Suppose that the map $t \mapsto A(t)$ from some open subset of $\mathbb{R}$ to the set of positive matrices is differentiable. It is known that the map $t \mapsto \sqrt{A(t)}$ is differentiable, where $\sqrt{A(t)}$ is the unique positive root of $\sqrt{A(t)}$ (and not the componentwise root of the matrix). I am looking for a proof of this fact and the expression of the derivative.
Asked
Active
Viewed 680 times
1
-
One nice method is to use the power series – Ben Grossmann May 27 '16 at 04:31
-
Or perhaps it's possible to use the derivative of the map $t \mapsto [A(t)]^2$, which is $A(t)A'(t) + A'(t)A(t)$. – Ben Grossmann May 27 '16 at 04:47
-
Could you please elaborate on the power series method? Thanks. – kayencee May 27 '16 at 11:46
1 Answers
2
Use the implicit function theorem to show that the derivative exists.
As for computing it: we will have a very nice derivative if $A'(t)$ commutes with $A(t)$. In particular, we find that $$ \frac d{dt}\sqrt{A(t)} = \frac 12[A(t)]^{-1/2}A'(t) $$ this can be confirmed via the power series for $x \mapsto \sqrt{x}$ centered at $x = 1$, an applying the appropriate normalization to $A$ so that $\|A - I\|$ is within the radius of convergence.
Not sure about the more general case, though.
Ben Grossmann
- 234,171
- 12
- 184
- 355
-
I guess a (1/2) factor is missing. The remark about implicit function theorem should be useful, but I am wondering if the derivative could be obtained from first principles (as can be done for the map $t \mapsto A(t)^{-1}$). – kayencee May 27 '16 at 11:45