Is the square root of a 2 by 2 matrix unique? How would I know if the square root of a 2 by 2 matrix is unique?
 A: 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.
A: 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!
A: 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.
A: $$\left[
\begin{array}{cc}
\pm 1 & 0\\
0 &\pm 1
\end{array}
\right]
$$
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.
