More fun in 2016, Part 18

Algebra Level 5

How many real $2\times 2$ matrices $A$ are there such that $A^{2016}=-I_2$ , where $I_2$ represents the $2\times 2$ identity matrix.

Enter 666 if you come to the conclusion that infinitely many such matrices $A$ exist.

The answer is 666.

3 solutions

Mark Hennings
Jan 14, 2016

Any matrix of the form $A \; = \; B\left(\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right)B^{-1}$ will do, where $B$ is a nonsingular real matrix and $\cos2016\theta = -1$ ; putting $\theta = \tfrac{\pi}{32}$ and $B \,=\, \left(\begin{array}{cc} 1 & a \\ 0 & 1 \end{array} \right)$ gives an infinite family of matrices $A_a \; = \; \left( \begin{array}{cc} \cos\frac{\pi}{32} + a\sin\tfrac{\pi}{32} & -(a^2+1)\sin\frac{\pi}{32} \\ \sin\frac{\pi}{32} & \cos\frac{\pi}{32} - a\sin\frac{\pi}{32} \end{array} \right)$ with $A_a^{32} \,=\, -I_2$ , and so certainly $A_a^{2016}\,=\, -I_2$ .

This does not quite work out. The 2016th power of your matrix will be $I_2$ since you are using a rotation through $\frac{\pi}{2}$ . Also there is an error in the matrix product.

Otto Bretscher - 5 years, 5 months ago

Oops! Problem fixed.

Mark Hennings - 5 years, 5 months ago

Great! It's really to same approach as mine (great minds think alike!); you are just using another matrix, a shear, to conjugate. We both move along an ellipse... ;)

Otto Bretscher - 5 years, 5 months ago

Otto Bretscher
Jan 9, 2016

Herr @Andreas Wendler is writing a fine solution. For the sake of variety let me suggest another solution based on @Calvin Lin 's idea.

Let $R$ be the rotation matrix through $\frac{\pi}{2016}$ , with $R^{2016}=-I_2$ . If $S$ is any invertible $2\times 2$ matrix, then $(S^{-1}RS)^{2016}=S^{-1}R^{2016}S=S^{-1}(-I_2)S=-I_2$ , so that $A^{2016}=-I_2$ for $A=S^{-1}RS$ . We need to make sure that we can construct infinitely many solutions $A$ this way.

For example, if we let $S=\begin{bmatrix}x&0\\0&1\end{bmatrix}$ , where $x\neq 0$ and $R=\begin{bmatrix} a&b\\c&d\end{bmatrix}$ , then $A=S^{-1}RS=\begin{bmatrix} a&b/x\\cx&d\end{bmatrix}$ , an infinite family of solutions for $x\neq 0$ .

Thus the answer is $\boxed{666}$ .

Great! We are simply doing a change of basis before performing the rotation.

Calvin Lin Staff - 5 years, 5 months ago

Calvin Lin Staff
Jan 6, 2016

[This is not a complete solution]

We want a linear transformation on the plane that when performed 2016 times results in a 180 degree rotation. There are 2016 of them, consisting of rotations of the form $(2n\pi +\pi)/2016$ .

So, why are there infinitely many of them?

But there are only 2016 such rotation matrices

$R_{\theta}=\begin{bmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{bmatrix}$

with $\theta=(2n+1)\pi/2016$ for $0\leq n \leq 2015$ , corresponding to the 2016 complex roots of $-1$ . We need to find other solutions to show that there are infinitely many.

Otto Bretscher - 5 years, 5 months ago

Ah yes. My bad. Let me edit the solution.

Calvin Lin Staff - 5 years, 5 months ago

Note that for matrices the association rule is valid, so:

$A^{2016}=A^{1008}*A^{1008}:=B*B=-I_{2}$

Now it must be fulfilled: $det(B∗B)=det(B)∗det(B)=[det(B)]^2=det(-I_{2})=1$

$det(B)=[det(A)]^{1008}$

We see that det(B) must be 1 which is possible for an infinite count of matrices A having a determinant of -1 or 1.

q.e.d.

Andreas Wendler - 5 years, 5 months ago

@Andreas Wendler – This does not quite clinch it. It is necessary that $\det A=\pm 1$ but not sufficient.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – We now have to construct A. Note that $A^{32}=-I_{2}$ because simply to proof each matrix $A^{32(1+2k)}=-I_{2}$ for natural k. We can write:

$A^{32}=A^{16}*A^{16}:=B*B$ ,

$A^{16}=B=A^{8}*A^{8}:=C*C$ ,

$A^{8}=C=A^{4}*A^{4}:=D*D$ ,

$A^{4}=D=A^{2}*A^{2}:=E*E$ ,

$A^{2}=E$

This gives us the possibility to determine A by recursion beginning with B from $B^{2}=-I_{2}$ tracing C, D and E. Already the first step lets two parameters of B free in choice. With real numbers s and t we get:

$b_{11}=s$ , $b_{12}=t$ $b_{21}=-\frac{1+s^{2}}{t}$ , $b_{22}=-s$

So we find out that an infinite set of matrices A must exist.

Andreas Wendler - 5 years, 5 months ago

@Andreas Wendler – It is my role as the setter of this problem to play "des Teufels Advokaten" until everything is crystal clear.

I see how you are giving us infinitely many matrices $B$ such that $B^2=-I_2$ , cleverly constructed, but I don't quite see the next step of the recursion. How are we getting the matrices $C$ such that $C^2=B$ ?

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – It is very smple: Successively we have to solve equation systems for the matrix members and since already B is not unique so A will not be!!!

Andreas Wendler - 5 years, 5 months ago

@Andreas Wendler – But how do you know that there exists a $C$ such that $C^2=B$ for the $B$ 's you constructed? Not every $2\times2$ matrix has a "square root".

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – Matrix B (like C...E) has determinante equal 1. So the inverse is existing and we can write: $B^{-1}*C^{2}=B^{-1}*B=I_{2}$

So with necessity $C^{2}$ must be B!

Andreas Wendler - 5 years, 5 months ago

@Andreas Wendler – Given $B$ , you have to show that there exists a $C$ such that $B=C^2$ . No such $C$ is given "a priori."

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – A few minutes ago I got a real solution for matrix C ( $1^{st}$ row: a, b and $2^{nd}$ row: c, d) by manually calculation related to matrix B(s,t) given in the 6th last posting:

$d=\sqrt{\frac{-s+\sqrt{2s^{2}+1}}{2}}$

$a=\sqrt{2s+d^{2}}$

$b=\frac{t}{a+d}$

$c=-b\frac{1+s^{2}}{t^{2}}$

We see that all parameter values for s and t produce real matrices C ! That's why the root of B is existing. As second conclusion we note that (similiar to real numbers) then the further (infinite) roots exist till we finally reach wanted matrix A !

Andreas Wendler - 5 years, 5 months ago

@Andreas Wendler – Thank you for your careful work! This has become quite a project.

I cannot agree with your statement that "the further roots" will necessarily exist; that step requires some more reasoning. As a simple counter example, consider the identity matrix which has as one of its roots the diagonal matrix with diagonal entries 1 and -1... but that matrix will not have a real square root.