Obviously all these three conditions are not necessary, as both iteration methods can be used to solve the matrix equation $A0=0$ regardless of $A$ (as long as the iteration matrices exist).
Apparently, you copied the first two conditions from the Wikipedia entries on Gauss-Seidal method and Jacobi method. If you read the two articles careful enough, you should know that both conditions 1 and 2 are sufficient conditions for convergence. As to condition 3, the answer depends on the norm. If the matrix norm in question is a consistent norm (which is true for virtually all matrix norms we encounter in practice) and the iteration matrix is $E$, then by the spectral radius formula, $\rho(E)\le\|E^k\|^{1/k}$ for any $k\in\mathbb{N}$. In particular, $\rho(E)\le\|E\|$. Since $\|E\|<1$, we conclude that the spectral radius $\rho(E)$ is also smaller than $1$ and hence both iteration methods converge.
Yes. From pg. 233 of A Friendly Introduction to Numerical Analysis by Brian Bradie, we have the following:
Theorem. Suppose $A$ is a an $n \times n$ matrix. If $ a_{ii} > 0$ for each $i$ and $a_{ij} \leq 0$ whenever $ i \neq j$, then one and only one of the following statements holds:
$0 \leq \rho (T_{gs}) < \rho (T_{jac}) < 1$
$1 < \rho (T_{jac}) < \rho (T_{gs})$
$\rho (T_{jac}) = \rho (T_{gs}) = 0$
$\rho (T_{jac}) =\rho (T_{gs}) = 1$
where $T$ is the iteration matrix that arises for each method, and $\rho (T)$ denotes the spectral radius of $T$.
Thus, for this larger class of matrices, the methods converge and diverge together. When they converge, Gauss-Seidel converges faster; when they diverge, Gauss-Seidel again does so faster.
If you want the proof of this, Bradie cites the following sources:
A. Ralston and P. Rabinowitz, A First Course in Numerical Analysis, 2nd edition, McGraw-Hill, New York, 1978.
D. M. Young, Iterative Solution of Large Linear Systems, Academic Press, New York, 1971.
Best Answer
Gauss-Seidel comes from a splitting of the matrix $A=M-N$. To solve $Ax=b$, the general form for such methods is
\begin{equation} x^{(k+1)} = M^{-1}Nx^{(k)} +M^{-1}b \end{equation}
and of course
\begin{equation} x = M^{-1}Nx +M^{-1}b \end{equation}
Combining theses
\begin{align} x - x^{(k+1)} &= M^{-1}N (x - x^{(k)}) \end{align}
So we have convergence for every initial guess $x^{(0)}$ iff the spectral radius of $M^{-1}N$ is less than 1.
If you are lucky and guess an eigenvector with eigenvalue less than 1 as a starting vector, then you can have convergence without the previous criterion.
Note : for Gauss-Seidel, the splitting is defined by $M$ being the lower triangular part of $A$ including the diagonal.