Wikipedia gives the following explicit formula for the functional calculus of $2\times2$ matrices: $$ f(A) = \frac{f(\lambda_+) + f(\lambda_-)}{2} I + \frac{\mathrm{tr}(A)/2 - \mathrm{adj}(A)}{\sqrt{\Delta(A)}} \frac{f(\lambda_+) - f(\lambda_-)}{2}, $$ where I am guessing that $\mathrm{tr}(A)/2 - \mathrm{adj}(A)$ actually means $(\mathrm{tr}(A)/2) I - \mathrm{adj}(A)$ and $\mathrm{adj}(A)$ stands for the adjugate of $A$ and where $\Delta(A)=(\mathrm{tr}(A)/2)^2 - \mathrm{det}(A)$ is the discriminant of the characteristic polynomial $\chi_A(t)=\mathrm{det}(tI-A)=t^2-\mathrm{tr}(A)t+\mathrm{det}(A)$ and $\lambda_\pm = \mathrm{tr}(A)/2 \pm \sqrt{\Delta(A)}$ are its roots. Unfortunately, the Wikipedia article is rather cryptic in this place without listing any assumptions whatsoever, and I have not been able to locate the formula in either Higham's book or in Bhatia's.

Is (my interpretation of) this formula really correct for any $A\in M_2(\mathbb{C})$ with $\Delta(A)\neq 0$ (suffices to assume $f\in\mathbb{C}[t]$ at the moment)? If not, then in what generality and where can I find a rigorous derivation of it?

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    $\begingroup$ The formula is equivariant under conjugation, so it suffices to check it for diagonal matrices. I'm too lazy to do that, but it shouldn't be hard (and, if it doesn't work as you interpret it, it might point you to what the notation means). $\endgroup$
    – LSpice
    Dec 13 '17 at 22:10
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    $\begingroup$ Also, I notice you ask if it is true for any $2\times2$ matrix, and the answer to that is trivially no: it doesn't work for $f$ the identity function and $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. The functional calculus can only be expected to work for normal matrices. $\endgroup$
    – LSpice
    Dec 13 '17 at 22:11
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    $\begingroup$ @LSpice The functional calculus "works" for all square matrices, but for $n \times n$ matrices with fewer than $n$ linearly independent eigenvectors the formulas will involve some derivatives of your function at eigenvalues rather than just the values there. $\endgroup$ Dec 13 '17 at 22:21
  • $\begingroup$ @LSpice: thanks for pointing this out, I have added the obvoius nonzero discriminant condition. The thing is Wikipedia does not say in what context this is supposed to be valid, polynomials at least can be evaluated at any square matrix. Perhaps "functional calculus" is a poor choice of wording here? $\endgroup$
    – M.G.
    Dec 13 '17 at 22:29

A general procedure for $f(A)$ for any $n \times n$ matrix $A$, where $f$ is an analytic function in a neighbourhood of the spectrum of $A$, is this. Let $p$ be a rational function such that $p(\lambda) = f(\lambda)$ for every eigenvalue $\lambda$ of $A$, and $p^{(k)}(\lambda) = f^{(k)}(\lambda)$ for $k \le d(\lambda) - 1$ where $d(\lambda)$ is the size of the largest Jordan block for eigenvalue $\lambda$ (and thus the multiplicity of $\lambda$ as a root of the minimal polynomial of $A$). Then $f(A) = p(A)$.

In this case, since we must assume $\Delta(A) \ne 0$, the eigenvalues are distinct so all $d(\lambda) = 1$. If $A$ is invertible, $\text{Adj}(A) = A^{-1} \det(A)$ and so the right side is $p(A)$ for some rational function $p$. We just have to check $p(\lambda_\pm) = f(\lambda_\pm)$. Then use continuity to deal with the case where $A$ is not invertible.

  • $\begingroup$ Thanks! A minor typo in your first paragraph: it seems $p(k)(\lambda)$ should be $p^{(k)}(\lambda)$. $\endgroup$
    – M.G.
    Dec 13 '17 at 22:43

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