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If $A$ is a positive semi-definite matrix and $B$ is a matrix such that $ A = B B $ then $B$ is called the square root of the matrix $A$. The approach to determine this is to write $A$ in the form $A = P^{-1} D P$ with $D$ a diagonal matrix and then you take the square root of the diagonal elements (write it as $D^{1/2}$) and identify $B$ as $P^{-1} D^{1/2} P$.

But what about the following example? Take $A$ as something really simple $$ A = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right). $$ This method would give $B$ in a straight forward manner. But, does the method give all solutions? For example what about the matrix $B$ given as $$ B = \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right). $$ This satisfies $$ B B = \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) $$ and so qualifies as the square root of $A$.

As a newbie to the field, is it known that there may be other solutions not given by looking at the method alluded to in the link?

jim
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    Note that $\begin{pmatrix} 0 & b \ 1/b & 0\end{pmatrix}$ also squares to identity for all $b\not=0$. (And more generally $\begin{pmatrix} a & b \ c & -a \end{pmatrix}$ for all reals $a,b,c$ with $a^2+bc=1$.) – Giulio R Apr 16 '21 at 15:03

2 Answers2

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Yes, there are other solutions. The set of all square roots of $I_2$ is given by $\{I_2\}\cup\{P\operatorname{diag}(1,-1)P^{-1}:P \text{ is invertible}\}$. This is an infinite set. E.g. every $B$ in the form of $\pmatrix{1&t\\ 0&-1}$ or $\pmatrix{\cos t&\sin t\\ \sin t&-\cos t}$ are square roots of $I_2$.

However, each positive semidefinite matrix $A$ possesses exactly one positive semidefinite square root $B$. In your case where $A=I_2$, its only positive semidefinite square root is itself.

user1551
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To complement user1551's answer, let me observe the special role of the identity matrix, which you want to extract the square root of.

The map $Mat_{2\times 2}(\mathbb C)\to Mat_{2\times 2}(\mathbb C):A\mapsto A^2$ can be "spelled out" as a polynomial map $\mathbb C^4\to\mathbb C^4$ $$ (a,b,c,d)\mapsto(a^2+bc,b(a+d),c(a+d),bc+d^2) $$ and remarkably the Jacobian (=det of Jacobian matrix) is $$ 4(a+d)^2(ad-bc)=4\,({\rm tr}\,A)^2\,\det A. $$ Thus any matrix $B$ which is the square of a traceless matrix (i.e. any scalar matrix $\lambda I$) is a singular value of the map. This includes the identity, and you indeed cannot expect to invert, not even locally in a neighborhood of $I$, this map. Other singular values are degenerate matrices $B$ (i.e. $\det B=0$).

For the other matrices you instead expect a discrete set of preimages. More details can be obtained in compact matrix spaces (e.g. orthogonal matrices...) where the number of preimages (=square roots) is constant (except of course for singular values, like the identity).

Giulio R
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