Question: Also, as a post question, if a matrix where each row and column individually sum to zero, is it guaranteed to have one of the eigen value as $0$ and one of the eigen vector as $(1,1,1,...1,1)$? What does this signify?

Answer: 
The result from multiplying a matrix with the all-ones vector is the sum of the matrix columns. Hence if this sum is the zero vector (i.e., all rows individually sum to zero) the all-ones vector is eigenvector with eigenvalue $0$.
Let's create a square matrix to test out your idea:

$$
 \begin{bmatrix}
 1 & -1 \\
 -1 & 1 \\
 \end{bmatrix}
$$

Now, if we find the eigenvalues:

$$
 \begin{vmatrix}
 1 - \lambda & -1 \\
 -1 & 1 - \lambda \\
 \end{vmatrix}
 = (1-\lambda)^2 -1 =0$$

Therefore, we get our eigenvalue to be $\lambda = 0$ and a corresponding eigenvector of $1$s.

Why does this work?

Well, mathematically, let's put any real number $k$:

$$
 \begin{bmatrix}
 k & -k \\
 -k & k \\
 \end{bmatrix}
$$

Now, let's get our eigenvalues:

$$
 \begin{vmatrix}
 k - \lambda & -k \\
 -k & k - \lambda \\
 \end{vmatrix}
 = (k-\lambda)^2 -k =0$$

As we can we, for any zero-sum matrix, the only eigenvalue that works, is when $\lambda = 0$