Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Whenever there is a transpose or inverse of a group of matrices, I just reverse their order. For eg: $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$ and $(ABC)^{T} = C^{T}B^{T}A^{T}$

But usually, I am taking this reverse "rule" for granted without really knowing why I have to reverse their order whenever there is an inverse or transpose.

What is the reason for reversing their order?

share|improve this question
1  
Well, if you don't reverse the factors the equalities are simply not true! –  Mariano Suárez-Alvarez Jul 4 '11 at 23:28
10  
In the morning, you put on your socks then your shoes. But when you come home, you have to take off your shoes before your socks! –  Corey Jul 4 '11 at 23:34
add comment

2 Answers

up vote 7 down vote accepted

For the inverse, you can just apply the definition and compute:

$(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=aa^{-1}=id$

and similarly $(b^{-1}a^{-1})(ab)=id$. Hence $b^{-1}a^{-1}$ is an (and by uniqueness of inverses, the) inverse of $ab$.

A popular way to 'explain' this, is by saying "the inverse of 'putting on socks and then shoes', is 'taking off shoes and then socks'".

For the transpose, you can also apply the definition: if $A$ has matrix elements $(a_{ij})_{ij}$, then $A^T$ has matrix elements $(a_{ji})_{ij}$. Then compute the matrix elements of $(AB)^T$ and of $B^TA^T$ and see that they are equal.

A more conceptual way however is the following: taking the dual is a contravariant functor. That is, a linear map $A:V\to W$ has dual map $A^*:W^*\to V^*$ between the dual spaces (in the other direction!), and this is compatible with composition: $(A\circ B)^*=B^*\circ A^*$. The definition is $A^*=-\circ A$, so just precompose the linear functional with $A$; the composition property is then immediate. Now if $A$ has matrix $a_{ij}$ with repsect to some bases of $V,W$, then the matrix of $A^*$ w.r.t. the corresponding dual bases of $V^*,W^*$ is $a_{ji}$ (this is an easy check by writing out the definitions).

share|improve this answer
3  
For transposes, consider rectangular matrices: $AB$ may be defined but not $BA$, if they are not square. Hence, $(AB)^T$ cannot be $A^T B^T$ because they do not "match", while if $AB$ is defined so is $B^T A^T$. Of course, this is not a proof that $(AB)^T=B^T A^T$ but it does help to remember the right equation. –  lhf Jul 5 '11 at 1:06
    
Nice interpretation, it is fresh to me. –  Sunni Jul 5 '11 at 1:19
    
Thanks a lot! :) –  xenon Jul 5 '11 at 6:33
add comment

For inverses, there is a "common sense" interpretation. To undo a multistep procedure, you undo each step in the reverse order in which you did it. Example

Procedure: Dress your feet
1.  Put on socks.
2.  Put on shoes.
3.  Tie shoes.

To undo:
1.  Untie shoes
2.  Remove shoes
3.  Remove socks.

This principle is applicable to composition of functions in general. My students find the explanation plausible and palatable.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.