Why not imaginary eigenvalues in the eigenvalue equation $A\psi=\lambda\psi$? in quantum mechanics we always see the eigenvalue equation $\hat A\psi=\lambda\psi$ and $\lambda$ is the probability amplitude meaning $\lambda^2$ is the actual probability of finding the system in the state $\psi$. We are also told that $\lambda$ must be real, but what would be wrong with having $\lambda$ imaginary because it's square is real as a physical quantity must be?
So why can we never have a complex $\lambda$?
 A: $\lambda$ is not the probability amplitude. And your equation $\hat A\psi=\lambda\psi$ means that $\psi$ is an eigenstate of $A$ with eigenvalue $\lambda$ where $\lambda$ is a physical quantity associated with $\hat A$ upon measurement of $\hat A$. Hence $\lambda$ itself must be real.
For example, if $\hat A$ was the momentum operator, then $\lambda$ is a value of momentum. And in the ubiquitous equation, $$\hat H\Psi=E\Psi$$ where $\hat H$ is the Hamiltonian, $E$ is energy.
You have a few misconceptions. Say we have a system described by a state
$$\Psi=\sqrt{\frac{1}{3}}\psi_1  +\sqrt{\frac{2}{3}}\psi_2$$ Then the coefficients $\sqrt{\frac{1}{3}}$ and $\sqrt{\frac{2}{3}}$ are probability amplitudes and the corresponding
square of these coefficients i.e., $\frac 13$ and $\frac 23$ are probabilities (corresponding to the
likelihood of finding the system in states $\psi_1$ and $\psi_2$ respectively).
When you look at the equation $\hat A\psi=\lambda\psi$ (with Hermitian or self-adjoint $\hat A$) by definition we need a coefficient $\lambda$ that will change the length of the
state $\psi$ and not a complex $\lambda$ since
multiplication by imaginary numbers will rotate a state.
Also, if the eigenvalues were imaginary, this would mean the operator $\hat A$ will not be Hermitian. This is because $\hat A$ will have imaginary elements on the diagonal. This means it cannot possibly be
Hermitian since the rule $$A^\dagger=(A^T)^*=A$$ cannot possibly hold.
Remember that the rule is that $\hat A$ must be Hermitian since it is also a requirement that $\hat A$ has a complete set of eigenvalues.
