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Square root of a matrix A is not a unique matrix, generally.

How to find all the square roots of a matrix, in particular the following matrix B:

$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$

other than forming 9 equations in 9 variables as:

$$ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$

In particular, is there any square root of B that has all elements as real numbers.

Truth-seek
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  • Well, if you have one square root, the opposite is also a square root; so square roots of matrices are almost never unique, strictly speaking. – Arnaud D. May 29 '17 at 10:00
  • For diagonal matrices, just take square roots of diagonal elements and construct a square root as another diagonal matrix. Of course because you can take either square root for each diagonal element, immediately you have $8$ square roots... – Macavity May 29 '17 at 10:00
  • @ArnaudD. What is the opposite of a Matrix? – Truth-seek May 29 '17 at 10:02
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    For example, any reflection at any plane or any rotation by 180° around any axis is a square root of the identity – Hagen von Eitzen May 29 '17 at 10:03
  • @Hagen von Eitzen reflection at any Plane and rotation by 180, what do you mean? Are you talking about the geographical interpretation of matrices as rotations in three dimensions? – Truth-seek May 29 '17 at 10:06
  • In real numbers, the square root is almost never unique $\sqrt 4 = \pm 2$, but we usually put a convention to consider only the positive one. In matrix analysis it is the same: every positive semidefinite matrix has a uniqe positive semidefinite square root, but in general is far to be the only square root possible – Exodd May 29 '17 at 10:07
  • @Truth-seek The opposite of $A$ is $-A$, the matrix with all elements multiplied by $-1$. – Arnaud D. May 29 '17 at 10:16
  • @Arnaud D. Ok got it. I call it the negative of the matrix. So, thought opposite was something else – Truth-seek May 29 '17 at 10:18
  • @Exodd What about $\begin{pmatrix}0 & -1 \ 1 & 0\end{pmatrix}^2=\begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix}$? – Arnaud D. May 29 '17 at 10:27
  • yup, totally forgot that the matrices of base change can be complex.. – Exodd May 29 '17 at 10:56

2 Answers2

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Square roots of matrices, where they exist, are far from unique. Given the 2×2 identity matrix, one square root is the identity matrix itself. Another is the negative of the identity matrix. Or just make either one element -1. Or the "interchange marrix" where the diagonal elements are zero and the nondiagonal elements 1.

And so on.

Oscar Lanzi
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In general there is no guarantee that a real matrix has a square root with real coefficients. In this case however you may find for e.g. $$M = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & \mp1 \\ 0 & \pm1 & 0 \end{bmatrix}$$

Now $M^2=(-M)^2=B$. This is easier to see if you consider the $B$ matrix as decomposed diagonally into identity matrix $I_1$ and $-I_2$. Now $-I_2$ can be considered a rotation by $\pi$, hence it has a real square root which is the rotation matrix by $\pi/2$.

Macavity
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