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}
Best Answer
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.