Let $\lambda_{i}(A) = i$-th eigenvalue of $A$. Show that $A \mapsto \lambda_{i}(A)$ is differentiable in $\mathcal{M}_{2 \times 2}(\mathbb{R})$ 
Let $\mathcal{L}_{\text{sim}}(\mathbb{R}) \subset \mathcal{M}_{2 \times 2}(\mathbb{R})$ the subspace of symmetric matrix of order $2$. Consider the aplication
  $$\lambda_{i}: \mathcal{L}_{\text{sim}}(\mathbb{R}) \to \mathbb{R},\quad i=1,2,\;\; \lambda_{i}(A) = i\text{-th eigenvalue of }A.$$
  Show that $A \mapsto \lambda_{i}(A)$ is differentiable in $\mathcal{M}_{2 \times 2}(\mathbb{R})$ and to calculate $\lambda'_{i}(A)$, for every $A \in \mathcal{L}_{\text{sim}}(\mathbb{R})$.

I couldn't develop much, but I started thinking about dividing into two cases: $A$ is invertible and $A$ is not invertible.
I can to use $\operatorname{tr}(A) = \lambda_{1} + \lambda_{2}$ and $\det(A) = \lambda_{1}\lambda_{2}$. I tried to write $\lambda_{i}(A + H) = \lambda_{i}(A) + L(H) + o(H)$ where $L(H)$ is a linear function and $\displaystyle \frac{o(H)}{||H||} \to 0$. My problem is: I couldn't to expand $\lambda_{i}(A + H)$ so that it was viable. There are probably some properties that should be used that I cannot remember. Can anybody help me?
 A: Let's take a general approach with any $n \times n$ matrix $A$ and see how far we get. Then take a step back and apply this to $2\times 2$ symmetric matrices.
Part 1: Generic
Note that for any $n \times n$ matrices $A$ and $B$ $$\det(A + h B) \equiv \det(A) + h \operatorname{tr}(\operatorname{Adj}(A)\cdot B) \pmod{h^2}$$ where $\operatorname{Adj}(A)$ is the adjugate matrix. Let $P \in \mathbb{R}[x]$ be the characteristic polynomial of $A$
$$P(x) = \det(x - A)$$ and let $\lambda_B$ be the directional derivative of $\lambda$ in direction $B$ (which is what we want to compute). Then the formula for the derivative of a determinant above leads to
$$
\begin{eqnarray}
0 &=& \det(\lambda(A+ h B) - A - hB) \\
&\equiv& \det(\lambda(A) - A + h (\lambda_B(A) - B)) \\
&\equiv& h \operatorname{tr}(\operatorname{Adj}(\lambda(A) - A)\cdot(\lambda_B(A) - B)) \\
&\equiv& h \ (\lambda_B(A) P'(\lambda(A)) - \operatorname{tr}(\operatorname{Adj}(\lambda(A) - A)\cdot B)) 
\end{eqnarray}$$
and therefore
$$ \lambda_B(A) = \frac{\operatorname{tr}(\operatorname{Adj}(\lambda(A) - A)\cdot B)}{P′(λ(A))}.$$
Note that $\lambda_B$ is well defined except where $\lambda(A)$ is a double root of $P(x)$, in other words, if $\lambda(A)$ is an eigenvalue of $A$ with multiplicity greater than one.
Part 2: Specific
Now for $2 \times 2$ symmetric matrices $A$ the formula above simplifies quite a bit. Suppose that $A$ has eigenvalues $\lambda_1$ and $\lambda_2$ and decomposes as $$A = \lambda_1(A) \pi_1(A) + \lambda_2(A) \pi_2(A)$$ where $\pi_1(A)$ and $\pi_2(A)$ are orthogonal projections onto the eigenspaces. Then $$\operatorname{Adj}(\lambda_1(A) - A) = (\lambda_1(A) - \lambda_2(A))\ \pi_1(A)$$ and $$P'(\lambda_1(A)) = \lambda_1(A) - \lambda_2(A).$$ So we get as directional derivative $\lambda_{1B}(A)$ of $\lambda_1(A)$ in the direction $B$ $$\lambda_{1B}(A) = \operatorname{tr}(\pi_1(A) \cdot B).$$
