[R] Matlab inv() and R solve() differences

[Cleber Nogueira Borges]

Hello,

is there a upper limit to kappa value where I can consider a matrix 
well-conditioned?


Cleber



Kingsford Jones wrote:
> I suppose the solution is unstable because x is ill-conditioned:
>
>   
>> x
>>     
>        [,1]   [,2]   [,3]  [,4]
> [1,]  0.133  0.254 -0.214 0.116
> [2,]  0.254  0.623 -0.674 0.139
> [3,] -0.214 -0.674  0.910 0.011
> [4,]  0.116  0.139  0.011 0.180
>   
>> cor(x)
>>     
>            [,1]       [,2]       [,3]       [,4]
> [1,]  1.0000000  0.9963557 -0.9883690  0.8548065
> [2,]  0.9963557  1.0000000 -0.9976663  0.8084090
> [3,] -0.9883690 -0.9976663  1.0000000 -0.7663847
> [4,]  0.8548065  0.8084090 -0.7663847  1.0000000
>
>   
>> kappa(x)
>>     
> [1] 2813.326
>
> hth,
>
> Kingsford Jones
>
> On Thu, Jan 29, 2009 at 7:00 PM, Joseph P Gray <jpgray at uwm.edu> wrote:
>   
>> I submit the following matrix to both MATLAB and R
>>
>> x=  0.133 0.254 -0.214 0.116
>>    0.254 0.623 -0.674 0.139
>>   -0.214 -0.674 0.910 0.011
>>    0.116 0.139  0.011 0.180
>>
>> MATLAB's inv(x) provides the following
>>  137.21 -50.68 -4.70 -46.42
>> -120.71  27.28 -8.94 62.19
>> -58.15   6.93  -7.89  36.94
>>  8.35   11.17 10.42 -14.82
>>
>> R's solve(x) provides:
>> 261.94 116.22 150.92 -267.78
>> 116.22 344.30 286.68 -358.30
>> 150.92 286.68 252.96 -334.09
>> -267.78 =358.30 -334.09 475.22
>>
>> inv(x)*x = I(4)
>> and solve(x)%*%x = I(4)
>>
>> Is there a way to obtain the MATLAB result in R?
>>
>> Thanks for any help.
>>
>>
>> Pat Gray
>>
>> ______________________________________________
>>