On 10/14/13 23:31, Šárka Rusá wrote:
Dear R,
I have a problem concerning the inverse of a matrix. My matrix is quite
large and I extracted just linearly independent columns from the matrix.
However, the function solve() still cannot invert the matrix - I get an
error that it is computationally s
Dear R,
I have a problem concerning the inverse of a matrix. My matrix is quite
large and I extracted just linearly independent columns from the matrix.
However, the function solve() still cannot invert the matrix - I get an
error that it is computationally singular. Is there any way how to extrac
e Cramer's Rule, but it will give you a feel for the
issue.
HTH
Rex
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Berend Hasselman
Sent: Wednesday, March 16, 2011 1:33 PM
To: r-help@r-project.org
Subject: Re: [R] Si
On 16-03-2011, at 21:11, David Winsemius wrote:
>
> On Mar 16, 2011, at 1:32 PM, Berend Hasselman wrote:
>> .
>> svd(a) indicates the problem.
>>
>> largest singular value / smallest singular value=1e17 (condition number)
>> --> reciprocal condition number=1e-17
>> and the standard solve c
; Sent: Wednesday, March 16, 2011 1:12 PM
> To: Berend Hasselman
> Cc: r-help@r-project.org
> Subject: Re: [R] Singularity problem
>
>
> On Mar 16, 2011, at 1:32 PM, Berend Hasselman wrote:
>
> >
> > Peter Langfelder wrote:
> >>
> >> On Wed, Mar
[mailto:r-help-boun...@r-project.org] On
Behalf Of Berend Hasselman
Sent: Wednesday, March 16, 2011 1:33 PM
To: r-help@r-project.org
Subject: Re: [R] Singularity problem
Peter Langfelder wrote:
>
> On Wed, Mar 16, 2011 at 8:28 AM, Feng Li <m...@feng.li> wrote:
>> Dear R,
>>
&
On Mar 16, 2011, at 1:32 PM, Berend Hasselman wrote:
Peter Langfelder wrote:
On Wed, Mar 16, 2011 at 8:28 AM, Feng Li wrote:
Dear R,
If I have remembered correctly, a square matrix is singular if and
only
if
its determinant is zero. I am a bit confused by the following co
Peter Langfelder wrote:
>
> On Wed, Mar 16, 2011 at 8:28 AM, Feng Li wrote:
>> Dear R,
>>
>> If I have remembered correctly, a square matrix is singular if and only
>> if
>> its determinant is zero. I am a bit confused by the following code error.
>> Can someone give me a hint?
>>
On Wed, Mar 16, 2011 at 8:28 AM, Feng Li wrote:
> Dear R,
>
> If I have remembered correctly, a square matrix is singular if and only if
> its determinant is zero. I am a bit confused by the following code error.
> Can someone give me a hint?
>
>> a <- matrix(c(1e20,1e2,1e3,1e3),2)
>> det(a)
> [1]
Numeric underflow. Try qr.solve(a)
Allan
On 16/03/11 15:28, Feng Li wrote:
Dear R,
If I have remembered correctly, a square matrix is singular if and only if
its determinant is zero. I am a bit confused by the following code error.
Can someone give me a hint?
a<- matrix(c(1e20,1e2,1e3,1e3),
Dear R,
If I have remembered correctly, a square matrix is singular if and only if
its determinant is zero. I am a bit confused by the following code error.
Can someone give me a hint?
> a <- matrix(c(1e20,1e2,1e3,1e3),2)
> det(a)
[1] 1e+23
> solve(a)
Error in solve.default(a) :
system is compu
On Jan 8, 2010, at 4:50 PM, Moohwan Kim wrote:
Dear R family
I have a problem with invertibility in a matrix.
m1 <- ar(x, method='mle')
Error in solve.default(res$hessian * length(x)) :
Lapack routine dgesv: system is exactly singular
How could I avoid this problem?
Take out the colline
Dear R family
I have a problem with invertibility in a matrix.
m1 <- ar(x, method='mle')
Error in solve.default(res$hessian * length(x)) :
Lapack routine dgesv: system is exactly singular
How could I avoid this problem?
Best
Moohwan
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