A real $2 \times 2 $ matrix $M$ such that $M^2 = \tiny \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$ , then : 
A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&0 \\ 0&-1-\epsilon  \\   \end{pmatrix}$$
(a) exists for all $\epsilon > 0$.
(b) does not exist for any $\epsilon > 0$.
(c) exists for some $\epsilon > 0$.
(d) None of the above is true.

Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix.
Could someone please give me a hint on how to go about this problem?
Thank you very much for your help in this regard.
 A: (I assume $M$ is supposed to be real; if it is allowed to be complex, then the exercise is trivial.)
Hint


*

*What are the possible eigenvalues of $M$?

*What can one say about the eigenvalues of real $2 \times 2$ matrices...?



Continued hint ...in particular what can one say about the eigenvalues of real $2 \times 2$ matrices when those eigenvalues are nonreal?

A: no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$ 
for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$
pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = \epsilon \tag 1$$ 
that means if $\epsilon \neq 0,$  then $a+d \neq0\implies b = c = 0.$ from $(1), a^2 = -1, d^2 = -1 - \epsilon$   there are no real solutions as $a^2 = -1.$
if $\epsilon = 0, a^2 - d^2 = 0$ it forces $a = -d, b = c = 0$ but then (1) forces $d^2 = -1$ which cannot be satisfied. the other case is $a = d.$ the constraints are $$a^2 + bc = -1, 2ab = 0 = 2ac, bc + d^2 = -1 $$ two choices: (i) $a = 0$ gives $bc = -1, d = 0$
(ii) $a \neq 0$ which gives $b = 0, c = 0, a^2 = -1$ not possible.
A: Let $M=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. Then we want $M^2=\left(\begin{matrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{matrix}\right)=\left(\begin{matrix}-1&0\\0&-1-\epsilon\end{matrix}\right)$.
Now we can't have $b=0$ or $c=0$ (why?). Therefore $a=-d$ and $$M^2=\left(\begin{matrix}a^2+bc&0\\0&a^2+bc\end{matrix}\right)=(a^2+bc)I$$
Then ?
A: Assume $v$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then $v$ is eigenvector of $M^2$ wih eigenvalue $\lambda^2\ge0$. Since $M^2$ has only negative eigenvalues $-1$ and $-1-\epsilon$, $M$ has no eigenvectors. Thus the matrix $A$ that maps $e_1\mapsto e_1$ and $e_2\mapsto Me_1$ is invertible and since $M$ maps $Me_1$ to $-e_1$ we find that 
$$M=A\begin{pmatrix}0&-1\\1&0\end{pmatrix}A^{-1}$$
so that 
$$ M^2=A\begin{pmatrix}0&-1\\1&0\end{pmatrix}^2A^{-1}=A\begin{pmatrix}-1&0\\0&-1\end{pmatrix}A^{-1}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$
