Find matrix $Q\in \mathbb{C}^{n \times n}$ such that $Q^2=A$. Let $A \in \mathbb{C}^{n \times n}$ such that the characteristic Polynomial $P_A=(\lambda - X)^n$ with $\lambda \in \mathbb{C}\setminus \{0\}$ and $(A-\lambda I_n)^2=0$ Show that there exists a $Q\in \mathbb{C}^{n \times n}$ such that $Q^2=A$.
My idea was that since A is triangularizable there exists an upper triangle Matrix $B \in \mathbb{C}^{n \times n} $ and an invertible Matrix $S\in \mathbb{C}^{n \times n}$ such that $A=S^{-1}BS$. Now since $A$ and $B$ are similar I thought that their characteristic Polynomial must be the same and therefore $(B-\lambda I_n)^2=0$. With that I would have nilpotent matrix $N:=B-\lambda I_n, N^2=0$ which could be useful to construct such a $Q$. My issue is how do I show that $(B-\lambda I_n)^2=0$  and would $Q$ be unique?
 A: Let $\mu \in \mathbb{C}$ such that $\mu^2 = \lambda$ (obviously, $\mu \neq 0$).
Let $Q = \dfrac{1}{2 \mu}(A + \lambda I_n)$.
Then $Q^2 = \dfrac{1}{4\lambda}(A+\lambda I_n)^2 = \dfrac{1}{4\lambda}\left[(A-\lambda I_n)^2 + 4\lambda A\right] = A$, since $(A-\lambda I_n)^2=0$, and you are done.
A: Conceptually: think of the Jordan normal form of $A$. The equation $P_A = (\lambda-x) ^n$ tells you that all Jordan blocks correspond to the same eigenvalue $\lambda$. The equation $(A-\lambda I)^2=0$ implies that all Jordan blocks are either 1 or 2 dimensional (check this if it's not evident!). It suffices to construct $Q$ such that $Q^2=A$ block by block. The case of 1 dimensional blocks is trivial. Two dimensional blocks require finding $Q$ such that:
$$
Q^2 = 
\begin{pmatrix}
\lambda & 1 \\
0 & \lambda
\end{pmatrix}
$$
Exercise: find such $Q$ (you can start with general $Q$ with entries $a,b,c,d$ and multiply or guess the solution).
Regarding the second question, $Q$ is obviously not unique as $Q'=-Q$ satisfies the same equation. More generally, you have two choices for $Q$ on every Jordan block of $A$, so the number of possible choices of $Q$ will be $2 ^k$, wkere $k$ is the number of Jordan blocks.
