$$$\left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right]$$$ 的奇异值分解

求$$$\left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right]$$$的奇异值分解。

求该矩阵的转置：$$$\left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right]^{T} = \left[\begin{array}{cc}1 & 0\\1 & 1\end{array}\right]$$$（步骤详见矩阵转置计算器）。

将矩阵与其转置相乘：$$$W = \left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right]\cdot \left[\begin{array}{cc}1 & 0\\1 & 1\end{array}\right] = \left[\begin{array}{cc}2 & 1\\1 & 1\end{array}\right]$$$（步骤详见矩阵乘法计算器）。

现在，求$$$W$$$的特征值与特征向量（步骤参见特征值与特征向量计算器）。

特征值：$$$- \frac{-3 + \sqrt{5}}{2}$$$，特征向量：$$$\left[\begin{array}{c}- \frac{-1 + \sqrt{5}}{2}\\1\end{array}\right]$$$。

特征值：$$$\frac{\sqrt{5} + 3}{2}$$$，特征向量：$$$\left[\begin{array}{c}\frac{1 + \sqrt{5}}{2}\\1\end{array}\right]$$$。

求非零特征值 ($$$\sigma_{i}$$$) 的平方根：

$$$\sigma_{1} = \frac{\sqrt{2} \sqrt{3 - \sqrt{5}}}{2}$$$

$$$\sigma_{2} = \frac{\sqrt{2} \sqrt{\sqrt{5} + 3}}{2}$$$

矩阵 $$$\Sigma$$$ 是一个对角线上为 $$$\sigma_{i}$$$、其余元素为零的矩阵：$$$\Sigma = \left[\begin{array}{cc}\frac{\sqrt{2} \sqrt{3 - \sqrt{5}}}{2} & 0\\0 & \frac{\sqrt{2} \sqrt{\sqrt{5} + 3}}{2}\end{array}\right]$$$。

矩阵 $$$U$$$ 的列是归一化（单位）向量：$$$U = \left[\begin{array}{cc}\frac{- \sqrt{10} + \sqrt{2}}{2 \sqrt{5 - \sqrt{5}}} & \frac{\sqrt{2} + \sqrt{10}}{2 \sqrt{\sqrt{5} + 5}}\\\frac{\sqrt{2}}{\sqrt{5 - \sqrt{5}}} & \frac{\sqrt{2}}{\sqrt{\sqrt{5} + 5}}\end{array}\right]$$$（关于求单位向量的步骤，参见 单位向量计算器）。

现在，$$$v_{i} = \frac{1}{\sigma_{i}}\cdot \left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right]^{T}\cdot u_{i}$$$：

$$$v_{1} = \frac{1}{\sigma_{1}}\cdot \left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right]^{T}\cdot u_{1} = \frac{1}{\frac{\sqrt{2} \sqrt{3 - \sqrt{5}}}{2}}\cdot \left[\begin{array}{cc}1 & 0\\1 & 1\end{array}\right]\cdot \left[\begin{array}{c}\frac{- \sqrt{10} + \sqrt{2}}{2 \sqrt{5 - \sqrt{5}}}\\\frac{\sqrt{2}}{\sqrt{5 - \sqrt{5}}}\end{array}\right] = \left[\begin{array}{c}\frac{1 - \sqrt{5}}{2 \sqrt{5 - 2 \sqrt{5}}}\\\frac{3 - \sqrt{5}}{2 \sqrt{5 - 2 \sqrt{5}}}\end{array}\right]$$$ (步骤详见 矩阵数乘计算器 和 矩阵乘法计算器).

$$$v_{2} = \frac{1}{\sigma_{2}}\cdot \left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right]^{T}\cdot u_{2} = \frac{1}{\frac{\sqrt{2} \sqrt{\sqrt{5} + 3}}{2}}\cdot \left[\begin{array}{cc}1 & 0\\1 & 1\end{array}\right]\cdot \left[\begin{array}{c}\frac{\sqrt{2} + \sqrt{10}}{2 \sqrt{\sqrt{5} + 5}}\\\frac{\sqrt{2}}{\sqrt{\sqrt{5} + 5}}\end{array}\right] = \left[\begin{array}{c}\frac{1 + \sqrt{5}}{2 \sqrt{2 \sqrt{5} + 5}}\\\frac{\sqrt{5} + 3}{2 \sqrt{2 \sqrt{5} + 5}}\end{array}\right]$$$ (步骤详见 矩阵数乘计算器 和 矩阵乘法计算器).

因此，$$$V = \left[\begin{array}{cc}\frac{1 - \sqrt{5}}{2 \sqrt{5 - 2 \sqrt{5}}} & \frac{1 + \sqrt{5}}{2 \sqrt{2 \sqrt{5} + 5}}\\\frac{3 - \sqrt{5}}{2 \sqrt{5 - 2 \sqrt{5}}} & \frac{\sqrt{5} + 3}{2 \sqrt{2 \sqrt{5} + 5}}\end{array}\right]$$$。

矩阵$$$U$$$、$$$\Sigma$$$和$$$V$$$使得初始矩阵满足$$$\left[\begin{array}{cc}1 & 1\\0 & 1\end{array}\right] = U \Sigma V^T$$$。

$$$U = \left[\begin{array}{cc}\frac{- \sqrt{10} + \sqrt{2}}{2 \sqrt{5 - \sqrt{5}}} & \frac{\sqrt{2} + \sqrt{10}}{2 \sqrt{\sqrt{5} + 5}}\\\frac{\sqrt{2}}{\sqrt{5 - \sqrt{5}}} & \frac{\sqrt{2}}{\sqrt{\sqrt{5} + 5}}\end{array}\right]\approx \left[\begin{array}{cc}-0.525731112119134 & 0.85065080835204\\0.85065080835204 & 0.525731112119134\end{array}\right]$$$A

$$$\Sigma = \left[\begin{array}{cc}\frac{\sqrt{2} \sqrt{3 - \sqrt{5}}}{2} & 0\\0 & \frac{\sqrt{2} \sqrt{\sqrt{5} + 3}}{2}\end{array}\right]\approx \left[\begin{array}{cc}0.618033988749895 & 0\\0 & 1.618033988749895\end{array}\right]$$$A

$$$V = \left[\begin{array}{cc}\frac{1 - \sqrt{5}}{2 \sqrt{5 - 2 \sqrt{5}}} & \frac{1 + \sqrt{5}}{2 \sqrt{2 \sqrt{5} + 5}}\\\frac{3 - \sqrt{5}}{2 \sqrt{5 - 2 \sqrt{5}}} & \frac{\sqrt{5} + 3}{2 \sqrt{2 \sqrt{5} + 5}}\end{array}\right]\approx \left[\begin{array}{cc}-0.85065080835204 & 0.525731112119134\\0.525731112119134 & 0.85065080835204\end{array}\right]$$$A