Is the square root of a 2 by 2 matrix unique?

Question

How would I know if the square root of a 2 by 2 matrix is unique?

Have you tried to write down the square roots of the identity matrix? of the zero matrix? — ancient mathematician, Mar 07 '17 at 17:08
Diagonal $2\times 2$-matrices with positive diagonal elements have $4$ distinct square roots, so no, the square root is not unique in general. — Peter, Mar 07 '17 at 17:12
@Peter Is there a way to find all of the 4 square roots ? I only found one — Wonder Women, Mar 07 '17 at 17:25
If you have a square root of a diagonal matrix, you can multiplay one or both diagonal elements with $-1$ and still have a square root. — Peter, Mar 07 '17 at 17:26

score 2 · Accepted Answer · answered Mar 09 '17 at 02:39

Seeing that square roots of a matrix are never unique is actually fairly simple. If $\sqrt{A}$ is a square root of $A$ such that

$$ \sqrt{A}\sqrt{A}=A $$

then $ (-\sqrt{A})$ is also a square root:

$$ (-\sqrt{A})(-\sqrt{A}=\sqrt{A}\sqrt{A}=A\ .$$

So the square root of a matrix is never unique. To work out what the square root actually is we can write

$$ \sqrt{A}= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\ . $$

Then

$$ A = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \end{pmatrix} = \begin{pmatrix} w & x \\ y & z \end{pmatrix}\ . $$

So for a given matrix $A$, $\sqrt{A}$ is given by the solution of the set of simultaneous equations

$$\begin{gather} a^2 + bc = w,\\ (a + d)b = x, \\ (a + d)c = y,\\ bc + d^2 = z. \end{gather}$$

However, as (almost) pointed out in another answer, the matrix where $w = y = z = 0$ does not have a square root at all, real or otherwise. This is because trying to solve the system of equations in this case does not yield a solution: we end up with a contradiction. I'd be interested to see if there is a way to know which matrices this is true for in general.

score 1 · Answer 2 · answered Mar 07 '17 at 17:28

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If $A^2 = B$ and you take $G = cI + dA$ where $c,d$ are almost-arbitrary numbers (only requirement is that $G$ has non-zero determinant) then the square of the matrix $G^{-1}AG$ is also $B$. So square-roots are far from unique.

answered Mar 07 '17 at 17:28

Mark

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Careful here. Since $G$ is a polynomial in the matrix $A$, we know that $G$ and $A$ commute. Therefore, $G^{-1}AG = A$ for any of the matrices $G$ that you proposed, and we can't prove nonuniqueness this way. – Jorge Arce Garro Aug 24 '18 at 21:57
Oops, I should probably delete my "answer". – Mark Aug 27 '18 at 01:17

score 1 · Answer 3 · answered Mar 07 '17 at 17:30

Let's try $$\left( \begin{array}{ccc} a & b \\ c & d \end{array} \right) \cdot \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right)$$

We find $b=c=0$ and $a= \pm 1$, $d =\pm 1$ as the only four possible diagonal solutions of this kind. Any generic solution must diagonalize in such a way, so any conjugate of these 4 diagonal matrices are solutions of our problem because if $AA=I$ then $(CAC^{-1})(CAC^{-1}) = I$. The solution is of course not unique!

score 0 · Answer 4 · answered Mar 07 '17 at 17:29

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$$\left[ \begin{array}{cc} \pm 1 & 0\\ 0 &\pm 1 \end{array} \right] $$ are four square roots of the identity.

$$\left[ \begin{array}{cc} 0 & t\\ 0 & 0 \end{array} \right] $$ yields an infinite set of square roots of the zero matrix.

answered Mar 07 '17 at 17:29

ancient mathematician

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Is the square root of a 2 by 2 matrix unique?

4 Answers4