Why is the 2-norm of A^k equal to 2-norm of T^k?

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The discussion centers on the equality of the 2-norms of A^k and T^k, where T is derived from the Schur decomposition of matrix A. The Schur decomposition expresses A as Q^H * A * Q = T, with D as a diagonal matrix of eigenvalues and N as a strictly upper triangular matrix. The key point is that the 2-norm of a matrix is invariant under unitary transformations, which explains why ||A^k|| = ||T^k|| for k>=0. The participants emphasize the importance of understanding this property in relation to matrix norms and eigenvalues. Clarifying this concept can enhance comprehension of matrix behavior in linear algebra.
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Let Q^H * A * Q = T = D + N be Schur decomposition of A. D is a diagonal matrix with eigenvalues of A and N is strictly upper triangular matrix. This sentence is used in a proof which I am not going to state but at some point it states that ||A^k|| 2 norm = ||T^k|| 2 norm for k>=0. I don't understand
where this equality comes from. I tried to work this equality out with an example of taking
 
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The problem is simple: why is the 2-norm of A^k equal to 2-norm of T^k?
 
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...

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