resposta: Alternativa E
Resolução:
$\,I_2\,$ é representação da matriz identidade de ordem 2, a saber $\;\begin{pmatrix} 1& 0\; \\ 0& 1 \end{pmatrix}\phantom{X}$.

---

$\,A^{\large 2}\,= \;\begin{pmatrix} 1& 1\; \\ 1& 1 \end{pmatrix}\,\centerdot\,\begin{pmatrix} 1& 1\; \\ 1& 1 \end{pmatrix}\;\Rightarrow\;$ $\;\begin{pmatrix} 1+1& 1+1\; \\ 1+1& 1+1 \end{pmatrix}\phantom{X}\;\Rightarrow\;$ $\;\begin{pmatrix} 2& 2\; \\ 2& 2 \end{pmatrix}$

---

$\,{\large \lambda}I_2\;=\;{\large \lambda}\centerdot\begin{pmatrix} 1& 0\; \\ 0& 1 \end{pmatrix}\;=\;$ $\begin{pmatrix} {\large \lambda}& 0\; \\ 0& {\large \lambda} \end{pmatrix}\;$

---

Então
$\,A^2 \,-\,\lambda I_2\;=\;\begin{pmatrix} 2& 2\; \\ 2& 2 \end{pmatrix}\, - \,\begin{pmatrix} {\large \lambda}& 0\; \\ 0& {\large \lambda} \end{pmatrix}\,=$ $\,\begin{pmatrix} 2-{\large \lambda}& 2\; \\ 2& 2-{\large \lambda} \end{pmatrix}$

---

O determinante de $\,\begin{pmatrix} 2-{\large \lambda}& 2\; \\ 2& 2-{\large \lambda} \end{pmatrix}\,$ é $\,(2\,-\,{\large \lambda})^{\large 2}\,-\,2^{\large 2}\,=$ $\,(2\,-\,{\large \lambda})^{\large 2}\,-\,4\,=\,0\Rightarrow\;$ $\,2^2\,-\,4{\large \lambda}\,+\,{\large \lambda}^2\,-\,4\,=\,0\;\Rightarrow\;$ ${\large \lambda}^2\,-\,4{\large \lambda}\,=\,0\;\Rightarrow\,$ $\,\left\{\begin{array}{rcr} &{\large \lambda}\,=\,0\;\mbox{ ou } \\ &{\large \lambda}\,=\,4\phantom{XX} \\ \end{array} \right.\,$
×