{array}{rc} -2&0\\ 0&2 \end{array}\!\!\right],~T(V_4)=\left[\!\!\begin{array}{cr} 0&-2\\ 0&0 \end{array}\!\!\right]$

$\displaystyle \begin{bmatrix} T \end{bmatrix}_{\boldsymbol{\beta}}=\begin{bmatrix} \begin{bmatrix} T(V_1) \end{bmatrix}_{\boldsymbol{\beta}}& \begin{bmatrix} T(V_2) \end{bmatrix}_{\boldsymbol{\beta}}& \begin{bmatrix} T(V_3) \end{bmatrix}_{\boldsymbol{\beta}}& \begin{bmatrix} T(V_4) \end{bmatrix}_{\boldsymbol{\beta}} \end{bmatrix}=\left[\!\!\begin{array}{ccrr} 0&0&-2&0\\ 2&0&0&-2\\ 0&0&0&0\\ 0&0&2&0 \end{array}\!\!\right]$

$\displaystyle [Z]_{\boldsymbol{\beta}}=\frac{\mathbf{u}_1^T\mathbf{b}}{\mathbf{u}_1^T\mathbf{u}_1}\mathbf{u}_1+\frac{\mathbf{u}_2^T\mathbf{b}}{\mathbf{u}_2^T\mathbf{u}_2}\mathbf{u}_2=\frac{1}{1}\begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix}+\frac{1}{2}\left[\!\!\begin{array}{r} -1\\ 0\\ 0\\ 1 \end{array}\!\!\right]=\left[\!\!\begin{array}{r} -\frac{1}{2}\\ 1\\ 0\\ \frac{1}{2} \end{array}\!\!\right]$

$\hat{X}=\begin{bmatrix} a&b\\ c&d \end{bmatrix}$ 使得 $T(\hat{X})=\left[\!\!\begin{array}{rc} -\frac{1}{2}&1\\ 0&\frac{1}{2} \end{array}\!\!\right]$。比較 $T(\hat{X})$ 的矩陣表達式，可得 $2c=1/2$$2a-2d=1$，解出 $c=1/4$$d=(2a-1)/2$，所求為 $\hat{X}=\begin{bmatrix} a&b\\ \frac{1}{4}&\frac{2a-1}{2} \end{bmatrix}$，其中 $a$$b$ 是任意數。

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