Eigenvalues solve the equation $\det A-\lambda I = 0$, which means that the eigenvalues $\lambda$ are precisely the values where the matrix $A-\lambda I$ is singular, as any matrix with determinant zero is singular.
Whenever a matrix is singular, it has a null space of dimension greater than zero, which is another way of saying "row reduction induces a fully-zero row."
Similarly, eigenvectors solve $Ax = \lambda x$, or $(A-\lambda I)x = 0$. This is another way of saying that eigenvectors are precisely the vectors in the null space of $A-\lambda I$. If the null space of $A-\lambda I$ has dimension 1, then when solving $(A-\lambda I)x=0$, you will get a single zero row. If the null space has dimension 2, you'll get two such rows, and so on.
In fact, eigenvectors form a basis for the null space of $A-\lambda I$. So if you know what the eigenvalues are, then finding the eigenvectors is nothing more than computing the basis of the null space of $A-\lambda I$ for each eigenvalue $\lambda$.