sufficient conditions for the values to make M diagonalizable. My first post was close so i tried it again:
This is a question from a university test in wich i have half of the score, so i post here the development i made for you to comment possible errors:
Be $M \in M_{2}[\mathbb{R}]$,$M=\left(\begin{matrix} a & b \\        
                     c & d  \end{matrix}\right)$ such that $Det(M)=1$ stablish conditions over the values of $a,b,c,d \in \mathbb{R}  $   such that $M$ is diagonalizable.
what i do is to take the matrix $ \left(\begin{matrix} a-\lambda & b \\        
                     c & d-\lambda  \end{matrix}\right)$ and evaluating the characteristic polynomial $(a-\lambda)(d-\lambda)-bc=\lambda^2 - \lambda(a+d) + 1$ we have two distinc roots if and only if the discriminant $(a+d)^2-4>0$ so if $a+d<-2$ or $a+d >2$ there exist two real distincts eigenvalues and $M$ is diagonalizable.
After that we analyse the case $a+d=2$ or $a+d=-2$ in the first case we have for the characteristic polynomial $\lambda^2-2\lambda+1=(\lambda+1)^2=0$ so an eigenvalue $\lambda=-1$ of multiplicity 2 , for the second case the characteristic polynomial $(\lambda-1)^2=0$ give us a second eigenvalue $\lambda=1$ again with miltiplicity 2.
reemplazing in the orginal matrix we have, for the value $\lambda=1$ $$ \left(\begin{matrix} a-1 & b \\        
                     c & d-1  \end{matrix}\right)   \left(\begin{matrix} x_{1} \\  x_{2} \\        
                      \end{matrix}\right) =  \left(\begin{matrix} 0 \\  0 \end{matrix}\right) $$
we have that the geometric multiplicity of the solution is two and so $M$ is diagonalizable only if $0=a-1=b=c=d-1$ and so $M=I$
in an analogous form we find that $a+1=d+1=b=c=0$ permits the diagonalization of $M$ that is $M=-I$.
In the final case that $-2<a+d<2$ we have no real eigenvalues.
I have tried to redact my answer in the form of my test so i thank your answers.
 A: The analysis is correct. On the other hand, it's perhaps simpler to prove that a $2\times2$ real matrix over the reals is diagonalizable if and only if it has two distinct real eigenvalues or is diagonal.
Indeed, the characteristic polynomial can be written in the form
$$
\lambda^2-(a+d)\lambda+(ad-bc)
$$
If the polynomial has distinct real roots, then the matrix is diagonalizable. Such condition is
$$
(a+d)^2-4(ad-bc)>0
$$
that is
$$
(a-d)^2+4bc>0
$$
(in your case $ad-bc=1$). In the case when the discriminant is $0$, the matrix has a single real eigenvalue of algebraic multiplicity $2$ and it is $\lambda=(a+d)/2$ (no need to use the quadratic formula). In order to find the geometric multiplicity, we have to consider
$$
\begin{pmatrix} a-\lambda & b \\ c & d-\lambda \end{pmatrix}=
\begin{pmatrix} (a-d)/2 & b \\ c & (d-a)/2 \end{pmatrix}
$$
and the geometric multiplicity is $2$ if and only if this is the zero matrix, that is, $a=d$ and $b=c=0$. Thus the matrix must be diagonal to begin with. In the case of determinant $1$, the condition is that $a=d=1$ or $a=d=-1$, $b=c=0$.
