AxB=B ????

yazid12111980

Hello, This my first post .. I hope to find someone can help me.

My question:
Assume that we have a 3x3 matrix B, is there a matrix A such that

AxB=B

??

Thanks

micromass

Take A the identity matrix. That is, take

[tex]A=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)[/tex]

yazid12111980

I tried that .. it doesnt work for cross product !!

yazid12111980

Its not a multiplication .. its a cross product. For example:

A =

1 2 3
4 5 6
7 8 9

>> B=eye(3)

B =

1 0 0
0 1 0
0 0 1

>> cross(B,A)

ans =

0 8 -6
-7 0 3
4 -2 0

???

micromass

So, how do you define the cross product for matrices??

Bloodthunder

This is a pretty good website to show you how to do matrix multiplication. Put in the matrices and they'll show you the working.

http://easycalculation.com/matrix/ma...iplication.php

D H

Quote by yazid12111980

Its not a multiplication .. its a cross product.

I would double check on that if I were you. The standard definition of A×B is that of matrix multiplication:

[tex](A\times B)_{ij} = \sum_k A_{ik} B_{kj}[/tex]

Interpreting A×B to mean Matlab's cross(A,B) when A and B are matrices is extremely non-standard.

Sillerpojken

Cross products for matrices can be done in Matlab, and it seems to be defined as several cross products in the matrix. Are you sure this is what you want? I have never seen this before. AxB has always been ordinary matrix multiplication.

Simon_Tyler

@Yazid: I'm not sure there is even a sensible way to define a cross product for matrices in 3 dimensions. The closest you can probably get to it is just the matrix commutator. It's a antisymmetric product that takes two matrices and returns a matrix.

Then [A,B] = B is basically an eigenvalue problem for the adjoint representation of A...
where you've been given an eigenvector (B) and need to find an operator (ad(A)) with that eigenvector.

Simon_Tyler

Actually, matlab does cross products on arrays of 3d vectors.
It works something like (I don't have matlab, so I can't be sure)

So if A = (a₁,a₂,...,a_n) and B=(b₁,b₂,...,b_n) with a_i and b_i 3d vectors, then

cross(A,B) = ( cross(a₁,b₁), cross(a₂,b₂), ..., cross(a_n,b_n) )

In Mathematica, the example provided can be written as

Code:

A = {{1, 2, 3}, 
     {4, 5, 6}, 
     {7, 8, 9}};
B = IdentityMatrix[3];

Transpose[Table[Cross[Transpose[B][[i]], Transpose[A][[i]]], {i, 3}]]
(* Output: 
{{ 0,  8, -6}, 
 {-7,  0,  3}, 
 { 4, -2,  0}}
*)

So for n=3, the original problem becomes

cross(A,B)=B ==> (a₁xb₁, a₂xb₂, a₃xb₃) = (b₁, b₂, b₃)

which requires that a_ixb_i=b_i, something that is never true...

pwsnafu

Quote by Simon_Tyler

In Mathematica, the example provided can be written as

I'd use the Thread command myself.

Simon_Tyler

@pwsnafu

I couldn't get Thread to work without Cross complaining about its arguments - although the error messages can be suppressed. Overall, I found that a combination of Apply at level 1 (@@@) the fastest, but not by much. Table is definitely the clearest to someone who is not used to functional programming.

Here's some timings using my custom TimeAv function. The FullForm of Transpose, instead of its typeset form, definitely makes the whole thing less readable...

Code:

In[1]:= A = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        B = IdentityMatrix[3];

In[3]:= SetOptions[TimeAv, Method -> {"MinNum", 10000}, "BlockSize" -> 1000];

In[4]:= Transpose[Table[Cross[Transpose[B][[i]], Transpose[A][[i]]], 
                  {i, 3}]] // TimeAv

During evaluation of In[4]:= Total wall time is 4.817133, 
total cpu time is 4.07 and total time spent evaluating the expression is 4.07

During evaluation of In[4]:= The expression was evaluated 10000 times, 
in blocks of 1000 runs. This yields a mean timing of 0.000407 
with a blocked standard deviation of 7.81025*10^-6.

Out[4]= {0.000407, {{0, 8, -6}, {-7, 0, 3}, {4, -2, 0}}}

In[5]:= Transpose[Apply[Cross, Transpose[{Transpose[B], Transpose[A]}], 
                        {1}]] // TimeAv

During evaluation of In[5]:= Total wall time is 3.847493, 
total cpu time is 3.85 and total time spent evaluating the expression is 3.85

During evaluation of In[5]:= The expression was evaluated 10000 times, 
in blocks of 1000 runs. This yields a mean timing of 0.000385 
with a blocked standard deviation of 5.*10^-6.

Out[5]= {0.000385, {{0, 8, -6}, {-7, 0, 3}, {4, -2, 0}}}

In[6]:= Quiet[Transpose[Thread[Cross[Transpose[B], Transpose[A]]]], 
              Cross::nonn1] // TimeAv

During evaluation of In[6]:= Total wall time is 4.462055, 
total cpu time is 4.46 and total time spent evaluating the expression is 4.46

During evaluation of In[6]:= The expression was evaluated 10000 times, 
in blocks of 1000 runs. This yields a mean timing of 0.000446 
with a blocked standard deviation of 9.16515*10^-6.

Out[6]= {0.000446, {{0, 8, -6}, {-7, 0, 3}, {4, -2, 0}}}

yazid12111980	AxB=B ???? Hello, This my first post .. I hope to find someone can help me. My question: Assume that we have a 3x3 matrix B, is there a matrix A such that AxB=B ?? Thanks

micromass Mentor Blog Entries: 8	Take A the identity matrix. That is, take [tex]A=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)[/tex]

yazid12111980	AxB=B ???? Its not a multiplication .. its a cross product. For example: A = 1 2 3 4 5 6 7 8 9 >> B=eye(3) B = 1 0 0 0 1 0 0 0 1 >> cross(B,A) ans = 0 8 -6 -7 0 3 4 -2 0 ???

Bloodthunder	This is a pretty good website to show you how to do matrix multiplication. Put in the matrices and they'll show you the working. http://easycalculation.com/matrix/ma...iplication.php

Sillerpojken	Cross products for matrices can be done in Matlab, and it seems to be defined as several cross products in the matrix. Are you sure this is what you want? I have never seen this before. AxB has always been ordinary matrix multiplication.