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对SVD唯一性的理解

feng posted @ Wed, 17 Oct 2012 00:12:52 +0800 in 未分类 , 10923 readers

 

这篇文章主要讨论一下对Singular value decomposition的理解。
 
SVD告诉我们,对于任何一个$m\times n$的矩阵$A$,都存在这样的一个分解:
\[A=U\Sigma V'\]
其中$U$是$m\times m$的酉矩阵,也就是$UU^* = I$;$V$是一个$n\times n$的酉矩阵;$\Sigma$是一个$m\times n$的矩阵,非对角上的元素都是0,对角线上的元素都是非负的实数,不管$A$是实数矩阵,还是酉空间中的矩阵,或者说有虚数元素。
 
定理告诉了我们总是存在一个这样的分解的,但并不是说这样的分解是唯一的。比如说有一个permutation matrix,$J$。permutation matrix就是把单位矩阵的行进行重排列,或者列进行重排列。比如说把单位矩阵的第一行和第二行进行交换,第四行和第九行进行交换,交换后的矩阵就是一个permutaion matrix。一个矩阵$A$,左乘一个$J$,相当于对$A$的相应行进行行交换。右乘的话,相当于列交换。两个$J$相乘为一个单位矩阵,相当于对单位矩阵交换了两行之后,再交换一次,不变。
 
回到正题,任何一个$J$拿到了之后,我们有下面的式子成立
\[A=(UJ_m)(J_m\Sigma J_n)(VJ_n)'\]
$UJ_m$依然是一个酉矩阵,$VJ_n$也依然是一个酉矩阵。这里$J$的小角标表示$J$的大小,是$m\times m$的还是$n\times n$的。如果说$m\ge n$,并且$J_n$是$J_m$左上角的,$J_m$右下角是一个单位阵,右上角和左下角都是0矩阵的话,那么$J_m\Sigma J_n$依然也是一个对角阵。之前对$J_m$和$J_n$的描述,相当于是说,如果$J_m$使得$\Sigma$的第$i$行和第$j$行交换的话,那么$J_n$应该使得$\Sigma$的第$i$列和第$j$列进行交换。从而使得$J_m\Sigma J_n$依然是一个非对角线上元素都是0的矩阵,相当于是将$\Sigma$的对角线上的元素进行了一个重新排列。这是事实告诉我们,SVD的分解是不唯一的。
 
接下来再考虑一点,如果说$\Sigma$唯一确定之后,这个$U$和$V$是否能够唯一确定的呢?我们再引入一个矩阵,用$K_m$表示,是一个$m\times m$的对角方阵,对角线上的元素具有$e^{i\phi}$的形式,其中$\phi$可是互相不相同。根据定义,我们有$K_mK_m^* = I$。那么
\[A=UK_m K_m^*\Sigma K_n(VK_n)^*\]
其中$U K_m$和$VK_n$分别仍然是酉阵,如果说$K_m$的对角线上第$i$个元素和$K_n$对角线上第$i$个元素相同的话,那么$K_m^*\Sigma K_n = \Sigma$。也就是说,如果$U$的第$i$列乘以$e^{i\phi}$,并且$V$的第$i$列也乘以$e^{i\phi}$,那么结论依然成立。
 
然后,我们继续疑问:如果$\Sigma$给定,并且允许$U$和$V$差一个$e^{i\phi}$的话,那么$U$和$V$是否能够唯一确定?
这一种情况,比较麻烦。如果说$\sigma_i$都是non-degenerate的话,那么是唯一确定的。否则的话,依然不能够唯一确定。
 
对degenerate的定义是这样子的。如果$\sigma_i$是degenerate的,那么它有两个互相独立的singular vector。
在这里补充一下,singular value就是$\sigma_i$,而$\mathbf{u}_i$和$\mathbf{v}_i$分别是singular vector,其中要求$i<=\min{m, n}$。并且有下面的式子成立
\[A\mathbf{v}_i = \sigma \mathbf{u}_i\]
并且
\[A^*\mathbf{u}_i = \sigma \mathbf{v}_i\]
 
如果说$\sigma_i$都是不一样的,那么所有的$\sigma_i$都是non-degenerate的,也就是说这个时候,$U$和$V$都是唯一确定的,在一定意义下。如果说$A$是一个实数矩阵,$U$和$V$可以也是实数的,这样的话,对于其唯一确定的理解是,差一个符号。
 
这里对于degenerate的讨论比较少,以后有时间补上。
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