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(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?

Sorry
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  • That's a nonlinear system of equations. You can't solve them with Gaussian elimination. Of course $J'=AJA^{-1}$ will also square to $-I$ whenever $A$ has an inverse. – Angina Seng Nov 17 '19 at 11:42
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    This post answers your question (though perhaps with more advanced tools than you've learned about). In particular, every $J$ solving the equation must have the form $$ J = \pmatrix{y&x\-(y^2 + 1)/x&-y} $$ for some $x,y \in \Bbb R$ with $x \neq 0$. – Ben Grossmann Nov 17 '19 at 11:47
  • @LordSharktheUnknown I have not had my first lecture in LA so I'm very novice. I can't solve it using the elimination method. That I get. From $J^{\prime}=A J A^{-1}$ and to the end of your comment I do not understand. I think I will have to look I couple of things up. What concepts do you suggest that I look into? – Sorry Nov 17 '19 at 11:47
  • @Omnomnomnom Ok, I will try to see whether I understand it. – Sorry Nov 17 '19 at 11:49
  • If you haven't had your first lecture in LA, how did you come across this problem? – Ben Grossmann Nov 17 '19 at 11:54
  • @Omnomnomnom It is in the syllabus plan for week two in the course. – Sorry Nov 17 '19 at 11:58
  • Is it in the problem as an example that you're supposed to be able to solve by the end of week 2? If so, then that seems very ambitious for an introductory LA course. – Ben Grossmann Nov 17 '19 at 11:59
  • @Omnomnomnom It is on a list of challenging problems. These problems are not mandatory! I'm glad that you indicate it's a bit overkill because I find it very hard. – Sorry Nov 17 '19 at 12:10

2 Answers2

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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, $$ A=\begin{bmatrix}a & b \\ -\dfrac{a^2+1}{b} & -a\end{bmatrix} $$

lhf
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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.)

user1551
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