Matrices A and B are shown to be not unitarily similar by examining their properties. A is defined as |1 2| |0 3| and B as |1 0| |0 3|. The argument hinges on the fact that if B were unitarily similar to A, then A would also have to be symmetric, which it is not. Additionally, the columns of B are orthogonal, while those of A are not, reinforcing their dissimilarity. Thus, A and B cannot be unitarily equivalent due to these fundamental differences in their properties.
#1
chuy52506
77
0
Let A=
|1 2|
|0 3|
and B=
|1 0|
|0 3|
Show that A and B are not unitarily similar?
A = A*, implying A is symmetric as well, which is false.
#4
g_edgar
606
0
The same thing said in another way: The columns of B are orthogonal, unitary transformation preserves orthogonality, but the columns of A are not orthogonal. Therefore A and B are not unitarily equivalent.
indeed, it all depends on how you define "unitary", by inner product properties, or matrix properties. there is a deep connection betwen inner products and matrix multiplication.
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime
Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...