<s>[INST] How to find matrix $B$? 
let $A$ be  a $2 \times 2$  matrix  and $I$ be the identity matrix . Assume that  the   null spaces  of $A-4I$ and $A-I$ respectively spanned  by $\begin{bmatrix} 3 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1\end{bmatrix}$ respectively.Find a matrix  $B $ such that $B^2= A$

My attempt :
According to question  , it tell that $Au= 4u$  and $Av= v$
Now take u = $\begin{bmatrix} 3 \\ 2\end{bmatrix}$ and $v = \begin{bmatrix} 1 \\ 1\end{bmatrix}$
$A\begin{bmatrix} 3 \\ 2\end{bmatrix}=4\begin{bmatrix} 3 \\ 2\end{bmatrix}$
$A\begin{bmatrix} 1 \\ 1\end{bmatrix}= \begin{bmatrix} 1 \\ 1\end{bmatrix}$
After that im not able to proceed further
any hints/solution
 [/INST] You know that $A.\left[\begin{smallmatrix}3\\2\end{smallmatrix}\right]=4\left[\begin{smallmatrix}3\\2\end{smallmatrix}\right]$ and that $A.\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]=\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$. So, if $P=\left[\begin{smallmatrix}3&1\\2&1\end{smallmatrix}\right]$, then$$P^{-1}.A.P=\begin{bmatrix}4&0\\0&1\end{bmatrix}=\begin{bmatrix}2&0\\0&1\end{bmatrix}^2.$$So, take $B=P.\left[\begin{smallmatrix}2&0\\0&1\end{smallmatrix}\right].P^{-1}$.
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