Square root of a 2 by 2 matrix

A square root of a 2×2 matrix ''M'' is another 2×2 matrix ''R'' such that ''M'' = ''R''², where ''R''² stands for the matrix product of ''R'' with itself. In general, there can be zero, two, four, or even an infinitude of square-root matrices. In many cases, such a matrix ''R'' can be obtained by an explicit formula.

Square roots that are not the all-zeros matrix come in pairs: if ''R'' is a square root of ''M'', then −''R'' is also a square root of ''M'', since (−''R'')(−''R'') = (−1)(−1)(''RR'') = ''R''² = ''M''.
A 2×2 matrix with two distinct nonzero eigenvalues has four square roots. A positive-definite matrix has precisely one positive-definite square root.

A general formula

The following is a general formula that applies to almost any 2 × 2 matrix. Let the given matrix be
$M = \begin A & B \\ C & D \end,$
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 ''s''² = ''δ'', and ''t'' be such that ''t''² = ''τ'' + ''2s''. That is,
$s = \pm\sqrt, \qquad t = \pm\sqrt.$
Then, if ''t'' ≠ 0, a square root of ''M'' is
$R = \frac\begin A + s & B \\ C & D + s \end = \frac\left(M + sI\right).$

Indeed, the square of ''R'' is
\begin
R^2 &= \frac\begin
A^2 + B C + 2 s A + s^2 & A B + B D + 2 s B \\
C A + D C + 2 s C & C B + D^2 + 2 s D + s^2
\end \\ ex &= \frac\begin
A^2 + B C + 2 s A + A D - BC & A B + B D + 2 s B \\
A C + C D + 2 s C & B C + D^2 + 2 s D + A D - B C
\end \\ ex &= \frac\begin
A(A + D + 2 s) & B(A + D + 2 s) \\
C(A + D + 2 s) & D(A + D + 2 s)
\end = M.
\end

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.

The general case of this formula is when ''δ'' is nonzero, and ''τ''² ≠ 4''δ'', in which case ''s'' is nonzero, and ''t'' is nonzero for each choice of sign of ''s''. Then the formula above will provide four distinct square roots ''R'', one for each choice of signs for ''s'' and ''t''.

Special cases of the formula

If the determinant ''δ'' is zero, but the trace ''τ'' is nonzero, the general formula above will give only two distinct solutions, corresponding to the two signs of ''t''. Namely,
$R = \pm\frac\begin A & B \\ C & D \end = \pm\frac M,$
where ''t'' is any square root of the trace ''τ''.

The formula also gives only two distinct solutions if ''δ'' is nonzero, and ''τ''² = 4''δ'' (the case of duplicate eigenvalues), in which case one of the choices for ''s'' will make the denominator ''t'' be zero. In that case, the two roots are
$R = \pm\frac\begin A + s & B \\ C & D + s \end = \pm\frac \left(M + s I \right),$
where ''s'' is the square root of ''δ'' that makes ''τ'' − 2''s'' nonzero, and ''t'' is any square root of ''τ'' − 2''s''.

The formula above fails completely if ''δ'' and ''τ'' are both zero; that is, if ''D'' = −''A'', and ''A''² = −''BC'', so that both the trace and the determinant of the matrix are zero. 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'', as is any matrix
$R = \begin 0 & 0 \\ c & 0 \end \quad \text \quad R = \begin 0 & b \\ 0 & 0 \end,$

where ''b'' and ''c'' are arbitrary real or complex values. Otherwise ''M'' has no square root.

Formulas for special matrices

Idempotent matrix

If ''M'' is an idempotent matrix, meaning that ''MM'' = ''M'', then if it is not the identity matrix, its determinant is zero, and its trace equals its rank, which (excluding the zero matrix) is 1. Then the above formula has ''s'' = 0 and ''τ'' = 1, giving ''M'' and −''M'' as two square roots of ''M''.

Exponential matrix

If the matrix ''M'' can be expressed as real multiple of the exponent of some matrix ''A'', $M = r \exp(A)$, then two of its square roots are $\pm\sqrt\exp\left(\tfracA\right)$. In this case the square root is real.

Diagonal matrix

If ''M'' is diagonal (that is, ''B'' = ''C'' = 0), one can use the simplified formula
$R = \begin a & 0 \\ 0 & d \end,$

where ''a'' = ±√''A'', and ''d'' = ±√''D''. This, for the various sign choices, gives four, two, or one distinct matrices, if none of, only one of, or both ''A'' and ''D'' are zero, respectively.

Identity matrix

Because it has duplicate eigenvalues, the 2×2 identity matrix $\left(\begin 1 & 0 \\ 0 & 1 \end\right)$ has infinitely many symmetric rational square roots given by
$\frac \begin s & r\\ r & -s\end \text \begin \pm 1 & 0\\ 0 & \pm 1\end,$
where are any complex numbers such that $r^2 + s^2 = t^2.$

Matrix with one off-diagonal zero

If ''B'' is zero, but ''A'' and ''D'' are not both zero, one can use
$R = \begin a & 0 \\ \frac & d \end.$

This formula will provide two solutions if ''A'' = ''D'' or ''A'' = 0 or ''D'' = 0, 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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Matrices