# 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 $B=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

-
More than 18,000 years ahead, and we already know what question 5 will be. Wow. –  Did Nov 8 '12 at 19:07
Funny, that's the course number :-) –  Robert S. Barnes Nov 8 '12 at 19:11

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.

-
Great answer, really elegant. –  Robert S. Barnes Nov 9 '12 at 7:24

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.

-
\pm produces $\pm$. –  Brian M. Scott Nov 8 '12 at 18:52
I think I can follow the general idea of your part A but it certainly can be clearer (You don't need to use your 2). Your part B looks good. For part C, it is not true in general that having the same eigenvalues with the same multiplicities means that the two matrices are similar. This is further complicated by the fact that the eigenvalues/eigenspaces are complex while your similarity is (I assume) real. In any case, part C definitely requires more elaboration. –  EuYu Nov 8 '12 at 19:17
It should say they are both diagonalizable because they both have a full set of linearly independent eigenvectors, and then having the same eigenvectors with the same geometric multiplicities makes them similar. –  Robert S. Barnes Nov 8 '12 at 20:27
You can use \det for the determinant operator. –  Asaf Karagila Nov 8 '12 at 20:32
Christian Blatter has given you an excellent answer for part C. An alternative is this. The matrix $A$ is indeed diagonalizable over $\mathbb{C}$ because the minimal polynomial splits into distinct linear factors. It follows that the matrix $B$ will be similar to $A$ in $\mathbb{C}$ for the reasons you stated. Matrices over $\mathbb{R}$ are similar in $\mathbb{C}$ if and only if they are also similar in $\mathbb{R}$ (see here) so your statement holds. –  EuYu Nov 8 '12 at 23:14