minimum eigenvalue interpolation

Question

Suppose we have two symmetric positive definite matrices $A,B$ (not simultaneously diagonalizable). How can I find a matrix function $f(t), t\in [0,1]$, such that $f(0)=A, f(1)=B$, and the minimum eigenvalue of $\lambda_{min}(f(t))$ is smoothly interpolating the minimum eigenvalues of $A,B$ such that $$\min(\lambda_{min}(A),\lambda_{min}(B)) \leq \lambda_{min}(f(t)) \leq \max(\lambda_{min}(A),\lambda_{min}(B))?$$ If it is possible to find $f(t)$ such that this is interpolation $\lambda_{min}(f(t))$ is concave, that would be even better.

The naive solution to interpolation would be linear, i.e. $f(t) = (1-t)A + tB$, but that does not satisfy that condition above even though it is concave, i.e. the minimum eigenvalue can easily get larger than the $\max(\lambda_{min}(A),\lambda_{min}(B))$.

Answer 1

Let $\lambda_1\leq \lambda_2\leq \ldots \leq \lambda_n$ denote the eigenvalues of $A$, and let $\mu_1\leq \mu_2 \leq \ldots \leq \mu_n$ denote the eigenvalues of $B$. We can assume without loss of generality that $A$ is the diagonal matrix $\text{diag}(\lambda_1,\ldots,\lambda_n)$. There is an orthogonal matrix $Q$ such that $QBQ^{-1} = D = \text{diag}(\mu_1,\ldots,\mu_n)$. We can assume that the determinant of $Q$ equals 1, i.e. $Q\in SO(n)$. Since $SO(n)$ is connected there is some smooth function $q(t)$ with values in $SO(n)$ such that $q(0)=I$ and $q(1)=Q$. Now let $f(t) = q(t)^{-1}((1-t)A+tD)q(t)$. Then $f(0)=A$ and $f(1)=B$, moreover, the minimum eigenvalue of $f(t)$ equals $(1-t)\lambda_1 + t\mu_1$.