线性核是正定的吗? [英] Is linear kernel positive definite?

问题描述

在很多文章中，线性核(两个矩阵的内积)被列为正定性，但是当我用玩具数据集尝试线性正核时，正定性测试将返回负结果.我检查了MATLAB SVM函数的线性核.

In a lot of articles, the linear kernel (inner product of two matrices) is listed as positive definite however when I try it with a toy dataset, positive definiteness test returns negative result. I checked the MATLAB SVM function for linear kernel.

线性核函数是一行命令，

The linear kernel function is one line command,

K=(u*v')

但是，在执行主svm_train函数的此步骤之后，它会使用K进行另一项操作，

However after this step in the main svm_train function it does another operation using K,

kx= (kx+kx')/2 + diag(1./boxconstraint) 

其中kx是K，diag(1./boxconstraint)只是大小为kx的对角矩阵，所得的kx通过了正定性检验.作为对此步骤的解释，它说 '% ensure function is symmetric.'我也检查了libsvm，但在那儿找不到此附加操作.

Where kx is K and diag(1./boxconstraint) is just a diagonal matrix of size kx and resultant kx pass the positive definiteness test. As an explanation of this step it says '% ensure function is symmetric.' I also checked libsvm but I could not find this additional operation there.

但是，内积已经是对称的，此步骤通常用于将不定矩阵转换为正定矩阵.我对内部乘积内核为什么不通过正定性测试感到困惑?

However the inner product is already symmetric and this step is usually used to turn indefinite matrices into positive definite matrices. I am a bit confused about why inner product kernel does not pass the positive definiteness test?

推荐答案

如果u，v都是行向量(即K计算为标量)，则仅当K = dot(u，v)时，K才是正定的. 0.

If u,v are both row vectors (i.e. K evaluates to a scalar), K will only be positive definite if K=dot(u,v)>0.

如果u和v是更通用的矩阵(不是行向量)并且u〜= v，则K = u * v'通常不会对称，更不用说正定了.即使当u = v时，除非u具有完整的行级，否则K将是非负定数，但不会严格地是正定数.但是，假设所有boxconstraint(i)> 0，则附加矩阵1./diag(boxconstraint)严格为正定.将非负定矩阵加到严格正定矩阵上，总是会得到严格正定结果.

If u and v are more general matrices (not row vectors) and u~=v, then K=u*v' will not generally be symmetric, let alone positive definite. Even when u=v, K will be non-negative definite, but will not be strictly positive definite unless u has full row rank. However, the additional matrix 1./diag(boxconstraint) is strictly positive definite assuming all boxconstraint(i)>0. Adding a non-negative definite matrix to a strictly positive definite matrix produces a strictly positive definite result always.

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