I've been given a 2x2 matrix $\mathbf A$, its eigenvalues $\lambda_1$ and $\lambda_2$, its eigenvectors $\mathbf v_1$ and $\mathbf v_2$, and a diagonal matrix $\mathbf D = \text{diag}(\lambda _1, \lambda _2)$, and a matrix $\mathbf K = [\mathbf v_1 | \mathbf v_2]$ such that $\mathbf D = \mathbf K^{-1} \mathbf A \mathbf K$.
How can I find a matrix $\mathbf B$, such that $\mathbf B^2 = \mathbf A$?
I was given a hint that it's easy to find a diagonal matrix $\mathbf E$, such that $\mathbf E^2 = \mathbf D$ and I've found $\mathbf E$.
I'm looking for the strategy and methods needed to solve the problem, but if you need some numbers to work with, let me know and I'll add them to my question.
I'm not sure what you're looking for and this might just be a reiteration of that you did, but here it goes anyway.
Let $\mu _1$ and $\mu_2$ be such that $\mu _1^2=\lambda _1$ and $\mu _2^2=\lambda _2$.
Assuming $A=K\text{diag}(\lambda _1, \lambda _2)K^{-1} = KDK^{-1}$, it follows that $$\begin{align} A&=K\text{diag}(\mu_1, \mu_2)\text{diag}(\mu_1, \mu_2)K^{-1} \\ &=K\text{diag}(\mu_1, \mu_2)K^{-1}K\text{diag}(\mu_1, \mu_2)K^{-1}\\ &=(K\text{diag}(\mu_1, \mu_2)K^{-1})^2 \\ &=(KEK^{-1})^2.\end{align}$$