3
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ 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
    Commented Dec 13, 2017 at 22:10
  • 2
    $\begingroup$ 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
    Commented Dec 13, 2017 at 22:11
  • 4
    $\begingroup$ @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. $\endgroup$
    – Robert Israel
    Commented Dec 13, 2017 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.
    Commented Dec 13, 2017 at 22:29

1 Answer 1

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks! A minor typo in your first paragraph: it seems $p(k)(\lambda)$ should be $p^{(k)}(\lambda)$. $\endgroup$
    – M.G.
    Commented Dec 13, 2017 at 22:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .