Proving Matrix Symmetries of A with Inverse Matrix

[shomey]

I have a matrix A which satisfies: A_ij(+B) = A_ji (-B)
(the matrix is a 4x4 matrix)
EDIT: i also know that each row and line summes to 1.
i want to prove that the inverse matrix of A sattisfies the same symmetry property. (But, with no success)

Do you have an idea how to do that?

thanks allot,
Ron

 

[marlon]

What is this (+B) and (-B) ?

do you mean $$A_{ij} + B = A_{ji} - B$$ ?

marlon

 

[shomey]

What is this (+B) and (-B) ?

do you mean $$A_{ij} + B = A_{ji} - B$$ ?

marlon

sorry, i really didn't explain that...
all matrix elements are functions of B...
so - in the left side you enter +B, and in the right side, you enter -B

 

[matt grime]

Let D be the inverse, what do we know?

i will use summation convention, ie a repeated index is summed

A_ij D_jk = d_ik (delta 1 if i=k 0 otherwise)


now if i sum over i too it follows that the sum D_jk over j is 1, so that the sum of each column is 1 and similialy, by considering DA rather than AD it follows that the sum over each row of D is one.

as for the other part, I know that

A(b)D(b)=Id =D(b)A(b)

so i can set b as -b if i feel like it and we still know

D(-b)A(-b)=Id

and i can transpose it

A^t(-b)D^t(-b)=Id

and i know that i can replace A^t(-b) with A(b), now i can reach the result i want.

 

[shomey]

Let D be the inverse, what do we know?

i will use summation convention, ie a repeated index is summed

A_ij D_jk = d_ik (delta 1 if i=k 0 otherwise)


now if i sum over i too it follows that the sum D_jk over j is 1, so that the sum of each column is 1 and similialy, by considering DA rather than AD it follows that the sum over each row of D is one.

as for the other part, I know that

A(b)D(b)=Id =D(b)A(b)

so i can set b as -b if i feel like it and we still know

D(-b)A(-b)=Id

and i can transpose it

A^t(-b)D^t(-b)=Id

and i know that i can replace A^t(-b) with A(b), now i can reach the result i want.

That's great! So ellegant...
thanks allot!