Decomposing the "sign flip" matrix in terms of Pauli matrices

Question

Can the matrix:

$A=\left[\begin{matrix}0 & -1 \\ -1 & 0\end{matrix}\right]$

be somehow expressed as a product of the $3$ standard Pauli matrices?

I'm being able to diagonalize $A$, but not sure if it can be expressed in terms of Pauli-X, Y, Z.

Answer 1

I start by knowing that the flip operator is $F=-\sigma_1$.

Then to express as a product of all the sigma matrices we just find a combination which reduces to the negative identity.

The relation

$$\sigma_1^2 = \sigma_2^2=\sigma_3^2=-i\sigma_1\sigma_2\sigma_3 = 1$$

lets us construct $(\sigma_1\sigma_2\sigma_3)^2 = -1$ to get

$$F=-\sigma_1(\sigma_1\sigma_2\sigma_3)^2$$ $$\require{cancel} F=-\cancel{\sigma_1\sigma_1}\sigma_2\sigma_3\sigma_1\sigma_2\sigma_3$$

We can also recover the solution provided by @euler-is-alive because $\sigma_1\sigma_2\sigma_3=i$ is just a number and so we can freely rearrange as $\sigma_1\sigma_2\sigma_3\sigma_2\sigma_3$

Answer 2

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