Show that $\log |\Sigma|+\text{Tr}(\Sigma^{-1}V)$ is uniquely minimized Let $V$ be an $n\times n$ fixed positive definite matrix and let $f$ be the function defined by
$$f(\Sigma)=\log |\Sigma|+\text{Tr}(\Sigma^{-1}V)$$
over the set  $P^+$ of $n\times n$ positive definite matrices $\Sigma$. Show that $f$ is uniquely minimized at $V$ over $P^+$.
My attempt:
Let $V=RR'$ be the Cholesky decomposition of $V$, and let $H:=R'\Sigma^{-1}R$. Then $H$ is positive definite and $\Sigma= RH^{-1}R'$. Hence $H\mapsto RH^{-1}R'$ is bijection from $P^+$ onto $P^+$ and, using the properties of determinant and trace, we see that the minimization problem is equivalent to minimizing
$$f(H):=\log |RH^{-1}R'|+\text{Tr}((R')^{-1}HR^{-1}V)$$
$$=\log |V|-\log |H|+\text{Tr}(H)$$
over $H\in P^+$. Now, Since every $H\in P^+$ has a unique Cholesky decomposition $LL'$ we see that the above problem is equivalent to minimizing
$$f(L)=\log |V|-\log |LL'|+\text{Tr}(LL')$$
$$=\log |V|-\sum_{i=1}^n\log(L_{ii}^2)+\sum_{i=1}^nL_{ii}^2+\sum_{i<j}L_{i,j}^2$$
$$=\log |V|+\sum_{i=1}^n\big(L_{ii}^2-\log(L_{ii}^2)\big)+\sum_{i<j}L_{i,j}^2$$
over $L$ lower triangular with positive diagonal entries. Since the map $x\mapsto x-\log x$ is uniquely minimized at $x=1$ it follows that $f$ is uniquely minimized at $L=I_n$, i.e. $\Sigma=V$.
Is this correct? Am I missing something?
Thanks a lot for your help.
 A: Another argument you can use the concavity of $\Sigma \to \log \left|\Sigma\right|$ (see here). Since $$f\left(\Sigma^{-1}\right) = -\log \left|\Sigma\right| + \text{Tr}\left(\Sigma V\right)$$ then $\Sigma \to f\left(\Sigma^{-1}\right)$ is strictly convex function (sum of a strictly convex function and a convex function see here). Since every convex function has at most one minimiser on a convex set (see here), $f$ is at most uniquely minimized.
Now let's prove that $f$ has a local minimum (this will be a global minimum as well).
Let $g(t) = f\left(\left(V^{-1}+tH\right)^{-1}\right)$ for fixed $H \in \mathbb S^n$ this function is well defined for $t\in (-\delta, \epsilon)$ with $\delta, \epsilon > 0$. $g(t)$ is convex
\begin{align}
g(t) &= -\log \left|V^{-1}+tH\right| + \text{Tr}\left(\left(V^{-1} + tH\right)V\right)\\
&= -\log \left|V^{-1}\right| - \log \left|I + tHV\right| + \text{Tr}(I) + \text{Tr}(tHV)\\
&= \log \left|V\right| - \left(\text{Tr}\left(tHV\right) + o\left(t^2\right)\right) + \text{Tr}(I) + \text{Tr}(tHV)\\
&= \log \left|V\right| + \text{Tr(I)} + o\left(t^2\right)
\end{align}
This proves that $g'(0) = 0$ and $0$ is a local minimum of $g$ and hence global minimum of this function (since it is convex).
$$f\left(\Sigma^{-1}\right) = f\left(\left(V^{-1} + \left(\Sigma - V^{-1}\right)\right)^{-1}\right) = g(1) \ge g(0) = f(V)$$
for $H = \Sigma - V^{-1}$. Thus $V$ is a minimum of $f$.
