如何在Matlab中查找矩阵是否为奇异 [英] How to find if a matrix is Singular in Matlab

问题描述

我使用下面的函数从优化器中为给定的一组猜测lambda生成beta.

I use the function below to generate the betas for a given set of guess lambdas from my optimiser.

运行时，我经常收到以下警告消息:

When running I often get the following warning message:

警告:矩阵对于工作精度是唯一的. 在9的NSS_betas中 在德拉姆达(Dlambda)，19岁 在36岁的Individual_Lambdas中

Warning: Matrix is singular to working precision. In NSS_betas at 9 In DElambda at 19 In Individual_Lambdas at 36

我希望能够从解决方案集中排除形成奇异矩阵的任何beta，但是我不知道如何进行测试?

I'd like to be able to exclude any betas that form a singular matrix form the solution set, however I don't know how to test for it?

我一直在尝试使用rcond()，但是我不知道在单数和非单数之间进行截断的位置是什么?

I've been trying to use rcond() but I don't know where to make the cut off between singular and non singular?

当然，如果Matlab生成警告消息，它已经知道矩阵是否为奇异，所以如果我能找到该变量的存储位置，我可以使用它吗?

Surely if Matlab is generating the warning message it already knows if the matrix is singular or not so if I could just find where that variable was stored I could use that?

function betas=NSS_betas(lambda,data)

mats=data.mats2'; 
lambda=lambda;
yM=data.y2';
nObs=size(yM,1);
G= [ones(nObs,1) (1-exp(-mats./lambda(1)))./(mats./lambda(1)) ((1-exp(-mats./lambda(1)))./(mats./lambda(1))-exp(-mats./lambda(1))) ((1-exp(-mats./lambda(2)))./(mats./lambda(2))-exp(-mats./lambda(2)))];

betas=G\yM;
r=rcond(G);

end

感谢您的建议:

在将lambda值设置为相等后，我测试了下面的所有三个示例，因此求出了奇异矩阵

I tested all three examples below after setting the lambda values to be equal so guiving a singular matrix

 if (~isinf(G))
  r=rank(G);
  r2=rcond(G);
  r3=min(svd(G)); 
 end

r = 3，r2 = 2.602085213965190e-16; r3 = 1.075949299504113e-15;

r=3, r2 =2.602085213965190e-16; r3= 1.075949299504113e-15;

因此，在此测试中，rank()和rcond()假设我采用下面给出的基准值，则可以正常工作.

So in this test rank() and rcond () worked assuming I take the benchmark values as given below.

但是当我有两个接近但不完全相等的值时会发生什么?

However what happens when I have two values that are close but not exactly equal?

如何确定太近的地方?

推荐答案

rcond是前往此处的正确方法.如果它接近机器精度零，则您的矩阵是奇异的.我通常会选择:

rcond is the right way to go here. If it nears the machine precision of zero, your matrix is singular. I usually go with:

if( rcond(A) < 1e-12 )
    % This matrix doesn't look good
end

您可以根据自己的需要尝试使用一个值，但是使用MATLAB甚至接近奇异值的矩阵求逆也会产生垃圾结果.

You can experiment with a value that suites your needs, but taking the inverse of a matrix that is even close to singular with MATLAB can produce garbage results.

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