Given Square Matrices Given square matrices $A$ and $B$. The matrix $B$ is the square root of matrix $A$ if $A=B^2$.
Question: Under what condition is $√B$ real and unique?
I'm not sure where to start with this? What does is mean under what condition is the square root real and unique?
 A: Consider the matrices $$I = \begin{bmatrix}1&0\\0&1\end{bmatrix};\;X=\begin{bmatrix}0&1\\1&0\end{bmatrix};\;K  = \frac{1-i}2\begin{bmatrix}1&i\\i&1\end{bmatrix};\;S=\begin{bmatrix}23& -6\\9& 2\end{bmatrix}.$$It turns out that $I^2 = I,$ so we can say that the identity matrix is its own square root. That's great! However, its square root is not unique because it turns out that also $X^2 = I.$ It has many (linearly independent) square roots. Another family is $\begin{bmatrix}\cos t&\sin t\\\sin t&-\cos t\end{bmatrix},$ for any $t$.
On the other hand, $X$ has (I think) only 4 square roots, with one being $K^2 = X.$ They are related to $K$ by complex conjugation and negation. We don't have any hope to lift this sign degeneracy. However: you can see the components of $K$ listed above, and they are hopelessly complex numbers. Apparently something about $X$ is "negative" so that when you try to take its "square root" this number $i = \sqrt{-1}$ can't help but show up!
What about $S$? Well it turns out $S$ has four square roots which are $$\pm \sqrt{\frac15}\begin{bmatrix}~~11&~~-2\\ 3&4\end{bmatrix};\;\pm\sqrt{\frac15}\begin{bmatrix}13&-6\\9&-8\end{bmatrix}.$$But it's only those 4, and they are all-real matrices.
So they are asking you: how do you tell? Which ones have only a couple real square roots, like $S,$ which ones have only complex ones, like $X$, and which ones have many linearly independent square roots to choose from, like $I$? And why is this number "4" so popular for the number of square roots of a 2x2 matrix, with 8 being the associated value for many 3x3 matrices?
To figure this out I would recommend that you calculate the eigenvectors of both $S$ and its square root, and its eigenvalues, and then think long and hard about what that means as regards its Jordan form, before calculating the eigenvalues of $I$ and $X$. 
