Is there a $2 \times 2$ real matrix $A$ such that $A^2=-4I$?

Does there exist a $2 \times 2$ matrix $A$ with real entries such that $A^2=-4I$ where $I$ is the identity matrix?

Some initial thoughts related to this question:

• The problem would be easy for complex matrices, we could simply take identity matrix multiplies by $2i$.
• There is another question on this site showing that this has no solution for $3\times3$ matrices, since the LHS has determinant $\det(A^2)=\det^2(A)$ which is a square of real number, but determinant of $-4I_3$ is negative. But the same argument does not work for $2\times2$ matrices, since $\det(-4I_2)=4$ is positive.
• Send $e_1$ to $-2e_2$ and $e_2$ to $2e_1$! – André Caldas Sep 28 '17 at 6:00
• @MartinSleziak Thank you Martin for your kind attempts to save this question, you are elevating my belief in human generosity... – Widawensen Feb 19 '18 at 13:19
• This post might be useful, too: If $A^2 = I$ (Identity Matrix) then $A = \pm I$. – Martin Sleziak Feb 22 '18 at 4:32

My previous answer to this question was the "quick and easy" one, merely showing the existence of a matrix $A$ with

$A^2 = -4, \tag 1$

as per our OP user485215's request, without going into the deeper theory, other than to note that the matrix

$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \tag 2$

satisfies

$J^2 = -I, \tag 3$

and may be used to build $A$ as follows:

$A = 2J \tag 4$

so in a sense $J$ corresponds to the imaginary unit $i$ (or $-i$ in the light of the comments of zwim to my other answer). But there are many other solutions for $J$, and hence for $A$; my intention here, in the spirit of the MMM's answer, is to paremetrically present all such matrices $J$. This is in fact a favorite topic of mine; I have been thinking about it and related matters for a long time, so I am glad to have a chance to post my remarks here.

Suppose we look for the most general $J$ satisfying (3); we may write

$J = \begin{bmatrix} j_{11} & j_{12} \\ j_{21} & j_{22} \end{bmatrix}, \tag 5$

with the $j_{kl} \in \Bbb R$, so that

$J^2 = \begin{bmatrix} j_{11} & j_{12} \\ j_{21} & j_{22} \end{bmatrix} \begin{bmatrix} j_{11} & j_{12} \\ j_{21} & j_{22} \end{bmatrix} = \begin{bmatrix} j_{11}^2 + j_{12}j_{21} & j_{11}j_{12} + j_{12} j_{22} \\ j_{21}j_{11} + j_{22}j_{21} & j_{21}j_{12} + j_{22}^2\end{bmatrix}$ $= \begin{bmatrix} j_{11}^2 + j_{12}j_{21} & (j_{11} + j_{22})j_{12} \\ (j_{11} + j_{22})j_{21} & j_{21}j_{12} + j_{22}^2\end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}, \tag 6$

from which we extract the four equations for the $j_{kl}$:

$j_{11}^2 + j_{12}j_{21} = -1, \tag 7$

$(j_{11} + j_{22})j_{12} = 0, \tag 8$

$(j_{11} + j_{22})j_{21} = 0, \tag 9$

$j_{21}j_{12} + j_{22}^2 = -1. \tag{10}$

(7)-(10) suggest that $\text{Tr} J = j_{11} + j_{22}$ may provide a useful grip on the succeeding analysis. If

$j_{11} + j_{22} \ne 0, \tag{11}$

(8) and (9) imply

$j_{12} =j_{21} = 0, \tag{12}$

and then (7) and (10) become

$j_{11}^2 = j_{22}^2 = -1, \tag{13}$

clearly impossible for real $j_{11}$, $j_{22}$. Ruling out (11), we must have

$j_{11}+j_{22} = 0, \tag{14}$

which suggests introducting a parameter $\alpha$ with

$j_{11} = \alpha = -j_{22}; \tag{15}$

then (7) aand (10) each become

$\alpha^2 + j_{12}j_{21} = -1, \tag{16}$

so if we set

$j_{12} = \beta, \tag{17}$

we see that

$j_{21} = \dfrac{-1 - \alpha^2}{\beta} = -\dfrac{1 + \alpha^2}{\beta}; \tag{18}$

now we may present the family of all $J$ parametrically:

$J = \begin{bmatrix} \alpha & \beta \\ -\dfrac{1 + \alpha^2}{\beta} & -\alpha \end{bmatrix}; \tag{19}$

here $\alpha \in \Bbb R$ may be taken arbitrarily but we must have $0 \ne \beta \in \Bbb R$. It is easily verified that any $J$ defined by (19) satisfies (2); thus we have presented all such $J$ in parametric form in (19).

It follows that every $A$ satisfying (1) is of the form

$A = 2J, \tag{20}$

with $J$ as in (19). For example, following up on MMM's example

$A = \begin{bmatrix} 1 & 1 \\ -5 & -1 \end{bmatrix}, \tag{21}$

we take

$J = \begin{bmatrix} \dfrac{1}{2} & \dfrac{1}{2} \\ -\dfrac{5}{2} & -\dfrac{1}{2} \end{bmatrix}, \tag{21}$

which corresponds to $\alpha = \beta = 1/2$. Widawensen's example

$A = \begin{bmatrix} 2 & -1 \\ 8 & -2 \end{bmatrix} \tag{22}$

yields

$J = \begin{bmatrix} 1 & -\dfrac{1}{2} \\ 4 & -1 \end{bmatrix}, \tag{23}$

with $\alpha = 1$ and $\beta = -1/2$; we can obviously go a long way in this direction.

Finally, it is engaging to observe, in the light of Widawensen's remark on matrices $A$ with prime entries, that there are many $J$ with prime entries as well; we simply choose $\alpha$ prime such that $\alpha^2 + 1$ is the product of precisely two prime factors, $\alpha^2 + 1 = pq$; then set $\beta = p$ and so $q = (1 + \alpha^2)/p$, for instance with $\alpha = 5$, we may take $\beta = 2$ and find

$J = \begin{bmatrix} 5 & 2 \\ -13 & -5 \end{bmatrix}; \tag{24}$

the list goes on, but whether it has any number-theoretic use or significance, I do not know . . .

• Solutions only in prime numbers are interesting cases, which can be further developed, but it's meaning I suppose is only important as an some kind of exercise for number theory. For example, interesting is equation $A^2=-5I$, where we have prime 5, and in the matrix $A$ are present 2,3,(-3),-7 so we have a set of consecutive primes. I suppose there is no many such sets.. – Widawensen Sep 29 '17 at 10:04

Every matrix satisfies its characteristic polynomial. So, any matrix with characteristic polynomial $\lambda^2+4$ would work. Taking the diagonal to be zero and the off diagonal entries to be $4$ and $-1$ would give you a solution.

A 2-by-2 matrix is not too bad to work out the algebraic equation for its coefficients:

Let $A=\begin{bmatrix} a & b \\c & d\end{bmatrix}$. We end up with the matrix equation: $$A^2 = \begin{bmatrix} a^2+bc & b(a+d) \\c(a+d) & d^2 + bc\end{bmatrix} = -4 \, I = \begin{bmatrix} -4 & 0 \\0 & -4\end{bmatrix}.$$

So we have 4 equations $$\begin{equation} a^2+bc = -4 \quad (1)\\ b(a+d) = 0 \quad (2) \\ c(a+d) = 0 \,\quad (3) \\ d^2+bc = -4 \quad (4). \end{equation}$$

It is easy to see that $a + d = 0$. Since if otherwise $a+d \not=0$, equations (2) and (3) it would imply that: $b=c=0$, then from equation (1): $a^2 = -4$ which can't happen.

Since $a+d=0$, we have $a^2 = d^2$. Thus equations (1) and (4) are identical. Therefore, the conclusion is that there are infinitely many solutions, as long as $a,b,c,d$ satisfy that $$\begin{cases} a+d=0 \quad &(5)\\ a^2 = -(4+bc) \quad &(6). \end{cases}$$

Since there are 4 variables with 2 equation, one can choose 2 free variables and derive the others from the free ones. The simplest is from suggestion of user "zwim":

$$A=\begin{bmatrix} a & -\frac{a^2+4}{c} \\c & -a\end{bmatrix},$$

where $a$ and $c$ are any real numbers.

Example: choose a = 1, c = -5 we have:

$A=\begin{bmatrix} 1 & 1 \\-5 & -1\end{bmatrix}.$

• I think it is simpler to have free variables being $a,c$ and deduce $b,d$ from them. $\begin{bmatrix}a&-\frac{a^2+4}c\\c&-a\end{bmatrix}$ – zwim Sep 28 '17 at 17:29
• @zwim, I agree, I modified my answer with your suggestion. – MMM Sep 30 '17 at 17:47

Yes.

$A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}; \tag 1$

There are others. Note that if

$J =\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \tag 2$

then

$J^2 = -I, \tag 3$

so $J$ is a real $2 \times 2$ matrix "version" of $i \in \Bbb C$. Here,

$A = 2J. \tag 4$

• If we consider the standard mapping of 2d-rotations by matrices $\begin{bmatrix}\cos \theta & -\sin \theta\\\sin \theta & \cos \theta\end{bmatrix}$ then for $\theta=\frac{\pi}2$ we should have $J=\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$ also verifying $J^2=-I$ as the "version" of complex $i$. Your $J$ is more the version of $-i$. – zwim Sep 28 '17 at 6:22

We have two matrix equations $A^2=-4I$ i.e.
$A^2+4I=0$

and general equation from Cayley-Hamilton theorem for $2 \times 2$ matrices $A^2-\text{tr}(A)A+\det(A)I=0$.

Comparing both equations we obtain

$\text{tr(A)}=0$ , $\det(A)=4$.

So if we denote $A=\begin{bmatrix} a & b \\c & d\end{bmatrix}$ then $d=-a$ and consequently $-a^2-bc=4$.
These conditions are sufficient to obtain an infinite number of solutions, even with integer values.
Check for example $A=\begin{bmatrix} 2 & -1 \\8 & -2\end{bmatrix}$.

Factorizing $a^2+4$ (with assistance for example Number Empire site ) you can obtain even a less obvious integer solutions, for example $A=\begin{bmatrix} 23 & -13 \\41 & -23\end{bmatrix}$ where all absolute values of entries are prime numbers.

• Note that $\text{tr}(A)A=(\det(A)-4)I$ can be only satisfied when $\det(A)-4=0$, otherwise it would mean that $A$ is multiply of $I$ what is obviously impossible for real $A$ where $A^2=-4I$. – Widawensen Feb 19 '18 at 13:36

$A = \left [\begin{array}{ccc} 0 & 2 \\ -2 & 0 \\ \end{array} \right ]$.