Find all solutions $A \in M_{2}(\mathbb{R})$ and $B \in M_{2}(\mathbb{C})$ to $J^2+I=0$

Question

(a) Show $J=\left(\begin{array}{rr}{0} & {-1} \\ {1} & {0}\end{array}\right) \in M_{2}(\mathbb{R}$ is a solution to $J^{2}+I=O$.

By inserting the matrix in the matrix equation.

(b) Are there other solutions $A \in M_{2}(\mathbb{R})$ to the above equation?

Yes, $-J$.

(c) Find all solutions $A \in M_{2}(\mathbb{R})$ and $B \in M_{2}(\mathbb{C})$ to $J^2+I=0$.

Let $A \in M_{2}(\mathbb{R})$ be given by

$$A=\left(\begin{array}{rr}{a} & {b} \\ {c} & {d}\end{array}\right)$$

By inserting $A$ in the equation I get

$$\left[\begin{array}{cc}{a^{2}+b c+1} & {a b+b d} \\ {c a+d c} & {b c+d^{2}+1}\end{array}\right] = \left[\begin{array}{ll}{0} & {0} \\ {0} & {0}\end{array}\right]$$

I'm stuck here.

This means $a^{2}+b c+1=0$ etc. I get four equations in four variables. To my minds eye they look non linear. Via the Gauss-Jordan elimination method I get (calculated with Maple)

$$\left[\begin{array}{llll}{1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0}\end{array}\right]$$

According to Maple there are no solutions. But that does not make any sense since for example

$$J=\left(\begin{array}{cc}{0} & {-1} \\ {1} & {0}\end{array}\right)$$ is a solution.

What do I do now?

Answer 1

Continuing where you stopped. We have $a^{2}+b c+1=0$. We can't have $b=0$ or $c=0$ because $a^2+1=0$ cannot happen in $\mathbb R$. Therefore, $b\ne0$ and so $a b+b d=0$ implies $d=-a$. Thus, $$ A=\begin{bmatrix}a & b \\ -\dfrac{a^2+1}{b} & -a\end{bmatrix} $$

Answer 2

Suppose $u$ is a nonzero vector in $\mathbb R^2$. It cannot be an eigenvector of $J$ over $\mathbb R$, or else the equalities $J^2=-I$ and $Ju=\lambda u$ together would imply that $\lambda^2=-1$ for some real number $\lambda$, which is impossible. Thus $u$ and $v=Ju$ are linearly independent. Since $Ju=v$ and $Jv\,(=J^2u)=-u$, the matrix of the linear operator $x\mapsto Jx$ with respect to the ordered basis $\{u,v\}$ is given by $R=\pmatrix{0&-1\\ 1&0}$, meaning that $J=PRP^{-1}$ where $P$ is the augmented matrix $[u|v]$. Clearly, a matrix of this form always satisfies the equation $J^2=-I$. Hence the solutions to $J^2=-I$ are precisely matrices in this form.

(Alternatively, since $J$ is $2\times2$ and it is annihilated by the quadratic polynomial $x^2+1$ that is irreducible over $\mathbb R$, $J$ is similar to the companion matrix of $x^2+1$. Thus $J=PRP^{-1}$ where $R$ is the matrix mentioned before.)