Let $B$ and $C$ be self-adjoint (complex) $n\times n$ matrices. I was wondering if there is an expansion: $B + tC = U_t^\ast \Lambda(t) U_t$, a continuously differentiable unitary diagonalization, where $\Lambda(t)$ and $U_t$ are continuously differentiable, where $\Lambda(t)$ is diagonal and $U_t$ is unitary. See (1) below for why I care about this.

I looked at a few StackExchange/Math Overflow posts on this subject. E.g.

Continuously changing Null Space

Do eigenvalues/vectors change continuously?

how to find/define eigenvectors as a continuous function of matrix?

Conditions for smooth dependence of the eigenvalues and eigenvectors of a matrix on a set of parameters

I also tried reading up on some articles on this, but mainly they discussed matrices with elements in $C(X)$, the continuous (real? complex?) functions on a topological space $X$, but I did not get far with getting an application in this case.

Note especially the last reference. The question that I have is that my example is very simple, analytic since it is linear, but it is not symmetric which the last reference says that a result in Kato's Perturbation Theory of Linear Operators says that it works, but self-adjoint (i.e. Hermitian). I also was not able to reduce the Hermitian case to the symmetric case by applying that result to $B + tC + \overline{B+tC}$ and $B + tC - \overline{B+tC}$, but $B$ and $C$ might not commute so I don't know if it can extend in this way.

I was wondering what the result would be in this case and if anyone has the relevant reference/theorem statement for the question that I am wondering.

Thank you very much.

(1): See Perturbation Theory: Derivative of a trace. for context, but a particular trace forumula follows using the property of the trace $\operatorname{Tr}[PQ] = \operatorname{Tr}[QP]$, $U_t^\ast U_t = 1$ (and so that $(U_{0}')^\ast U_0 + U_0^\ast U_0' = 0$), $B = U_0\Lambda(0)U_0$, and $C = \frac{d}{dt}|_{t=0}(B + tC) = \frac{d}{dt}|_{t=0}(U_t^\ast \Lambda(t) U_t)$. The desired result follows after a little calculation.

  • $\begingroup$ Kato uses "symmetric" to mean Hermitian symmetric. $\endgroup$ – Keith McClary Aug 22 '18 at 4:21
  • $\begingroup$ Really? That is nice to hear. Thank you. $\endgroup$ – 4-ier Aug 22 '18 at 4:35

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