Let \(A\) be an invertible matrix. Show that there exists a unique differentiable function \(f\) such that \[det(f(A)) = f(tr(A)).\]

**Solution**

If \(A\) is invertible, then there exists some diagonal matrix \(D\) and similar matrix \(P\) such that \[A = PD{P}^{-1}.\]

\[det(f(A)) = det(Pf(D){P}^{-1})\]

\[det(f(A)) = det(P)det(f(D))det({P}^{-1})\]

\[det(f(A)) = det(P{P}^{-1})det(f(D))\]

\[det(f(A)) = det(f(D))\]

Since the diagonal matrix is composed of eigenvalues along its main diagonal,

\[det(f(A)) = \prod _{ i=1 }^{ n }{ { f\left( { \lambda }_{ i } \right) } } .\]

There are two important properties of square matrices: for some square matrix \(M\), the sum of its eigenvalues equals its trace; the product of its eigenvalues equals its determinant [Proof].

Hence, we need to find a function such that \[\prod _{ i=1 }^{ n }{ { f\left( { \lambda }_{ i } \right) } } =f\left( \sum _{ i=1 }^{ n }{ { \lambda }_{ i } } \right).\]

Only the exponential function satisfies this criteria.

Therefore, \[det(\exp{(A)}) = \exp{(tr(A))}.\]

Check out my other notes at Proof, Disproof, and Derivation

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## Comments

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TopNewestThe question does not make sense. What are the domains and ranges of \(f\) ? From the left hand side it appears that both the domain and range are space of square matrices. On the other hand from the right hand side it appears that the domain and range are sets of real numbers !

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You have to be careful about these operations. The determinant of a square matrix (with real number entries) is a real number. The trace of a square matrix(with real number entries) is a real number. So on both sides, there are two mappings. LHS f:M->M then det:M->R. RHS trace:M->R then f:R->R. Both are ultimately mappings from square matrices to reals, but the journey there is different.

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I just thought about it on a walk today. I think I will change the question and actually prove the only function that is valid for this identity. Stay tuned to see it.

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