Finding four square roots of $\small\begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ This question is about finding the 4(!!) square roots of a 2 by 2 matrix. The method I use is through Diagonalization (or actually doing the "opposite"). I have no problem finding the four roots of a 2 by 2 matrix that has real eigenvalues (and thus real eigen vectors). Heck, I found 8 roots of a 3 by 3 matrix that has real eigen values. The problem I am running into is if the to be square rooted matrix has complex eigenvalues. In my view, the method should work the same, but somehow the algebra is letting me down. Example: Find the four square roots of
$ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$. Note, I found this matrix by squaring $ \begin{bmatrix}2 & -1\\1 & 2\end{bmatrix}$, so obviously this is one root as well as taking opposite of its entries. So here is my work: The characteristic equation of $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ is $(3-\lambda)^2+16=0$ from which we find $\lambda=3+4i$ and $\lambda=3-4i$. An eigenvector for $\lambda=3+4i$ is $ \begin{bmatrix}1 \\-i  \end{bmatrix}$ and an eigenvector for $\lambda=3-4i$ is $ \begin{bmatrix}1 \\i  \end{bmatrix}$. So the given matrix can now be diagonalized: $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$=$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}$$ \begin{bmatrix}3+4i & 0\\0 & 3-4i\end{bmatrix}$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$. I found$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix}^{-1}$=$\frac{1}{2i} \begin{bmatrix}i & -1\\i & 1\end{bmatrix}$ which is $ \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$. It's easy to find out that the square roots of $3+4i$ are $2+i,-2-i$, and of $3-4i$ are $2-i,-2+i$. So now I would arrive at my 4 solutions:$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}2+i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & 2-i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
$$ \begin{bmatrix}1 & 1\\-i & i\end{bmatrix} \begin{bmatrix}-2-i & 0\\0 & -2+i\end{bmatrix} \begin{bmatrix}0.5 & 0.5i\\0.5 & -0.5i\end{bmatrix}$$
When I work out my first candidate, I get a matrix, but when squared it doesn't give me $ \begin{bmatrix}3 & -4\\4 & 3\end{bmatrix}$ and so something isn't right. I don't know what I did wrong and ANY help from the math community is greatly appreciated. I feel that my line of thinking is correct, where did I go wrong? And does anyone happen to know online software that can do matrix multiplications with complex entries, because the TI doesn't work
 A: Let $A = [\begin{smallmatrix}3 & -4\\4 & 3\end{smallmatrix}]$ and $\chi_A(\lambda)$ be its characteristic polynomial. Since
$$\chi_{A}(\lambda) = \det(\lambda I_2 - A) = \lambda^2 - 6\lambda + 25$$
By Cayley Hamiltonian, we have
$$A^2 - 6A + 25 I_2 = 0
\iff (A+5I_2)^2 = 16 A \iff (A - 5I_2)^2 = -4A
$$
Taking "square root" on both sides and rearrange, one find $A$ has following four square roots.
$$\pm \frac14(A + 5I_2) = \pm \begin{bmatrix}2 & - 1\\1 & 2\end{bmatrix},\quad \pm \frac{i}{2}(A - 5 I_2) = \pm i\begin{bmatrix}-1 & -2\\2 & -1\end{bmatrix}$$
A: Your work seems correct.  Doing the matrix multiplications for your four candidates
in Wolfram Alpha gives these square roots:  $\pm\begin{bmatrix}2&-1\\1&2\end{bmatrix}$ and $\pm\begin{bmatrix}i&2i\\-2i&i\end{bmatrix}$.
A: There should only be two square root matrices.  $\pm\begin{bmatrix} 2&-1\\1&2\end{bmatrix}$
I have been noodling with this a bit.
$\begin{bmatrix}0 & -i\\i & 0\end{bmatrix} \begin{bmatrix} 2&-1\\1&2\end{bmatrix} = \begin{bmatrix} i&2i\\-2i&i\end{bmatrix}$
Appears to work, and so would
$-\begin{bmatrix} i&2i\\-2i&i\end{bmatrix}$
