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How can I let Mathematica tell me that trace(Identity matrix)=n for n by n identity matrix? It's easy for human but what would a Mathematica code be?

A harder task: How can I let Mathematica tell me that Exp[I] is also an n by n diagonal matrix with all diagonal entries equal to Exp[1]?

(I simplified my actual question so that humans already know the answer. My actual matrix M is an (2n+1) by (2n+1) tridiagonal matrix with each entry one. I want the [[n,n]] matrix entry of matrix M^n, so that I will then compute the same entry of Exp[M])

Related: How to define and compute various properties of a variable size matrix. But the problem is not resolved there.

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    $\begingroup$ For finite dimensions, nmax = 20; And @@ Table[MatrixExp[ IdentityMatrix[n]] == E*IdentityMatrix[n], {n, 1, nmax}] $\endgroup$ Commented Oct 16, 2024 at 1:58
  • $\begingroup$ Thanks! It would be great if Mathematica has the symbolic power for any-size matrices though, that is, return an answer in terms of nmax. I am not sure if Mathematica has some of these features for operators or tensors. $\endgroup$ Commented Oct 16, 2024 at 4:42
  • $\begingroup$ FullSimplify[MatrixExp[IdentityMatrix[n]]] doesn't really give me E*IdentityMatrix[n] $\endgroup$ Commented Oct 16, 2024 at 4:43
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    $\begingroup$ In version 14.1 SymbolicIdentityArray is introduced as an EXPERIMENTAL function, but Tr cannot handle it for now. Perhaps next version? $\endgroup$ Commented Oct 16, 2024 at 5:31
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    $\begingroup$ Yeah, though experimental, those functions in 14.1 for symbolic arrays are a big step forward and can already do a lot of interesting things. For more info check reference.wolfram.com/language/guide/SymbolicArrays.html $\endgroup$ Commented Oct 16, 2024 at 6:25

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