<s>[INST] Help in proof involving matrices 
Let  $$ P=\begin{pmatrix}11&-6\\18&-10\end{pmatrix}. $$ Show that
  there is no $2\times2$ matrix $Q$ such that $Q^2=P$.

I don't know how to approach this question. Is brute force the only way to go? That is, take a general matrix 
$$
Q=\begin{pmatrix}a&b\\c&d\end{pmatrix}
$$
and get 
$$
Q^2=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+c)&bc+d^2\end{pmatrix}.
$$
Then equate element by element and try to find a contradiction? This seems tedious and not very instructive.
In a previous part of the question, we established that $P^2+P=2I$, $|P|=-2$, and $$P^{-1}=\frac12\begin{pmatrix}10&-6\\18&-11\end{pmatrix}.$$
But I don't see how these would be useful...
Any hint would be appreciated.
 [/INST] The Eigen values of $P$ are $2$ and $-1$. Now there exists a Matrix $Q$ such that $Q^2=P$. So $Det(Q^2)=Det(P)$ $\implies$ $Det(Q)=i\sqrt{2}$ or $-i\sqrt{2}$.
Now by Cayley hamilton theorem the characteristic equation of $Q$ is of the form
$$Q^2-Tr(Q)Q+Det(Q)I=0$$ $\implies$ Since $Q^2=P$
$$Tr(Q)Q= P+Det(Q)I=\begin{pmatrix}
11+i\sqrt{2} & -6\\ 
18 & -10+i\sqrt{2}
\end{pmatrix}$$
Taking Determinant on both sides we get
$$Det(Tr(Q)Q)=-4+i\sqrt{2}$$ Using $Det(kA)=k^nDet(A)$ we get
$$Tr(Q)^2Det(Q)=-4+i\sqrt{2}$$ So
$$Tr(Q)^2=\frac{-4+i\sqrt{2}}{i\sqrt{2}}=1+2i\sqrt{2}$$ So
$$Tr(Q)=\sqrt{1+2i\sqrt{2}}=\sqrt{2}+i$$  Finally
$$Q=\frac{1}{\sqrt{2}+i}\begin{pmatrix}
11+i\sqrt{2} & -6\\ 
18 & -10+i\sqrt{2}
\end{pmatrix}$$
$$Q=\begin{pmatrix}
 4\sqrt{2}-3i& -2\sqrt{2}+2i\\ 
6\sqrt{2}-6i & -3\sqrt{2}+4i
\end{pmatrix}
$$
</s>