Question about A\b

8 次查看（过去 30 天）
Stephen Wang
Stephen Wang 2017-12-8
编辑： Matt J 2017-12-9
Hi everyone,
I really like A\b (just 3 letters and it can do magic). However, I run into this problem.
>> A = [1 1; 1 1]
A =
1 1
1 1
>> b = [1; 1]
b =
1
1
>> A\b
Warning: Matrix is singular to working precision.
(Type "warning off MATLAB:singularMatrix" to suppress this warning.)
ans =
NaN
NaN
I understand that A is not invertible. However there is a solution and I kind of expect MATLAB will return [1; 0].
Or maybe I have the wrong expectation.
  3 个评论
显示 1更早的评论
John D'Errico
John D'Errico 2017-12-9
编辑：John D'Errico 2017-12-9
It looks like everyone in that same class will be asking this virtually identical question.
Stephen Wang
Stephen Wang 2017-12-9
Wow. What a coincidence. I am not in that class. Actually my question is more on MATLAB A\b than on linear algebra itself.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Matt J
Matt J 2017-12-8
编辑：Matt J 2017-12-8
It's really all a question of which linear solver you use. Each has its own idea of which of the infinite solutions to choose from. One alternative is,
>> pinv(A)*b
ans =
0.5000
0.5000
Another is,
>> lscov(A,b)
Warning: A is rank deficient to within machine precision.
> In lscov (line 200)
ans =
1.0000
0
  3 个评论
显示 1更早的评论
Stephen Wang
Stephen Wang 2017-12-9
Hi Matt,
Thanks for your suggestions. Thanks for introducing the matlab function lscov().
My apology. In my original post, I should have stated my question more clearly. So here it goes again.
I would like to solve the good old (you can even call it boring) least squares equation in MATLAB,
A*x = b
where A can be any matrix, i.e. square, skinny (eg. 5-by-3) or
fat (eg. 3-by-5 matrix). Both x and b are just column vectors
having the same number of rows as matrix A.
Since this is a least squares problem, this means that there is always a solution. If there is a unique solution for x, then it is great and we are done. However, if there are infinite number of solutions for x, then I want x to have maximum number of elements to be zero.
Based on what I want, I always think that (until now) the answer is the magic 3 letters in MATLAB:
A\b
And I thought that A\b always works regardless of A. So it caught me by surprise that in my toy example, A\b breaks down.
>> A = [1 1; 1 1], b = [1; 1], x = A\b
A =
1 1
1 1
b =
1
1
Warning: Matrix is singular to working precision.
(Type "warning off MATLAB:singularMatrix" to suppress this warning.)
x =
NaN
NaN
So I guess I should not use A\b anymore (this is what I have been memorizing so far).
So let's try lscov. It does not quite always work either. Acutally it is worst. LSCOV errors out in the following example
>> A = [1 1], b = [1], x = lscov(A, b)
A =
1 1
b =
1
Warning: A is rank deficient to within machine precision.
(Type "warning off MATLAB:lscov:RankDefDesignMat" to suppress this warning.)
> In lscov (line 200)
The logical indices in position 1 contain a true value outside of the array bounds.
Error in lscov (line 201)
R = R(keepCols,keepCols);
Any other suggestion?
Matt J
Matt J 2017-12-9
编辑：Matt J 2017-12-9
You may have found a corner case, but LSCOV should always return the solution with a maximum number of zeros. From the documentation:
x = lscov(A,B) returns the ordinary least squares solution to the linear system of equations A*x = B, i.e., x is the n-by-1 vector that minimizes the sum of squared errors (B - A*x)'*(B - A*x), where A is m-by-n, and B is m-by-1. B can also be an m-by-k matrix, and lscov returns one solution for each column of B. When rank(A) < n, lscov sets the maximum possible number of elements of x to zero to obtain a "basic solution".

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by