Disulfur difluoride

A square root of a 2 by 2 matrix M is another 2 by 2 matrix R such that M = R², where R² stands for the matrix product of R with itself. In many cases, such a matrix R can be obtained by an explicit formula.

The general formula

Let

$M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$

where A, B, C, and D may be real or complex numbers. Furthermore, let τ = A + D be the trace of M, and δ = AD - BC be its determinant. Let s be such that s² = δ, and t be such that t² = τ + 2s. That is,

$s=\pm \sqrt{\delta }t=\pm \sqrt{\tau +2s}$

Then, if t ≠ 0, a square root of M is

$R=\frac{1}{t}\left(\begin{array}{cc}A+s& B\\ C& D+s\end{array}\right)$

Indeed, the square of R is

$\begin{array}{rcl}{R}^2& =& \frac{1}{t^2}\left(\begin{array}{cc}(A+s{)}^2+BC& (A+s)B+B(D+s)\\ C(A+s)+(D+s)C& (D+s{)}^2+BC\end{array}\right)\\ & =& \frac{1}{A+D+2s}\left(\begin{array}{cc}A(A+D+2s)& (A+D+2s)B\\ C(A+D+2s)& D(A+D+2s)\end{array}\right)=M\end{array}$

Note that R may have complex entries even if M is a real matrix; this will be the case, in particular, if the determinant δ is negative.

Special cases of the formula

In general, the formula above will provide four distinct square roots R, one for each choice of signs for s and t. If the determinant δ is zero but the trace τ is nonzero, the formula will give only two distinct solutions. Ditto if δ is nonzero and τ² = 4δ, in which case one of the choices for s will make the denominator t be zero.

The general formula above fails completely if δ and τ are both zero; that is, if D = −A and A² = −BC. In this case, if M is the null matrix (with A = B = C = D = 0), then the null matrix is also a square root of M; othwerwise M has no square root.

Simplified formulas for special matrices

If M is diagonal (that is, B = C = 0), one can use the simplified formula

$R=\left(\begin{array}{cc}a& 0\\ 0& d\end{array}\right)$

where a = ±√A and d = ±√D; which, depending on the sign choices, gives four, two, or one distinct matrices, if none of, only one of, or both A and D are zero, respectively.

If B is zero but A and D are not both zero, one can use

$R=\left(\begin{array}{cc}a& 0\\ C/(a+d)& d\end{array}\right)$

This formula will provide two solutions if A = D, and four otherwise. A similar formula can be used when C is zero but A and D are not both zero.

References

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