Square root of a 2 by 2 matrix explained

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.

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 s² = δ, and t be such that t² = 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 $\begin R^2 &= \frac\begin A^2 + B C + 2 s A + s^2 & A B + B D + 2 s B \\ C A + D C + 2 s C & C B + D^2 + 2 s D + s^2 \end \\[1ex] &= \frac\begin A^2 + B C + 2 s A + A D - BC & A B + B D + 2 s B \\ A C + C D + 2 s C & B C + D^2 + 2 s D + A D - B C \end \\[1ex] &= \frac\begin A(A + D + 2 s) & B(A + D + 2 s) \\ C(A + D + 2 s) & D(A + D + 2 s) \end = M.\end$

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 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.

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 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 $R = \pm\frac\begin A + s & B \\ C & D + s \end = \pm\frac \left(M + s I \right),$ where s is the square root of δ that makes τ − 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 $R = \begin 0 & 0 \\ c & 0 \end\quad \text \quadR = \begin 0 & b \\ 0 & 0 \end,$

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{r}\exp\left(\tfrac{1}{2}A\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

\left(\begin{smallmatrix}1&0\ 0&1\end{smallmatrix}\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

r²+s²=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.