# Unique Matrix Function

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 Note by Steven Zheng
6 years ago

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

You jumped way too many steps: how does $f(A) = Pf(D)P^{-1}$??? You will need to justify how $f(PDP^{-1})$ 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?

- 1 year, 12 months ago

The 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 !

- 5 years, 11 months ago

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

- 1 year, 12 months ago

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.

- 5 years, 11 months ago

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.

- 5 years, 11 months ago