为什么矩阵F范数的平方等于奇异值平方的和

Others 2021-11-25 14:19:32 views: null

首先考虑一个M乘N的矩阵H，其 F范数的平方 定义为矩阵所有元素的模平方的和

\|\mathbf{H}\|_{F}^{2}=\sum_{n=1}^{N} \sum_{m=1}^{M}[\mathbf{H}]_{m n}^{2}

也可以等价表示为 H乘H的共轭转置的迹

\begin{array}{l} \operatorname{Tr}\left\{\mathbf{H} \mathbf{H}^{H}\right\}=\sum_{m=1}^{M}[\mathbf{H}]_{m,:}\left[\mathbf{H}^{H}\right]_{:, m}=\sum_{m=1}^{M}[\mathbf{H}]_{m,:}[\mathbf{H}]_{m,:}^{H} \\ =\sum_{m=1}^{M}\left\|[\mathbf{H}]_{m,:}\right\|_{2}^{2}=\sum_{m=1}^{M} \sum_{n=1}^{N}[\mathbf{H}]_{m, n}^{2} \end{array}

 

根据矩阵的SVD分解，矩阵H可以表示为

\mathbf{H}=\mathrm{U} \Lambda \mathrm{V}^{H}

其中 矩阵U和V是酉阵，即

\mathrm{U}\mathrm{U}^{H}=\mathrm{V}\mathrm{V}^{H}=\mathbf{I},

中间上尖矩阵是由奇异值组成的对角阵。（\mathbf{H} \mathbf{H}^{H}或\mathbf{H}^{H}\mathbf{H}的特征值=H的奇异值的平方）

 

接下来利用矩阵SVD分解和迹的循环等价性质，即可证明：矩阵F范数的平方等于奇异值平方的和

\begin{array}{l} \|\mathbf{H}\|_{F}^{2}=T r\left\{\mathbf{H} \mathbf{H}^{H}\right\} \\ =\operatorname{Tr}\left\{\mathrm{U} \Lambda \mathrm{V}^{H} \mathrm{~V} \Lambda^{H} \mathrm{U}^{H}\right\} \\ =\operatorname{Tr}\left\{\mathrm{U} \Lambda \Lambda^{H} \mathrm{U}^{H}\right\} \\ =\operatorname{Tr}\left\{\Lambda \Lambda^{H} \mathrm{U}^{H} \mathrm{U}\right\} \\ =\operatorname{Tr}\left\{\Lambda \Lambda^{H}\right\} \\ =\sum_{i=1}^{\min (M, N)}[\Lambda]_{i}^{2} \\ =\sum_{i=1}^{\min (M, N)} \lambda_{i} \end{array}

 

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Origin blog.csdn.net/jianti9962/article/details/111348967
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