Vtyjkyuk

Let $A$, $B$ be $d$ by $d$ matrices where $A$, $A+B$ are symmetric positive semi-definite matrices and $textdetB=0$. $f(t)=(textdet(A+tB))^frac1d$
is my function where $tin[0,1]$.

If $f(t)leq f(0)+tf'(0)$, then $f$ is concave over $[0,1]$.

I just asked To be concave on $[0,1]$, $f(t)leq f(0)+tf'(0)$ is enough?
for general function, but turns out there is a counterexample.

$f$ is special here, which is a composition of exponent function
and determinent, it could be true. Thanks in advance.

My trial : $f(t)=(textdet(A+tB))^frac1d$
looks like a exponent $frac1d$ of a polynomial function of order
$d$. then $f(t)=(a_1t^d+cdots)^frac1d$.

Second trial : Let $f(t)=(a_dt^d+cdots+a_0)^1/d$.

$tf(1/t)=(det(tA+B))^1/d$, LHS$=(a_d+cdots+a_0t^d)^1/d$
take $t=0$, then $a_d=textdetB=0$.

So $f(t)=(a_d-1t^d-1+cdots+a_0)^1/d$

We can see $f$ is smooth. Let's claim $f$ doesn't have zero on $(0,1)$.

Suppose not, i.e. $f(z)=0$ for some $zin(0,1)$. We know $f(1)=left(textdet(A+B)right)^1/dgeq0$.

Then, we can change $A$ into $A+zB$ then the condition $tildef(t)leqtildef(0)+ttildef'(0)$
doesn't make sence. So $f$
doesn't have any zero on $(0,1)$.

Here, I can't go farther.

Some headway... Let $C(t) = A+tB$ for $tin[0,1]$. Note that $C(t)$ is a convex combination of positive semi-definite matrices $A$ and $A+B$, so it is positive semi-definite. Thus, $det C(t) ge 0$ and thus $f(t)$ is well defined for every $tin[0,1]$.

One can attempt to employ the highschool second derivative test. Define $s := 1/d$, and $g(t):= det C(t)$ for notational brevity. Then we have

$$ f'(t) = fracddt (g(t)^s ) =sg(t)^s-1 g'(t)$$
and
$$ f''(t) = s(s-1) g(t)^s-2 g'(t)^2 + sg(t)^s-1 g''(t)$$

Now observe that

• $fracddt [detA(t)]=detA(t) operatornametr[A^-1(t), fracddt A(t)]$


• $fracd^2dt^2 det(A(t)) = det left(Aright) left( left(operatornametrleft(A^-1 A_tright)right)^2 + operatornametrleft(A^-1 A_ttright)-operatornametrleft(A^-1A_t A^-1 A_tright)right)$

so
beginalignf''(t)
&= s(s-1)g(t)^s [operatornametr(C^-1 B)]^2 + s g(t)^sleft([operatornametr(C^-1 B)]^2
+0-operatornametr(C^-1BC^-1B) right) \
&= sg(t)^s left( s [operatornametr(C^-1 B)]^2
-operatornametr((C^-1B)^2) right) endalign

Similarly to the linked question, it seems that the given inequality $ f(t) le f(0) + tf'(0)$,
i.e.,
$$g(t)^s le g(0)^s + tsg(0)^soperatornametr(A^-1B)$$
is not helpful. It seems like we need to know an inequality between traces of $D:=C^-1B$ and $D^2$. As the trace is a sum of eigenvalues, it may be useful to note that

$$ s(operatornametr D)^2 - operatornametr(D^2) = s(sum_i lambda_i )^2 - (sum_i lambda_i^2) $$

If one assumes that $A,A+B$ are better behaved, namely positive definite, then $ f > 0$ is smooth on $(0,1)$, so the above has a chance of working. If in addition the dimension is 2, then there is the equality
$$ det M = frac12((operatornametrM)^2 - operatornametr(M^2) ) $$
so we can notice that
$$ f''(t) = frac12g(t)^1/2 left( -frac12operatornametr((C^-1B)^2)right) < 0$$
which proves the result true in this special case, without using the inequality $f(t) le f(0) + tf'(0)$. There are also similar inequalities in dimensions $>2$ but I don't know how to use them.