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Consider the {$2x2$} identity matrix. What is its square root?

{$$ \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}$$}

This yields equations of the form:

{$$ {a_{11}}^2 + a_{12}a_{21} = 1 $$}

{$$ a_{12}a_{22} + a_{11}a_{12} = 0 $$}

The latter equation means {$ a_{12}=0$} or {$ a_{22} = -a_{11} $}.

Similarly, by symmetry, {$ a_{21}=0$} or {$ a_{22} = -a_{11} $}.

Combining, we have ({$ a_{12}=0$} and {$ a_{21}=0$}) or {$ a_{22} = -a_{11} $}.

Solving further, this yields the following two possibilities:

{$ \begin{pmatrix}\pm1 & 0\\ 0 & \pm1\end{pmatrix} $} or {$ \begin{pmatrix}\sqrt{1-ab} & a\\ b & -\sqrt{1-ab}\end{pmatrix} $} or {$ \begin{pmatrix}-\sqrt{1-ab} & a\\ b & \sqrt{1-ab}\end{pmatrix} $}

But the first case and the second case match when a or b = 0. Thus the answer is:

{$ \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} $} or {$ \begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix} $} or {$ \begin{pmatrix}\sqrt{1-ab} & a\\ b & -\sqrt{1-ab}\end{pmatrix} $} or {$ \begin{pmatrix}-\sqrt{1-ab} & a\\ b & \sqrt{1-ab}\end{pmatrix} $}

Note that {$a$} and {$b$} can be any complex number. However, if we want a real matrix, then we must have {$a$} and {$b$} real such that {$ab \leq 1$}.