# minimum eigenvalue interpolation

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 $$\max(\lambda_{min}(A),\lambda_{min}(B))$$.

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 $$(1-t)\lambda_1 + t\mu_1$$.