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We know that trace plays a very important role in representation theory in the form of characters or in lie algebra in the form of killing form among many other things. What's so special about trace that makes it such a strong player as opposed to say the sum of a product of eigenvalues taken two at a time, or the sum of entries of the other diagonal (from right up to left down )?

nobody
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    The sum of products of pairs of eigenvalues is a significant quantity that shows up in many places (though of course it is not as ubiquitous as the trace). Unlike the trace, he sum of the antidiagonal entries of the matrix representation of an endomorphism is not independent of the choice of basis. – Travis Willse Feb 10 '19 at 21:28
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    The trace and the sum of products of unordered pairs of eigenvalues are very similar in nature: they are both $\pm$ some coefficient of the characteristic polynomial. – Matt Samuel Feb 10 '19 at 21:35
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    If $M$ is an $n\times n$ square matrix, then I imagine that any symmetric polynomial in $n$ variables evaluated on the eigenvalues of $M$ could be an important invariant. It just so happens that the trace and determinant correspond to quite simple symmetric polynomials, and can be calculated directly from the matrix. – Adam Higgins Feb 10 '19 at 21:56
  • See also https://math.stackexchange.com/questions/2775848/intuition-behind-the-choice-of-the-trace-function-in-character-theory – D.R. May 29 '22 at 05:43

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Here are a couple examples of trace occurring without asking for it. Perhaps that will make it seem more special.

The trace map is the derivative of the determinant map $\mbox{det}:\mbox{GL}_n(\mathbb{R}) \to \mathbb{R}$ at the identity, which you would see in studying Lie algebras. For one, it tells you that $\mathfrak{sl}_n(\mathbb{R})$ is the traceless matrices.

It is also the very natural pairing map $V\otimes_\mathbb{R} V^* \to \mathbb{R}$ under the natural identification $\mbox{End}_\mathbb{R}(V) \cong V\otimes_\mathbb{R} V^*$. See here.

(If anyone knows more, please comment and let me know!)

user555203
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