Question 5 from Summer 2012 Exam Course 20109 Let $A$ be a $2\times2$ matrix over $\mathbb{R}$ such that $A^2=-I$
A) Show that $A$ has no real eigenvalues.
B) Let $v\in \mathbb{R}^2$, $v\neq0$.  Show that $\{v, Av\}$ is a basis for $\mathbb{R}^2$.
C) Show that $A$ is similar to 
$B=\begin{bmatrix}
0 & -1 \\ 
1 & 0
\end{bmatrix}$
 A: The matrix $A$ defines a linear map $\tilde A:\ X\to X$ where $X:={\mathbb R}^2$. To be exact: $A$ is the matrix of $\tilde A$ with respect to the standard basis $\bigl((1,0),(0,1)\bigr)$ of ${\mathbb R}^2$.
(a) Assume $\lambda\in{\mathbb R}$ is a real eigenvalue of $\tilde A$. Then there is a nonzero vector ${\bf e}\in X$ with $A{\bf e}=\lambda{\bf e}$ and therefore $A^2{\bf e}=A\bigl(A{\bf e}\bigr)=\lambda^2{\bf e}$. On the other hand, by assumption on $A$ we know that $A^2{\bf e}=-{\bf e}$, and as ${\bf e}\ne{\bf 0}$ we necessarily would have $\lambda^2=-1$, which is impossible for a real $\lambda$.
(b) Given any ${\bf v}\ne{\bf 0}$ the vector $A{\bf v}$ cannot be a scalar multiple of ${\bf v}$, by (a). It follows that the two vectors ${\bf v}$ and $A{\bf v}$ are linearly independent, and as $X$ is two-dimensional, they form a basis of $X$.
(c) Choose any ${\bf v}\ne{\bf 0}$ and use ${\bf e}_1:={\bf v}$, ${\bf e}_2:=A{\bf v}$ as basis of $X$. Then 
$$A{\bf e}_1={\bf e}_2\ ,\qquad A{\bf e}_2=A^2{\bf e}_1=-{\bf e}_1\ .$$
This means that with respect to the basis $({\bf e}_1,{\bf e}_2)$ of $X$ the map $\tilde A$ has matrix
$$\left[\matrix{0&-1\cr 1&0\cr}\right]=:B\quad.$$
By general principles of linear algebra it follows that the matrices $A$ and $B$ are similar.
A: A) I used the following:
1) $A^kv=\lambda^kv$ 
2) $(A+\beta I)v=(\lambda +\beta)v$
In all cases $v\neq0$ because $v$ is an eigenvector.
$A^2=-I \rightarrow A^2+I=0 \rightarrow 0$ is the only eigenvalue of $A^2+I$
It follows from that $2$ that $-1$ is the only eigenvalue of $A^2$, because $\lambda +1=0$.
It follows from $1$ that $\pm\sqrt{-1}$ is the only eigenvalue of $A$.
B) Assume $v,Av$ is not a basis. Then there exist $a,b\in \mathbb{R}$ not both zero such that $av+bAv=0$.  Assume $b$ is not zero. Then $Av=-\frac{a}{b}v$ and we have $-\frac{a}{b}\in\mathbb{R}$ is an eigenvalue of $A$ in contradiction to part A.
C) $\det(B-\lambda I)=\begin{bmatrix}
-\lambda & -1 \\ 
1 & -\lambda
\end{bmatrix}= \lambda^2+1$.  They are both diagonalizable because they both have a full set of linearly independent eigenvectors, and $B$ has the same eigenvalues of $A$ with the same geometric multiplicity, so they are similar.
