Characteristic polynomial of $2\times2$ matrix $A$ with $A^2v=-v$ This is a multiple select question, i.e., more than one answer can be correct:

If $A\ne0$ is a $2\times2$ real matrix and suppose $A^2v=-v$ for all vectors $v\in\Bbb R^2$, then

*

*$-1$ is an eigenvalue of $A$,

*The characteristic polynomisal of $A$ is $\lambda^2+1$,

*The map from $\Bbb R^2\to\Bbb R^2$ given by $v\to Av$ is surjective,

*$\det A=1$.


My try:

*

*-1 can't be eigenvalue: If it is, then $Av=-1v$ for some nonzero $v$, hence $A^2v=A(Av)=A(-v)=-(Av)=-(-v)=v$, which contradicts hypothesis.

*Minimal polynomial must divide $x^2+1$, and since $A$ is real matrix, $x^2+1$ is minimal polynomial. But how to see if it is characteristic polynomial as well?

*Suppose $A$ not surjective, then image of $A$ is one or zero dimensional, hence $A^2$ has one or zero dimensions image. But $A^2v=-v$ says that $A^2$ is surjective, i.e. has two dimensional image.

*If option 2 correct, then this ids true as well.

I sam having serious trouble with option 2 and 4. Please help!
 A: You can also argue without Cayley-Hamilton and without minimal polynomials as follows:


*

*$A^2 = -I \Rightarrow (-A)A = I \Rightarrow A^{-1} = -A$
It follows immediately that $A$ is invertible and hence surjective. For eigenvalues $\lambda_1, \lambda_2$ of $A$ now follows (i=1,2)


*

*$\frac{1}{\lambda_i}=-\lambda_i\Rightarrow \lambda_i^2 = -1\Rightarrow \lambda_i = \pm i$
Since $A$ is real, the characteristic polynomial of $A$ must have both $i$ and $-i$ as roots. Hence $p_A(\lambda) = \lambda^2 \color{blue}{+1}$, which gives also $\color{blue}{\det A = 1}$.
A: characteristic polynomial of a matrix of order $2×2$ is of degree atmost 2 and minimal polynomial divides it,hence $x^{2}+1$ is the characteristic polynomial and $det(A) =1$
A: As the rest is easy, I'll just indicate an alternative way to get at the characteristic polynomial, without using either the Cayley-Hamilton theorem or complex numbers. There can be no real eigenvalues, as $A^2v=-v$ for en eigenvector $v$ shows the corresponding eigenvalue$~\lambda$ should have $\lambda^2=-1$, which it cannot. So for any nonzero vector $v$, its image $Av$ is linearly independent, and therefore $[v,Av]$ forms a basis. Now $A^2v=-v$ shows that change of basis of $A$ to the basis $[v,Av]$ gives
$$ \pmatrix{0&-1\\1&0} $$
whose characterisitic polynomial, which conincides with that of $A$, is $X^2+1$.
