3 Eigenstates for $X$?

Question

Unless I have made an error, the Pauli $X$ operator has 3 eigenstates. Qiskit lists the two eigenstates for $X:|+\rangle= \frac 1{\sqrt2}(|1\rangle + |0\rangle)$ and $|-\rangle = \frac 1{\sqrt 2}|0\rangle -|1\rangle$. Attempting to prove this myself, I did the work behind finding eigenstates with $\lambda=\pm1$, found from finding the determinant: $$\det(X-\lambda I)$$ Yet, I found that there were two possible solutions for the eigenstate correlated with eigenvalue $-1$.
Assuming that we know our eigenvalues, we find eigenstates of $X$ through the form:
$$(X-\lambda I)\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0-\lambda&1\\1&0-\lambda \end{bmatrix}\begin{bmatrix} x_1\\x_2\end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix}$$.
When $\lambda=1$, our eigenstate is $\frac 1{\sqrt 2}(|0\rangle + |1\rangle)$, or $|+\rangle$. But when $\lambda=-1$, we are left with $x_1 + x_2 =0$, meaning that either $x_1$ or $x_2$ can be $-1$. This leaves our eigenvalue of $\lambda = -1$ having eigenstate of both $\frac 1{\sqrt 2}(|0\rangle - |1\rangle)$, or $\frac 1{\sqrt 2}(|1\rangle - |0\rangle)$. This is problematic considering that the $X$ basis of measurement is based on the eigenstates of $X: |+\rangle$ and $|-\rangle$. Please help me understand this dilemma. Thank you.

Answer 1

Remember that states differing by a global phase are physically equivalent. The two states mentioned differ by a global factor of $-1$.

By the same token, one can find other eigenstates with eigenvalue $+1$ such as $(-|0\rangle-|1\rangle)/\sqrt{2}$.