A square root of a 2×2 matrix M is another 2×2 matrix R such that M = R², where R² 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) = R² = 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.

The following is a general formula that applies to almost any 2 × 2 matrix.^[1] Let the given matrix be $${\displaystyle M={\begin{pmatrix}A&B\\C&D\end{pmatrix}},}$$ 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, $${\displaystyle s=\pm {\sqrt {\delta }},\qquad t=\pm {\sqrt {\tau +2s}}.}$$ Then, if t ≠ 0, a square root of M is $${\displaystyle R={\frac {1}{t}}{\begin{pmatrix}A+s&B\\C&D+s\end{pmatrix}}={\frac {1}{t}}\left(M+sI\right).}$$

Indeed, the square of R is $${\displaystyle {\begin{aligned}R^{2}&={\frac {1}{t^{2}}}{\begin{pmatrix}A^{2}+BC+2sA+s^{2}&AB+BD+2sB\\CA+DC+2sC&CB+D^{2}+2sD+s^{2}\end{pmatrix}}\\[1ex]&={\frac {1}{t^{2}}}{\begin{pmatrix}A^{2}+BC+2sA+AD-BC&AB+BD+2sB\\AC+CD+2sC&BC+D^{2}+2sD+AD-BC\end{pmatrix}}\\[1ex]&={\frac {1}{A+D+2s}}{\begin{pmatrix}A(A+D+2s)&B(A+D+2s)\\C(A+D+2s)&D(A+D+2s)\end{pmatrix}}=M.\end{aligned}}}$$

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 τ² ≠ 4δ, 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.

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, $${\displaystyle R=\pm {\frac {1}{t}}{\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\pm {\frac {1}{t}}M,}$$ where t is any square root of the trace τ.

The formula 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. In that case, the two roots are $${\displaystyle R=\pm {\frac {1}{t}}{\begin{pmatrix}A+s&B\\C&D+s\end{pmatrix}}=\pm {\frac {1}{t}}\left(M+sI\right),}$$ where s is the square root of δ that makes τ − 2s nonzero, and t is any square root of τ − 2s.

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 is any matrix $${\displaystyle R={\begin{pmatrix}0&0\\c&0\end{pmatrix}}\quad {\text{and}}\quad R={\begin{pmatrix}0&b\\0&0\end{pmatrix}},}$$

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

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.

If the matrix M can be expressed as real multiple of the exponent of some matrix A, ${\displaystyle M=r\exp(A)}$, then two of its square roots are ${\displaystyle \pm {\sqrt {r}}\exp \left({\tfrac {1}{2}}A\right)}$. In this case the square root is real.^[2]

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

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.

Because it has duplicate eigenvalues, the 2×2 identity matrix ${\displaystyle \left({\begin{smallmatrix}1&0\\0&1\end{smallmatrix}}\right)}$ has infinitely many symmetric rational square roots given by $${\displaystyle {\frac {1}{t}}{\begin{pmatrix}s&r\\r&-s\end{pmatrix}}{\text{ and }}{\begin{pmatrix}\pm 1&0\\0&\pm 1\end{pmatrix}},}$$ where (r, s, t) are any complex numbers such that ${\displaystyle r^{2}+s^{2}=t^{2}.}$^[3]

If B is zero, but A and D are not both zero, one can use $${\displaystyle R={\begin{pmatrix}a&0\\{\frac {C}{a+d}}&d\end{pmatrix}}.}$$

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.