MHB How Do You Prove (det A)aij = Cij(A) for an Orthogonal Matrix with det(A)=+1?

[ognik]

Hi, the question (from math methods for physicists) is: If A is orthogonal and det(A)=+1, show that (det A)a_ij = C_ij(A).

I know that if det(A)=+1, then we are looking at a rotation.
(Side question - I have seen that det(A) =-1 can be a reflection, but is 'mostly not reflections'; what does det(A)=-1 most often indicate then? A link to something simple that covers this would be nice :-) - I couldn't find anything)

But my main issue is to prove the above. I want to do this using indexing, so I tried:
By defn: det A = $ \sum_{}^{}{a}_{ij}({-1})^{i+j}{C}_{ij}(A) $
Then (det A)a_ij = $ {a}_{ij} \sum_{}^{}{a}_{ij}({-1}^{i+j}){C}_{ij}(A) $ ...

What struck me immediately is I don't know how to treat what happens when an indexed element is outside the summation? I am reasonably sure that a_ij is a single element here, so I can't include it in the summation. I can almost see an argument that by multiplying by the a_ijth element, I am 'selecting' only that element's cofactor out of the summation - but that seems too flimsy to me, would appreciate a better understanding.

Another side question is that I have the definition of a cofactor as: $ {C}_{ij}(A)=({-1})^{i+j}{M}_{ij}(A) $ Where {M}_{ij} is the minor...
It seems to me that $ ({-1})^{i+j} $ shouldn't be in both $ \sum_{}^{}{a}_{ij}({-1})^{i+j}{C}_{ij}(A) $ AND $ {C}_{ij}(A)=({-1})^{i+j}{M}_{ij}(A) $ ?

Thanks

 

[I like Serena]

Hi, the question (from math methods for physicists) is: If A is orthogonal and det(A)=+1, show that (det A)a_ij = C_ij(A).

I know that if det(A)=+1, then we are looking at a rotation.
(Side question - I have seen that det(A) =-1 can be a reflection, but is 'mostly not reflections'; what does det(A)=-1 most often indicate then? A link to something simple that covers this would be nice :-) - I couldn't find anything)

But my main issue is to prove the above. I want to do this using indexing, so I tried:
By defn: det A = $ \sum_{}^{}{a}_{ij}({-1})^{i+j}{C}_{ij}(A) $
Then (det A)a_ij = $ {a}_{ij} \sum_{}^{}{a}_{ij}({-1}^{i+j}){C}_{ij}(A) $ ...

What struck me immediately is I don't know how to treat what happens when an indexed element is outside the summation? I am reasonably sure that a_ij is a single element here, so I can't include it in the summation. I can almost see an argument that by multiplying by the a_ijth element, I am 'selecting' only that element's cofactor out of the summation - but that seems too flimsy to me, would appreciate a better understanding.

Another side question is that I have the definition of a cofactor as: $ {C}_{ij}(A)=({-1})^{i+j}{M}_{ij}(A) $ Where {M}_{ij} is the minor...
It seems to me that $ ({-1})^{i+j} $ shouldn't be in both $ \sum_{}^{}{a}_{ij}({-1})^{i+j}{C}_{ij}(A) $ AND $ {C}_{ij}(A)=({-1})^{i+j}{M}_{ij}(A) $ ?

Thanks

Hi ognik!

It seems to me we shouldn't worry much which summation applies to find $\det A$.
That's because we already know that $\det A = 1$.

Anyway, to answer your side questions first.
You are quite right that you have a superfluous $({-1})^{i+j}$ in your summation for $\det A$.
And if you have an indexed element outside of the summation, it should use different indices.

So it should be:
$$(\det A)a_{ij} = {a}_{ij} \sum_{k,l}^{}{a}_{kl}(-1)^{k+l}{M}_{kl}(A) = {a}_{ij} \sum_{k,l}^{}{a}_{kl}{C}_{kl}(A)$$To get back to your problem, perhaps you can use that an orthogonal matrix has the property that $A^{-1}=A^T$.
And that an inverse is given by $A^{-1}=\frac{1}{\det A}\big(C(A)\big)^T$.

 

[ognik]

Hi and thanks. I wasn't sure I could use that inverse relationship as it hasn't been covered in the text yet, but I can't see any other way. Having said that, i think I need some more understanding as you will see:
$$ If\: A^-1 =\frac{1}{\left| A \right|}.{C}^{T}\;then\:\left| A \right|\left({A}^{-1}\right)={C}^{T} $$
$$ Then \:\left| A \right|\left({A}^{-1}\right)_{ij} ={C}^{T}_{ij}\:\:but\:{C}^{T}_{ij}={C}_{ji}$$
So for what I am trying to prove, somehow $$\left({A}^{-1}\right)_{ij}\:must\:=A_{ji} $$
but I don't know why that would be true?

 

[ognik]

..sorry, of course I do! For an orthog matrix
$$ {A}^{-1}={A}^{T} \therefore ({A}^{-1})_{ij}={A}^{T}_{ij}={A}_{ji} $$
Thanks for your help (BTW, how do I add a thanks?)

 

[I like Serena]

Good! :)

Every post has a thanks button at the bottom right.

 

[ognik]

I also found that after I posted, one of those days ...
Would appreciate it if you had a look at http://mathhelpboards.com/advanced-applied-mathematics-16/confirm-equation-numerov-method-14831.html ... thanks again :-)