inv(matrix) takes shorter time than \ operator

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Bramer
Bramer 2013-2-8
评论： Christine Tobler 2020-7-31
A=magic(5)
A =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
>> tic
inv(A)
toc
ans =
-0.0049 0.0512 -0.0354 0.0012 0.0034
0.0431 -0.0373 -0.0046 0.0127 0.0015
-0.0303 0.0031 0.0031 0.0031 0.0364
0.0047 -0.0065 0.0108 0.0435 -0.0370
0.0028 0.0050 0.0415 -0.0450 0.0111
Elapsed time is 0.006097 seconds.
>> tic
B=eye(5);
A\B
toc
ans =
-0.0049 0.0512 -0.0354 0.0012 0.0034
0.0431 -0.0373 -0.0046 0.0127 0.0015
-0.0303 0.0031 0.0031 0.0031 0.0364
0.0047 -0.0065 0.0108 0.0435 -0.0370
0.0028 0.0050 0.0415 -0.0450 0.0111
Elapsed time is 0.008993 seconds.
Just a general question, shouldn't have been the other way round? inv(matrix) is supposed to be more computationally intensive than the \ operation.
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Matt J
Matt J 2020-7-30
We can only assume it is a vector, based on where Armin wrote: "where mu is the centroid of each cluster and x the current feature-vector"
Christine Tobler
Christine Tobler 2020-7-31
Based on the formula (x-mu)*cov^(-1)*(x-mu)', you'd expect the output here to be symmetric positive (semi-)definite. If cov is numerically full rank, you can use the following to compute this matrix in a way that will keep it symmetric positive semi-definite.
R = chol(cov);
A = (x-mu)/R;
A = A*A';
If cov is low-rank (positive semi-definite), you could consider doing the same using the ldl function instead of chol.

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采纳的回答

Shashank Prasanna
编辑：Shashank Prasanna 2013-2-8
While inv indeed does involve more operations than \ because it involves inverse as well as multiplication, \ is just one operation. It is safer as you can use it on non-full rank of the system which is determined by QR decomp. I persuade you to look at this literature from Cleve the creator of MATLAB: Linear Equation and of course this link in the documentation.
To support what James said, your example is a very bad one
Not the best but a supporting example is as follows. Using a larger matrix, and making sure the random numbers are the same
rng(5);tic, inv(rand(1000))*rand(1000,1); toc
rng(5);tic, rand(1000)\rand(1000,1); toc
Elapsed time is 0.321691 seconds.
Elapsed time is 0.142591 seconds.
You will have to run this a couple of times. The first run may result in MATLAB performing some optimizations and memory allocations.
  1 个评论
Hunter Colegrove
Hunter Colegrove 2020-7-30
Is there a reason inv(A) uses LU decomposition as the algorithm to solve the inverse of the matrix? Is there a computational advantage of some sort?

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更多回答（2 个）

James Tursa
James Tursa 2013-2-8
1) Your example is so small as to make the timing irrelevant
2) You burden the \ method with creating another variable, B
3) The \ operation doesn't know ahead of time that B is the identity matrix, so it may not be able to take computational advantage of that fact.
4) Why do you think inv is more computationally intensive than \ ?
  2 个评论
Matt J
Matt J 2013-2-8
编辑：Matt J 2013-2-8
5) Omitting semicolons and displaying to the screen is also adding unnecessary overhead.
6) You have to re-run the code multiple times, preferably inside a function to see a fully optimized execution. There is no way inv(A) on a 5x5 matrix takes 0.006097 seconds on any reasonable machine.
the cyclist
the cyclist 2013-2-8
编辑：the cyclist 2013-2-8
Regarding James' point #4, he may have gotten the idea from the documentation of the inv() function:
"In practice, it is seldom necessary to form the explicit inverse of a matrix. A frequent misuse of inv arises when solving the system of linear equations Ax = b. One way to solve this is with x = inv(A)*b. A better way, from both an execution time and numerical accuracy standpoint, is to use the matrix division operator x = A\b."
@Bramen, I think the larger worry with inv() is actually the numerical stability for arrays that are nearly singular. See, for example, this old blog post:
http://blogs.mathworks.com/loren/2007/05/16/purpose-of-inv/

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Matt J
Matt J 2013-2-8
It's not that inv(A) is supposed to be more computationally intensive than \. It's that A\b is faster than inv(A)*b, as demonstrated below.
N=2000;
A=rand(N);
b=rand(N,1);
tic;
inv(A)*b;
toc
Elapsed time is 0.509519 seconds.
tic
A\b;
toc
Elapsed time is 0.195016 seconds.
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Matt J
Matt J 2015-11-4
编辑：Matt J 2015-11-9
You shouldn't be comparing with (X'*X)\X'*y nor with an explicit call to qr(). You should be comparing with X\y, which is equivalent (in this case) to the QR method.
tic;
for i=1:10000
inv(X'*X)*X'*y;
end
toc;
tic;
for i=1:10000
(X'*X)\X'*y;
end
toc;
tic;
for i=1:10000
X\y;
end
toc;
Elapsed time is 0.082173 seconds.
Elapsed time is 0.157320 seconds.
Elapsed time is 0.086341 seconds.
That said, when you have a very tall X matrix, then using the normal equations directly can be faster, and also save memory. However, since X'*X has a higher condition number than X, it will also be less stable numerically and amplify noise, as demonstrated in the example below. So you need to be sure you have a well-conditioned X in order for this to be a good compromise.
d=30;
N=1e5;
X=diag(1:d).^2*rand(d,'single');
noise=randn(d,N);
z1 = X\noise;
Error1=std(z1(:)),
z2=(inv(X.'*X)*X.'*noise);
Error2=std(z2(:)),
Error1 =
1.6255
Error2 =
210.7583
Daniel
Daniel 2015-11-10
Oh, thanks for that answer. I wasn't aware of this. It should make the code faster and ease the problem of stability that I happened to have.

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