Calculating square root of a matrix

Question

I was trying to solve this equation over here.

$\begin{bmatrix}a&b\\c&d\end{bmatrix}^2 = \begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ I realize that there are some obvious ones like $\begin{bmatrix}i&0\\0&i\end{bmatrix}$, $\begin{bmatrix}-i&0\\0&i\end{bmatrix}$, $\begin{bmatrix}-i&0\\0&-i\end{bmatrix}$ $\begin{bmatrix}i&0\\0&-i\end{bmatrix}$ These solutions are also what I've found in wolframalpha.

However, I also found

$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$

If $a$, $b$, $c$, and $d$ are all complex numbers, are there any solutions that I missed?

Answer 1

We want to solve the matrix equation $X^2+I=0$ for complex $2\times 2$-matrices.

By Cayley Hamilton, $X^2-tr(X)X+\det(X)I=0$. Substituting $X^2=-I$, we obtain $tr(X)X=(\det(X)-1)I$, i.e., $b\cdot tr(X)=c\cdot tr(X)=0$ with $tr(X)=a+d$.

Case $1$: $b$ or $c$ is nonzero. Then $0=tr(X)=a+d$ and $a^2+bc+1=0$.

Case $2$: $b=c=0$. Then $a^2=d^2=-1$.

This yields all solutions.

Answer 2

The matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ has the Jordan form of $PJP^{-1}$ with

$$J = \begin{bmatrix} \lambda & 0 \\ 0 & \mu \end{bmatrix} \qquad \text{or} \qquad J = \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}$$

for some invertible matrix $P \in M_{2 \times 2}(\mathbb{C})$ and $\lambda, \mu \in \mathbb{C}$. Substituting into $M^2 = -I$ after easy transformations gives $J^2 = -I$.

In the first case this is equivalent to $\lambda^2 = \mu^2 = -1$ and that gives a family of solutions.

In the second case this gives $\lambda^2 = -1$ and $2 \lambda = 0$, which is impossible.

Therefore the solutions are exactly the matrices of the form

$$M = P \begin{bmatrix} \pm i & 0 \\ 0 & \pm i \end{bmatrix} P^{-1}.$$

This can be further simplified to $M = \pm i I$ or

$$M = iP \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} P^{-1}$$

(the symmetric placement of the $-$ sign can be achieved by swapping the eigenvectors in $P$). So essentially $M$ is either multiplication by $\pm i$, or a complex line symmetry scaled by $i$.

Answer 3

To full-proof this, we just have to make the computations in four unknowns for four equations:

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^2= \begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&bc+d^2\end{pmatrix}$$ or

$$\begin{align}a^2+bc&=-1\\d^2+bc&=-1\\b(a+d)&=0\\c(a+d)&=0\end{align}.$$ Then we arrive at the four cases $(b\neq0\neq c)\vee (b=0\neq c)\vee (c=0\neq b)\vee( b=0=c)$. Of course if $b=c=0$ we have $a^2=d^2=-1$ and the four cases given by $$\begin{pmatrix}\pm_1i&0\\0&\pm_2i\end{pmatrix},$$ as you had at first. If only $b=0$ we have $a+d=0$ and so our matrices are $$\alpha\in \mathbb C: \begin{pmatrix}\pm i&0\\\alpha&\mp i\end{pmatrix},$$ the transpose of this case is the base where $b$ is nonzero but $c$ is 0; and the final case is when both are nonzero so that $a=-d$; in this case our only condition becomes $a^2+bc=-1$ and so letting $a$ be our first free parameter and $c=\beta$ our second we have $$\alpha,\beta\in \mathbb C: \begin{pmatrix}\alpha&-\frac{1+\alpha^2}{\beta}\\\beta&-\alpha\end{pmatrix}.$$

Answer 4

One more solution for $c\neq 0$ is $$\begin{bmatrix}-d & \frac{-d^2-1}{c} \\ c & d\end{bmatrix}$$

Answer 5

You know how to calculate the square root of a real number by starting with an approximation and repeatedly performing Newton’s iteration.

You can try to apply the exact same method with matrices. And like with real numbers where you are not going to find a solution to x^2 = -2, this will not always succeed.

Answer 6

You're asking for matrices with the characteristic polynomial of $x^2+1=0$.

https://en.wikipedia.org/wiki/Characteristic_polynomial

In linear algebra, the characteristic polynomial of a square matrix is a polynomial which is invariant under matrix similarity and has the eigenvalues as roots. It has the determinant and the trace of the matrix among its coefficients. The characteristic polynomial of an endomorphism of a finite-dimensional vector space is the characteristic polynomial of the matrix of that endomorphism over any basis (that is, the characteristic polynomial does not depend on the choice of a basis). The characteristic equation, also known as the determinantal equation,[1][2][3] is the equation obtained by equating the characteristic polynomial to zero.

Once you have one solution, the set of solutions is the similarity class of that matrix.