Rank of sum of rank-$1$ matrices

Question

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?

Answer 1

If the vectors $\{u_k\}_{k=1}^N$ are linearly independent and $N\lt n$, then using the Gram-Schmidt Process, we can construct a basis $\{u_k\}_{k=1}^n$ for $\mathbb{R}^n$ containing the original vectors.

Consider $$ M_1=\sum_{k=1}^Nu_ku_k^T $$ and $$ M_2=\sum_{k=N+1}^nu_ku_k^T $$ Note that if $N=n$, then $M_2=0$.

Suppose that $(M_1+M_2)v=0$, that means that $$ \sum_{k=1}^nu_k\left(u_k^Tv\right)=0 $$ Since $u_k$ are independent, this means that $u_k^Tv=0$ for all $k$. Since $\{u_k\}_{k=1}^n$ is a basis, $v=0$. Thus, $$ (M_1+M_2)v=0\implies v=0 $$ Therefore, $M_1+M_2$ has rank $n$. Since the rank of $M_1+M_2$ is no greater than the sum of the ranks of $M_1$ and $M_2$, $M_1$ must have rank $N$, because the rank of $M_2$ is at most $n-N$.

So, yes: if $\{u_k\}_{k=1}^N$ are linearly independent, $\displaystyle\sum_{k=1}^Nu_ku_k^T$ has rank $N$.

Answer 2

Slightly more general: If $$ X = u_1 v_1^T + u_2 v_2^T + \cdots + u_N v_N^T $$ where $(u_1, u_2, \ldots, u_N)$ are linearly independent, and $(v_1, v_2, \ldots, v_N)$ are also linearly independent, then the rank of $X$ is equal to $N$.

Proof: We need two properties of the matrix rank:

• Subadditivity: $\operatorname{rank}(A+B) \le \operatorname{rank}(A) +\operatorname{rank}(B)$ .
• Sylvester’s rank inequality: If $A$ is an $m × n$ matrix and $B$ is $n × k$, then $\operatorname{rank} (A)+\operatorname{rank} (B)-n\leq \operatorname{rank} (AB)$.

$\operatorname{rank}(X) \le N$ follows from the subadditivity, since each $u_j v_j^T$ is a rank-one matrix.

And applying Sylvester’s rank inequality to $$ X = \begin{pmatrix} u_1 & \ldots & u_N\end{pmatrix} \begin{pmatrix} v_1^T \\ \vdots \\ v_N^T \end{pmatrix} =: U \cdot V $$ gives $$ N = \operatorname{rank}(U) + \operatorname{rank}(V) - N \le \operatorname{rank}(X) \, . $$

Answer 3

Assuming that $N$ (the number of vectors) is the same as $n$ (the dimension of the space) then yes. Otherwise it may need to be as involved as @robjohn's answer.

Consider nonzero $v_1\in\operatorname{span}\{u_2,\ldots,u_N\}^{\perp}$. Then $Xv_1=u_1(u_1^Tv_1)$ is a nonzero scalar multiple of $u_1$. (If it were zero, then we have that $v_1$ is also orthogonal to $u_1$, and so all of the $u_i$ are in $v_1^{\perp}$ which is $(n-1)$-dimensional, a contradiction to the linear independence assumption.)

Similarly, nonzero scalar multiples of each $u_i$ are in the range of $X$. So it is a rank $N$ matrix (since the $u_i$ are independent.)

Answer 4

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 $rank(X)=rank(UU^T)=rank(U)=N$.

Answer 5

Yes. If you express this linear transformation in the basis $u_1,\ldots,u_n$ itself, then the rows represent in the new basis the linear forms that were given in the old basis by $u_1^T,\ldots,u_n^T$. Since the new basis is not orthonormal it is not so easy to express what the new rows will be, but they certainly remain independent.

Answer 6

$X=UV^T$

The columns of $U$ are linearly independent so it has a left inverse. Thus
$\text{rank}\big(X\big)=\text{rank}\big(UV^T\big)=\text{rank}\big(V^T\big)$.

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extra details:
let the left inverse of $U$ be given by $Z$. Then
$\text{rank}\big(V^T\big)=\text{rank}\big(ZUV^T\big)\leq \text{rank}\big(UV^T\big)\leq \text{rank}\big(V^T\big)$
$\implies \text{rank}\big(UV^T\big)=\text{rank}\big(V^T\big)$