Unique Matrix Function

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

Solution

If AA is invertible, then there exists some diagonal matrix DD and similar matrix PP such that A=PDP1.A = PD{P}^{-1}.

det(f(A))=det(Pf(D)P1)det(f(A)) = det(Pf(D){P}^{-1})

det(f(A))=det(P)det(f(D))det(P1)det(f(A)) = det(P)det(f(D))det({P}^{-1})

det(f(A))=det(PP1)det(f(D))det(f(A)) = det(P{P}^{-1})det(f(D))

det(f(A))=det(f(D))det(f(A)) = det(f(D))

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

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

There are two important properties of square matrices: for some square matrix MM, 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 i=1nf(λi)=f(i=1nλi).\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)).det(\exp{(A)}) = \exp{(tr(A))}.

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

Note by Steven Zheng
5 years, 2 months ago

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The question does not make sense. What are the domains and ranges of ff ? 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 !

Abhishek Sinha - 5 years, 2 months ago

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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.

Steven Zheng - 5 years, 2 months ago

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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.

Steven Zheng - 5 years, 2 months ago

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I second this comment. The definition of f f is not clear by what is written; OP would really need to reconsider the definitions...

A Former Brilliant Member - 1 year, 2 months ago

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Whoa whoa whoa. You have a function ff acting on a matrix and then use the same ff to act on a number. Also, what are the outputs on ff?

You jumped way too many steps: how does f(A)=Pf(D)P1 f(A) = Pf(D)P^{-1} ??? You will need to justify how f(PDP1)f(PDP^{-1}) ends up being what you say it is... everything past that calculation is a blur to anyone who is going to read it.

One final note: you will need to justify why the exponential function is the ONLY function that satisfies this property. One one hand, you want existence of the function, but in the title you say this is unique. Which one?

A Former Brilliant Member - 1 year, 2 months ago

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