Let $A$ be an $n × n$ matrix. Then, we can create a family of matrices $A(t) = tB + D$ where $D$ is the same as $A$ with all the off-diagonal entries reduced to zero and $B$ is the same as $A$ with all the diagonal entries reduced to zero all along the interval $0 ≤ t ≤ 1$.
If the eigenvalues of $A$ are $\lambda_{1}\leq...\leq\lambda_{n}$ then what can be said about the eigenvalues of $A(t)$ based on eigenvalues of $A$?
Thanks
With the setup you gave, let $$D = \begin{bmatrix}d_1 & 0 & \cdots & 0 \\ \vdots & & \ddots & \\ 0 & 0 & \cdots & d_n \end{bmatrix} \mbox{.}$$ Let the absolute row sums of $B$ be $s_1, s_2, \ldots, s_n$.
The Gershgorin Circle Theorem states that the eigenvalues of $A$ are nearly the $d_i$, in the sense that $\lvert \lambda_i - d_i \rvert \leq s_i$.
By considering $A(t)$ instead, you are scaling the $s_i$, or shrinking the radii of the Gershgorin discs. Thus the eigenvalues of $A(t)$ satisfy $\lvert \lambda_i(t) - d_i \rvert \leq t s_i$.
Note: This doesn't say anything at all about where in the disc the eigenvalues are located, though I expect they lie along a nice curve with $d_i$ at one endpoint.