Assumes we have a real symmetric matrix $A_{n,n}$ and the corresponding eigen decomposition of it is
$ A = UDU^T$
where $U^TU = I_K$, $D \in R^{K\times K}$ is the diagonal matrix with non-zero eigenvalues.
The question is that if there are only k distinct rows in $A$, how can we prove that there are also only $K$ distinct rows in $U$?
Since $U^T U = I_k$, the $k\times n$ matrix $U^T$ has full rank, so its $k$ rows are linearly independent $n$-vectors. Since $D$ has nonzero diagonal it is invertible, so $DU^T$ also consists of $k$ independent rows.
By definition of matrix multiplication, each row of $A = UDU^T$ consists of a linear combination of the rows of $DU^T$, whose coefficients come from the corresponding row of $U$. Since the former’s rows are independent, a distinct combination will always yield a distinct row. The converse is of course also true: the same combination will yield the same row. Thus the number of distinct rows of $A$ exactly matches the number of distinct rows of $U$.