Can a symmetric positive definite matrix interpolation have linear trace and determinant? Definition.
Suppose $A_0, A_1 \in S_+^N$ are real symmetric positive definite $N \times N$ matrices.
An interpolation from $A_0$ to $A_1$ is a continuous / smooth function $A \colon [0, 1] \to S_+^N$ such that $A(0) = A_0$ and $A(1) = A_1$.
Example 1.
The linear interpolation $A_{\text{lin}}(t) := (1 - t) A_0 + t A_1$ has linear trace, that is, $\text{tr}(A_{\text{lin}}(t)) = (1 - t) \text{tr}(A_0) + t \text{tr}(A_1)$, but not linear determinant: $\det(A_{\text{lin}}(t)) \ne (1 - t) \det(A_0) + \det(A_1)$.
Example 2.
The logarithmic interpolation $A_{\log}(t) := \exp\left( (1 - t) \log(A_0) + t \log(A_1)\right)$ has linear determinant but not linear trace.
My question.
Does there exist an interpolation from $A_0$ to $A_1$ with linear trace and determinant?
My attempt.
I tried to show that this is not possible for $N = 2$ (it clearly is possible for $N = 1$).
Suppose $A_k = \begin{pmatrix} a_k & b_k \\ b_k & c_k \end{pmatrix}$ for $k \in \{ 0, 1 \}$ and $A(t) = \begin{pmatrix} a(t) & b(t) \\ b(t) & c(t) \end{pmatrix}$.
Then we require that
\begin{gather*}
a(t) + c(t)
\overset{!}{=} (1 - t) (a_0 + c_0) + t (a_1 + c_1) \\
a(t) c(t) - b(t)^2
\overset{!}{=} (1 - t) [a_0 c_0 - b_0^2] + t [a_1 c_1 - b_1^2].
\end{gather*}
We can't simply choose $a(t) = (1 - t) a_0 + t a_1$ and $c(t)$ similarly, because then we are in the case of Example 1 and thus don't have linear determinant.

Since trace and determinant both only depend on the eigenvalues, there might be an argument using diagonalization.
Plugging the first into the second equation of @AndreasLenz yields
$$
\lambda_1(t) \big((1 - t) (\lambda_1(0) + \lambda_2(0)) + t (\lambda_1(1) + \lambda_2(1)) - \lambda_1(t) \big) = (1 - t) \lambda_1(0) \lambda_2(0) + t \lambda_1(1) \lambda_2(1)
$$
and thus
\begin{align} 2 \lambda_1(t) & = (1 - t)(\lambda_1(0) + \lambda_2(0)) + t (\lambda_1(1) + \lambda_2(1)) \\ & \quad \pm \sqrt{((1 - t)(\lambda_1(0) + \lambda_2(0)) + t (\lambda_1(1) + \lambda_2(1)))^2 - 4 \lambda_1(0) \lambda_2(0) (1 - t) - 4 \lambda_1(1) \lambda_2(1) t}.\end{align}
 A: This is not always possible. E.g. let $A_0=I_2$ and $A_1=2I_2$. If we want the trace and determinant of $A(t)$ to vary linearly with $t$, we need
$$
\begin{aligned}
\operatorname{tr} A(t)&=(1-t)(2)+t(4)=2(1+t),\\
\det A(t)&=(1-t)(1)+t(4)=1+3t.\\
\end{aligned}
$$
The characteristic polynomial of $A(t)$ is therefore $p_t(x)=x^2-2(1+t)x+(1+3t)$. However, when $t\in(0,1)$, the discriminant of $p_t$, $\big(2(1 + t)\big)^2 - 4 (1 + 3 t) = 4 t (t - 1)$ is negative. Hence $A(t)$ cannot even be real symmetric on this open interval, not to say positive definite.
A: Some definitions first. Define $p_N:[0,1]\to[0,1],\,p_N^\max\in\mathbb R$ and $\tau,\delta:[0,1]\to\mathbb R_{>0}$ by
$$
\begin{aligned}
p_N(x)&=\dfrac{x(1-x)^{N-1}}{(N-1)^{N-1}}\quad\text{(with $p_1(x)=x$ by convention),}\\
p_N^\max&=\max\limits_{x\in[0,1]}p(x)=p\left(\dfrac1N\right)=\dfrac{1}{N^N},\\
\tau(t)&=(1-t)\operatorname{tr} A_0+t\operatorname{tr} A_1,\\
\delta(t)&=(1-t)\det A_0+t\det A_1.\\
\end{aligned}
$$
Here is a partial answer to your question:
$$ \bbox[10px,border:2px solid red]
{
\text{A continuous solution $A$ exists if and only if $\dfrac{\delta}{\tau^N}\le p_N^\max=\frac{1}{N^N}$ on $[0,1].\qquad(\dagger)\ $}
}
$$
Its necessity is an easy consequence of the AM-GM inequality for the eigenvalues of $A(t)$. We now prove its sufficiency.
Let $A_i=e^{K_i}S_ie^{-K_i}$ be an orthogonal diagonalisation, where $K_i$ is skew-symmetric and $S_i$ is a positive diagonal matrix. If we can find a continuous positive diagonal matrix function $S(t)$ such that $S(0)=S_0,\,S(1)=S_1,\,\operatorname{tr}S(t)=\tau(t)$ and $\det S(t)=\delta(t)$ where $\tau$ and $\delta$ are the linear interpolation functions defined at the beginning of this answer, a continuous solution $A(t)$ is obtained:
$$
\begin{cases}
K(t)=(1-t)K_0+tK_1\\
A(t)=e^{K(t)}S(t)e^{-K(t)}
\end{cases}.
$$
Let $X=\dfrac{1}{\tau}S$. Whenever $X(t)$ exists, we have $\operatorname{tr}X(t)=1$ and $0<f_N(t):=\det X(t)=\dfrac{\delta(t)}{\tau(t)^N}\le p_N^\max$ for all $t$, where the latter inequality is due to $(\dagger)$. Let $x_i(t)$ be the $i$-th diagonal element of $X(t)$. The existential problem of $S$ thus reduces to the following abstract problem:

*

*Given any continuous function $f_N:[0,1]\to\mathbb R_{>0}$ such that $f_N\le p_N^\max$ and given any $2N$ real numbers $x_1(0),x_2(0),\ldots,x_N(0),x_1(1),x_2(1),\ldots,x_N(1)$ such that for $T\in\{0,1\}$, we have
$$
0<x_i(T)\leq 1 \text{ for all $i$},\quad \sum_{i=1}^Nx_i(T)=1\quad \text{and}\quad\prod_{i=1}^Nx_i(T)=f_N(T),\tag{$\ast$}
$$
is it possible to extend each $x_i$ to a continuous functions $x_i:[0,1]\to(0,1]$ such that condition $(\ast)$ is satisfied when $T\in\{0,1\}$ is replaced by any $t\in[0,1]$?

We shall prove the feasibility of such extension by mathematical induction on $N$. The base case $N=1$ is solved by the constant function $x_1=1$. In the inductive case, suppose $N\ge2$. When $T=0,1$, by the AM-GM inequality, we have
$$
f_N(T)
=\prod_ix_i(T)
\le x_N(T)\left(\frac{\sum_{i<N}x_i(T)}{N-1}\right)^{N-1}
=x_N(T)\left(\frac{1-x_N(T)}{N-1}\right)^{N-1}
=p_N(x_N(T)).
$$
Since $f_N\le p_N^\max$, there exists some continuous function $h$ on $[0,1]$ such that $f_N\le h\le p_N^\max,\ h(T)=p_N(x_N(T))$ for $T\in\{0,1\}$ and $h(\frac1N)=p_N^\max$.
Note that $p_N$ is a continuous function that is strictly increasing on $\mathcal I=[0,\frac1N]$ and strictly decreasing on $\mathcal J=[\frac1N,1]$. So, for $T=0,1$, if we define
$$
q_T=\begin{cases}
(p_N|_{\mathcal I})^{-1}&\text{if }h(x_N(T))\in[0,\frac1N],\\
(p_N|_{\mathcal J})^{-1}&\text{if }h(x_N(T))\in(\frac1N,1],\\
\end{cases}
$$
then $q_T\circ h:[0,1]\to[0,1]$ is a continuous function such that $(q_T\circ h)(T)=x_N(T)$. Therefore we may extend $x_N$ to a continuous function on $[0,1]$ by defining
$$
x_N(t)=\begin{cases}
q_0\circ h&\text{if }t\in[0,\frac1N],\\
q_1\circ h&\text{if }t\in(\frac1N,1].\\
\end{cases}
$$
It follows that $p_N(x_N(t))=h(t)\ge f_N(t)>0$. Since $p_N(0)=p_N(1)=0$, we must have $0<x_N(t)<1$ on the unit interval. We may therefore define a continuous function $f_{N-1}:[0,1]\to\mathbb R_{>0}$ by
$$
f_{N-1}(t)=\frac{f_N(t)}{x_N(t)(1-x_N(t))^{N-1}}.
$$
Then
$$
f_{N-1}(t)
\le\frac{h(t)}{x_N(t)(1-x_N(t))^{N-1}}
=\frac{p(x_N(t))}{x_N(t)(1-x_N(t))^{N-1}}
=\frac{1}{(N-1)^{N-1}}
=p_{N-1}^\max.
$$
and hence $0<f_{N-1}\le p_{N-1}^\max$. Now for $T=0,1$ define $y_i(T)=\frac{x_i(T)}{1-x_N(T)}$ for $i=1,2,\ldots,N-1$. One may readily verify that
$0<y_i(T)\leq 1,\ \sum_{i=1}^{N-1}y_i(T)=1$ and $\prod_{i=1}^{N-1}y_i(T)=f_{N-1}(T)$. By induction hypothesis, we may extend the $y_i$s to to continuous functions on $[0,1]$ such that the three properties of the $y_i$s conditions are true when $T$ is replaced by any $t\in[0,1]$. Now define $x_1,x_2,\ldots,x_{N-1}$ by $x_i=y_i(1-x_N)$ and we are done.
