If $2i$ is an eigenvalue of $A_{2\times 2} \in \mathbb{R}^{2\times 2}$, find $A^2$ If $2i$ is an eigenvalue of $A_{2\times 2} \in \mathbb{R}^{2\times 2}$, find $A^2$. 
Attempt:
So I'm trying to find $A$ first. If $2i$ is an eigenvalue, then $\det (A-2iI) = 0$.
Solving for $\det(A-2iI) = 0$ we find $$ad-i2a-i2d-4-bc = 0$$
However, I'm not sure how this simplifies to anything useful to determine $A$.
 A: Since $A$ is a real matrix and the eigenvalues are roots of the characteristic polynomial which is with real coefficients, then these eigenvalues are conjugate so the other eigenvalue is $-2i$ and then $A$ is diagonalizable since it has two distinct eigenvalues: $$A=P\operatorname{diag}(2i,-2i)P^{-1},\,\text{ hence }\,A^2=-4I_2$$
A: Let $\chi_A$ be the characteristic polynomial of $A$. It's a real polynomial, and one of its (complex roots) is $2i$, therefore $\overline{2i} = -2i$ is also one of its roots. Since $\deg \chi_A = 2$, these are all of its roots.  It follows that $A$ is diagonalizable over $\mathbb{C}$, $A = PDP^{-1}$ where $P \in GL_2(\mathbb{C})$, $D$ is diagonal with entries $2i, -2i$. Squaring $A$ you get that $A^2 = P(D^2)P^{-1} = -4I_2$.
A: HINT: Your matrix is diagonalizable and you know all of its eigenvalues. (Why?) So you can write $$\tag{1}A=TDT^{-1}$$ for an invertible matrix $T$. Moreover, you explicitly know $D$. Use (1) to compute $A^2$.
A: If $2i$ is an eigenvalue then so is $-2i$. On an eigenvector for eigenvalue $\pm2i$, $A^2$ acts as multiplication with $(\pm2i)^2=-4$. As the eigenvectors are independent, they form a basis, hence $A^2$ is multiplication with $-4$ throughout.
