It is well know that any analytic function of an $n \times n$ real/complex matrix $f(A)$ can be expressed as linear combination of the first $n$ powers of $A$ by the Cayley-Hamilton theorem.

Is it possible to express the function (adjoint action of $SU(4)$ on $\mathfrak{su}(4)$ in a similar way. I.e. is it possible to write:

$UAU^{\dagger} = aI + bA + cA^2 + dA^3$.

and actually find formulas for $a,b,c,d$ in terms of $U$ and $A$?

Would it be possible in the case that $U=\exp(B)$ for some $B\in \mathfrak{su}(4)$ ?


If this were possible, $UAU^\dagger$ would always commute with $A$.

  • $\begingroup$ Oh yeah, should have thought about that a bit more! So where is my idiocy? Is this not an' analytic function' of $A$ in the correct sense, it's even a linear function. $\endgroup$ – Benjamin Sep 5 '15 at 3:21
  • $\begingroup$ @Benjamin A linear function in $A$ is not necessarily a linear polynomial in $A$, not to mention a polynomial, a power series or an analytic function. This is not the scalar case. $\endgroup$ – user1551 Sep 5 '15 at 3:29
  • $\begingroup$ What do you mean by the scalar case. For example, the matrix exponential $\exp(A)$ can be written as a polynomial in the first $n$ powers of $A$ by Cayley-Hamilton, there is nothing scalar about that case either. $\endgroup$ – Benjamin Sep 5 '15 at 14:53
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    $\begingroup$ @Benjamin: The matrix version of what you consider analytic function has its values commuting with its argument by construction. Yes, conjugation by $U$ is linear in $A$, but so is any map $y_{ij}=\sum_{k,h}c_{ijkh}a_{kh}$, and in general that does not produce a result that commutes with $A$. $\endgroup$ – ccorn Sep 5 '15 at 15:49
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    $\begingroup$ @Benjamin When we say that a matrix function $f(A)$ is "analytic", we actually mean this function is the primary matrix function associated with some scalar-valued analytic function $f(z)$, but not that $f(A)$, when viewed as a function on $\mathbb C^{n^2}$, is analytic. That's why a "linear" function is not necessary "analytic". Linear functions of $M_n(\mathbb C)$ are necessarily primary matrix function based on a scalar-valued linear function only when $n=1$ (i.e. when the matrices are essentially just scalars). $\endgroup$ – user1551 Sep 9 '15 at 2:19

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