TITLE: Rank of sum of rank-$1$ matrices
QUESTION [15 upvotes]: If you sum a certain number of rank-$1$ matrices:
$$X = u_1 u_1^T + u_2 u_2^T + \cdots + u_N u_N^T$$
Is the result guaranteed to be rank-$N$ assuming the individual $U$ vectors are linearly independent?

REPLY [3 votes]: The answer is yes as others have said. My 2 cents is a short argument which may be instructive.
Say $X$ is $n\times n$, then since $u_1,...,u_N$ are independent we have $N \le n$.
Let $U=[u_1,...,u_N]$ be the matrix who's columns are $u_1,...,u_N$, then
                                   $$X=UU^T.$$
By the svd decomposition we have $rank(X)=rank(UU^T)=rank(U)=N$.