Matlab, operator A\B

Question

What is the result of the operation A\B, where A(1, m) and B (1, m)?

In the manual it is written:

A\B returns a least-squares solution to the system of equations A*x= B.

So it means x = inv (A'*A)*A'*B? However, the matrix A'*A is singular...

Let us suppose:

A=[1 2 3]
B=[6 7 6]
A\B

0         0         0
0         0         0
2.0000    2.3333    2.0000

If ve use MLS:

C = inv (A'*A)   singular matrix
C = pinv(A'*A)

0.0051    0.0102    0.0153
0.0102    0.0204    0.0306
0.0153    0.0306    0.0459

D= C*A'*B

0.4286    0.5000    0.4286
0.8571    1.0000    0.8571
1.2857    1.5000    1.2857

So results A\B and inv (A'*A)*A'*B are different...

Solution 1

My MATLAB (R2010b) says quite a lot about what A\B does:

mldivide(A,B) and the equivalent A\B perform matrix left division (back slash). A and B must be matrices that have the same number of rows, unless A is a scalar, in which case A\B performs element-wise division — that is, A\B = A.\B.

If A is a square matrix, A\B is roughly the same as inv(A)*B, except it is computed in a different way. If A is an n-by-n matrix and B is a column vector with n elements, or a matrix with several such columns, then X = A\B is the solution to the equation AX = B. A warning message is displayed if A is badly scaled or nearly singular.

If A is an m-by-n matrix with m ~= n and B is a column vector with m components, or a matrix with several such columns, then X = A\B is the solution in the least squares sense to the under- or overdetermined system of equations AX = B. In other words, X minimizes norm(A*X - B), the length of the vector AX - B. The rank k of A is determined from the QR decomposition with column pivoting. The computed solution X has at most k nonzero elements per column. If k < n, this is usually not the same solution as x = pinv(A)*B, which returns a least squares solution.

mrdivide(B,A) and the equivalent B/A perform matrix right division (forward slash). B and A must have the same number of columns.

If A is a square matrix, B/A is roughly the same as B*inv(A). If A is an n-by-n matrix and B is a row vector with n elements, or a matrix with several such rows, then X = B/A is the solution to the equation XA = B computed by Gaussian elimination with partial pivoting. A warning message is displayed if A is badly scaled or nearly singular.

If B is an m-by-n matrix with m ~= n and A is a column vector with m components, or a matrix with several such columns, then X = B/A is the solution in the least squares sense to the under- or overdetermined system of equations XA = B.

Solution 2

x = inv (A'*A)*A'*B goes for over determined systems (i.e. which feature A as an n x m matrix with n>m; in these circumstances A'A is invertible).

In your case you have an under determined system.

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Thus, what may happen?

My opinion, although you can check, at least in your case:

when you do A\B matlab solves an optimization problem in the inverse sense w.r.t. the usual least squares, that is

 X = argmin_{X \in S} ||X||,

where S is the set of solutions. In other words, it gives you the solution of the system having minimum L^2 norm. (Consider that you can handle the problem by hands, at least in your case).

Comments

[justik]

@ FakeDIY. Thanks, but I had already read before I posted my question. However, the case three it is not clear for me...