Square root of a 2 by 2 matrix

The following is a general formula that applies to almost any 2 × 2 matrix.{{r|Bernard1}} Let the given matrix be

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,

$$s\; =\; \backslash pm\backslash sqrt\{\backslash delta\},\; \backslash qquad\; t\; =\; \backslash pm\backslash sqrt\{\backslash tau\; +\; 2s\}.$$

Then, if t ≠ 0, a square root of M is

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

$$\backslash begin\{align\}$$

R^2 &= \frac{1}{t^2}\begin{pmatrix}

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{pmatrix} \\[1ex]

&= \frac{1}{t^2}\begin{pmatrix}

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{pmatrix} \\[1ex]

&= \frac{1}{A + D + 2 s}\begin{pmatrix}

A(A + D + 2 s) & B(A + D + 2 s) \\

C(A + D + 2 s) & D(A + D + 2 s)

\end{pmatrix} = M.

\end{align}

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,

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

R = \pm\frac{1}{t}\begin{pmatrix} A + s & B \\ C & D + s \end{pmatrix}

= \pm\frac{1}{t} \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{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, $M\; =\; r\; \backslash exp(A)$, then two of its square roots are $\backslash pm\backslash sqrt\{r\}\backslash exp\backslash left(\backslash tfrac\{1\}\{2\}A\backslash right)$. In this case the square root is real.{{r|Harkin2}}

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

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 has infinitely many symmetric rational square roots given by

\frac{1}{t} \begin{pmatrix} s & r\\ r & -s\end{pmatrix} \text{ and }

\begin{pmatrix} \pm 1 & 0\\ 0 & \pm 1\end{pmatrix},

where {{math|(r, s, t)}} are any complex numbers such that $r^2\; +\; s^2\; =\; t^2.${{r|Mitchell}}

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

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.

The algebra M(2, R) of 2x2 real matrices has three types of planar subalgebras. Each subalgebra admits the exponential map. If $p\; =\; \backslash exp(q),\; \backslash text\{then\}\; \backslash pm\; \backslash exp(\backslash frac\{q\}\{2\})$ are square roots of p. The condition that the matrix is the image under exp limits it to half the plane of dual numbers, and to a quarter of the plane of split complex numbers, but does not constrain ordinary complex planes since the exponential mapping covers them. In the split-complex case there are two more square roots of p since each quadrant contains one.

{{reflist|refs=

{{citation

| last = Levinger | first = Bernard W.

| date = September 1980

| doi = 10.1080/0025570X.1980.11976858

| issue = 4

| journal = Mathematics Magazine

| jstor = 2689616

| pages = 222–224

| title = The square root of a $2\backslash times\; 2$ matrix

| volume = 53}}

{{citation

| last1 = Harkin | first1 = Anthony A.

| last2 = Harkin | first2 = Joseph B.

| doi = 10.1080/0025570X.2004.11953236

| issue = 2

| journal = Mathematics Magazine

| jstor = 3219099

| mr = 1573734

| pages = 118–129

| title = Geometry of generalized complex numbers

| url = https://people.rit.edu/harkin/research/articles/generalized_complex_numbers.pdf

| volume = 77

| year = 2004}}

{{citation

| last = Mitchell | first = Douglas W.

| date = November 2003

| doi = 10.1017/S0025557200173723

| issue = 510

| journal = The Mathematical Gazette

| jstor = 3621289

| pages = 499–500

| title = 87.57 Using Pythagorean triples to generate square roots of $I\_2$

| volume = 87| doi-access = free

}}

}}

Category:Matrices (mathematics)