Is A./B different from B.\A?

조회 수: 24 (최근 30일)
Oliver Woodford
Oliver Woodford 2015년 6월 17일
편집: James Tursa 2021년 10월 29일
Given two matrices, A and B, will A./B ever give a different answer from B.\A, or are these two expressions equivalent?
It seems that even for complex numbers they return the same thing. E.g.
>> A = sqrt(randn(3));
>> B = sqrt(randn(3));
>> isequal(A./B, B.\A)
ans = 1
  댓글 수: 3
Ingrid
Ingrid 2015년 6월 17일
Stephen, what he is asking is something else namely
>> 2./3
ans =
0.6667
>> 3.\2
ans =
0.6667
so I think they will always give the same result
Stephen23
Stephen23 2015년 6월 17일
According to the documentation A.\B and B./A are the same:
  • ldivide: " B.\A divides each element of A by the corresponding element of B"
  • rdivide: " A./B divides each element of A by the corresponding element of B"
Unless the definition of "divide" is different, then these should be the same.

댓글을 달려면 로그인하십시오.

채택된 답변

James Tursa
James Tursa 2015년 6월 17일
I can't think of any reason why one would ever get different results for numeric types. I suppose there might be speed differences if one form used multi-threading and the other form didn't, but in tests I just ran they both appeared to take about the same amount of time.
User defined classes could of course overload them differently.
  댓글 수: 6
Bruno Luong
Bruno Luong 2021년 9월 30일
편집: Bruno Luong 2021년 9월 30일
"The only way to get non-commutative objects such as quaternions is to create a class for them"
But they are available in some toolboxes, e.g., https://www.mathworks.com/help/robotics/ref/quaternion.html
I don't have any of those toolboxes to check how "A./B" and "B.\A" works, but I expect them NOT give the same results for general cases:
James Tursa
James Tursa 2021년 10월 29일
편집: James Tursa 2021년 10월 29일
Yes, your expectations are correct. For the MATLAB toolbox quaternion class objects, the q./p and p.\q operations are implemented as expected by multiplying by the inverse, and since multiplication is non-commutative you get different results.
>> x = rand(1,4)-0.5; x = x/norm(x); q = quaternion(x);
>> x = rand(1,4)-0.5; x = x/norm(x); p = quaternion(x);
>> q
q =
quaternion
-0.62168 + 0.46748i + 0.58112j + 0.23933k
>> p
p =
quaternion
0.64169 + 0.60532i - 0.26832j + 0.38709k
>> q./p
ans =
quaternion
-0.17923 + 0.38713i + 0.24217j + 0.87141k
>> p.\q
ans =
quaternion
-0.17923 + 0.96545i + 0.17j - 0.082977k
>> q*conj(p)
ans =
quaternion
-0.17923 + 0.38713i + 0.24217j + 0.87141k
>> conj(p)*q
ans =
quaternion
-0.17923 + 0.96545i + 0.17j - 0.082977k
>> which quaternion
C:\Program Files\MATLAB\R2020a\toolbox\shared\rotations\rotationslib\@quaternion\quaternion.m % quaternion constructor
Note that the / and \ operators are not implemented for this class:
>> q/p
Error using /
Arguments must be numeric, char, or logical.
>> p\q
Error using \
Arguments must be numeric, char, or logical.

댓글을 달려면 로그인하십시오.

추가 답변 (2개)

Alberto
Alberto 2015년 6월 17일
Both are pointwise, but A./B divides every element in A by the same element in B. A.\B divides every element in B by the same element in A.
  댓글 수: 1
Oliver Woodford
Oliver Woodford 2015년 6월 17일
I didn't ask about A.\B though

댓글을 달려면 로그인하십시오.


H. Sh. G.
H. Sh. G. 2021년 9월 28일
Hi every body.
I wonder what kind of calculations the division of a matrix (X) by a row vector (y), i.e. X/y, does, where both have the same number of columns.
The result is a column vector of the same number of rows that X has.
Recall that X./y divides all elements of each column in X by the element of y in the same column, resulting in a matrix with the same size of X.
  댓글 수: 4
Bruno Luong
Bruno Luong 2021년 9월 29일
편집: Bruno Luong 2021년 9월 30일
"Yes, I know that b/A gives pinv(A)*b."
It's incorrect. In some cases b/A is A*pinv(b) (and not the opposite as you wrote)
B=rand(2,4);
A=rand(3,4);
A/B
ans = 3×2
0.3160 0.6389 -0.4427 0.6439 2.4027 -0.5932
A*pinv(B)
ans = 3×2
0.3160 0.6389 -0.4427 0.6439 2.4027 -0.5932
but when rank(b) < size(b,1) such formula is not correct
B=rand(4,2);
A=rand(3,2);
A/B
ans = 3×4
1.2479 0 0 0.0022 -0.2076 0 0 1.0500 0.1178 0 0 1.4824
A*pinv(B)
ans = 3×4
1.1939 -0.2063 0.1426 0.0418 -0.1644 0.3918 0.3595 0.5231 0.1610 0.4850 0.5542 0.7518
Now to your question.
In case B = b is a row vector, let consider the system
% x*B = A;
x is column vector (since is a row vector). This system works row by row of x and A independenty. So consider a row equation
% x(i) * b = A(i,:).
So what you ask is which scalar x(i) that when multiplying with a vector (b) must be equal to another vector A(i,:). Such solution does not exist unless A(i,:) is proportional to b. In general MATLAB returns the least square solution:
% x(i) = dot(A(i,:),b) / dot(b,b)
Illustration test :
b=rand(1,4);
A=rand(3,4);
A/b
ans = 3×1
0.4635 0.8082 0.3771
A1=A(1,:); x1=dot(A1,b)/dot(b,b)
x1 = 0.4635
A2=A(2,:); x2=dot(A2,b)/dot(b,b)
x2 = 0.8082
A3=A(3,:); x3=dot(A3,b)/dot(b,b)
x3 = 0.3771
% Or all together
x=A*b'/(b*b')
x = 3×1
0.4635 0.8082 0.3771
Note that for a row vector b, pinv(b) is b'/(b*b').
And x*B will not match A, unless all the rows of A are proportional to b (which is a strong coincidence in general).
H. Sh. G.
H. Sh. G. 2021년 9월 29일
Thanks Bruno,
This explains my question well.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Coordinate Transformations에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by