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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.

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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 $$\begin{align} A&=K\text{diag}(\mu_1, \mu_2)\text{diag}(\mu_1, \mu_2)K^{-1} \\ &=K\text{diag}(\mu_1, \mu_2)K^{-1}K\text{diag}(\mu_1, \mu_2)K^{-1}\\ &=(K\text{diag}(\mu_1, \mu_2)K^{-1})^2 \\ &=(KEK^{-1})^2.\end{align}$$

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I edited my question to include a matrix K which can be used to find D. Does your matrix P refer to the same matrix? –  Knut Remi Mar 4 at 16:02
    
@KnutRemi Yes, my $P$ is your $K$. I'm still not sure what you want though. –  Git Gud Mar 4 at 16:06
    
I think you answered my question. Gonna do some calculations and I'll get back to you. –  Knut Remi Mar 4 at 16:10
    
I got the correct numbers when doing the calculations. I also renamed P to K and added D and E from the question. Not sure if I should add D and E to the solution process as well. –  Knut Remi Mar 4 at 17:07
    
People can always ask for more details if they don't understand something. I'd leave it like this. –  Git Gud Mar 4 at 17:08

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