Square root of a 2 by 2 matrix

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 general there can be no, two, four or even an infinitude of square root matrices. In many cases such a matrix R can be obtained by an explicit formula.

A 2 × 2 matrix with two distinct nonzero eigenvalues has four square roots. A positive-definite matrix has precisely one positive-definite square root.

Square roots of a matrix of any dimension 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) = R² = M.

One 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,

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

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

${\begin{array}{rcl}R^{2}&=&\displaystyle {\frac {1}{t^{2}}}\left({\begin{array}{cc}(A+s)^{2}+BC&(A+s)B+B(D+s)\\C(A+s)+(D+s)C&(D+s)^{2}+BC\end{array}}\right)\\[3ex]{}&=&\displaystyle {\frac {1}{A+D+2s}}\left({\begin{array}{cc}A(A+D+2s)&(A+D+2s)B\\C(A+D+2s)&D(A+D+2s)\end{array}}\right)\;=\;M.\end{array}}$

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. Also, note that R is positive when s>0 and t>0.

Special cases of the formula

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 $\tau$ = 1, giving M and -M as two square roots of M.

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. It also gives only two distinct solutions if δ is nonzero and τ² = 4δ (the case of duplicate eigenvalues), in which case one of the choices for s will make the denominator t be zero.

The formula above fails completely if δ and τ are both zero; that is, if D = −A and A² = −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 are

$R=\left({\begin{array}{cc}0&0\\c&0\end{array}}\right)\quad {\text{and}}\quad R=\left({\begin{array}{cc}0&b\\0&0\end{array}}\right)$

for any real or complex values of b and c. Otherwise M has no square root.

Simpler formulas for special matrices

Diagonal matrix

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.

Identity matrix

Because it has duplicate eigenvalues, the 2×2 identity matrix ${\bigl (}{\begin{smallmatrix}\\1&0\\0&1\end{smallmatrix}}{\bigr )}$ has infinitely many symmetric rational square roots given by

${\tfrac {1}{t}}{\bigl (}{\begin{smallmatrix}\\s&r\\r&-s\end{smallmatrix}}{\bigr )},$ ${\tfrac {1}{t}}{\bigl (}{\begin{smallmatrix}\\s&-r\\-r&-s\end{smallmatrix}}{\bigr )},$ ${\tfrac {1}{t}}{\bigl (}{\begin{smallmatrix}\\-s&r\\r&s\end{smallmatrix}}{\bigr )},$ ${\tfrac {1}{t}}{\bigl (}{\begin{smallmatrix}\\-s&-r\\-r&s\end{smallmatrix}}{\bigr )},$ ${\bigl (}{\begin{smallmatrix}\\1&0\\0&\pm 1\end{smallmatrix}}{\bigr )},$ and ${\bigl (}{\begin{smallmatrix}\\-1&0\\0&\pm 1\end{smallmatrix}}{\bigr )},$

where (r, s, t) is any Pythagorean triple—that is, any set of positive integers such that $r^{2}+s^{2}=t^{2}.$ In addition, any non-integer, irrational, or complex values of r, s, t satisfying $r^{2}+s^{2}=t^{2}$ give square root matrices. The identity matrix also has infinitely many non-symmetric square roots.

Matrix with one off-diagonal zero

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.