MHB Proving the Inverse of the Adjoint Matrix Property for nxn Matrices

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Inverse
Yankel
Messages
390
Reaction score
0
Hello

I need some help proving the next thing, I can't seem to be able to work it out..

Let A be an nxn matrix.

Prove that:

(adj A)^{-1} = adj(A^{-1})

Thanks...
 
Physics news on Phys.org
$A = IA = A^*(A^*)^{-1}A$

so:

$A^* = (A^*(A^*)^{-1}A)^* = A^*((A^*)^{-1})^*A$

therefore:

$A^*A^{-1} = A^*((A^*)^{-1})^*$

and multiplying on the left by $(A^*)^{-1}$ we get:

$A^{-1} = ((A^*)^{-1})^*$

so

$(A^{-1})^* = ((A^*)^{-1})^{**} = (A^*)^{-1}$
 
thanks, took me some time to understand your proof, but now I see it, nice one !
(Yes)
 
The world of 2\times 2 complex matrices is very colorful. They form a Banach-algebra, they act on spinors, they contain the quaternions, SU(2), su(2), SL(2,\mathbb C), sl(2,\mathbb C). Furthermore, with the determinant as Euclidean or pseudo-Euclidean norm, isu(2) is a 3-dimensional Euclidean space, \mathbb RI\oplus isu(2) is a Minkowski space with signature (1,3), i\mathbb RI\oplus su(2) is a Minkowski space with signature (3,1), SU(2) is the double cover of SO(3), sl(2,\mathbb C) is the...
Back
Top