What's the rank of $\sum_{i=1}^{m}x_{i}y_{i}^{T} \leq m$?

Question

I have two sets of vectors: $\{x_{1}, x_{2}, ... , x_{n}\}$ and $\{y_{1}, y_{2}, ... , y_{n}\}$, I have to show that : rank $[\sum_{i=1}^{m}x_{i}y_{i}^{T}]\leq m$. They say to first show that rank of the $x_{i}y_{i}^{T}$ square matrix is equal to 1.

I did it saying that :

First we see that the $x_{i}y_{i}^{t}$ matrix has $rank = 1 $. Indeed, \begin{equation} \begin{aligned} A = x_{i}y_{i}^{t} = \begin{pmatrix} x_{1}y_{1} && x_{1}y_{2} && \cdots && x_{1}y_{n}\\ x_{2}y_{1} && x_{2}y_{2} && \cdots && x_{2}y_{n}\\ \vdots && \vdots && \ddots && \vdots\\ x_{n}y_{1} && x_{n}y_{2} && \cdots && x_{n}y_{n}\\ \end{pmatrix} \end{aligned} \end{equation} can easily be reduced as : \begin{equation} \begin{aligned} \begin{pmatrix} x_{1}y_{1} && x_{1}y_{2} && \cdots && x_{1}y_{n}\\ 0 && 0 && \cdots && 0\\ \vdots && \vdots && \ddots && \vdots\\ 0 && 0 && \cdots && 0\\ \end{pmatrix} \end{aligned} \end{equation} because each row is a linear combination of the first one. Thus $rank = 1$.

But I don't understand what the rank of a sum means... And I tried with a simple example and It didn't worked. [EDIT: this is wrong] I see that this sum is like the trace of the matrix, but I don't see the purpose to calculate the rank of a trace and even if there is a purpose I don't know how should I do it. Thanks for your help ;)

Answer 1

Hint: First we have the fact that the rank of the matrix $x_{i}y_{i}^{T}$ is 1.

Then we use the rank inequality $r(A+B) \leq r(A)+r(B)$

Answer 2

The rank of a matrix $A$ equals the dimension of the space generated by the images $Az$. If $A$ has the form $A = \sum_{i = 1}^{m}x_{i}y_{i}^{t}$, then $Az = \sum_{i = 1}^{m}(y_{i}^{t}z)x_{i}$ is a linear combination of the $m$ vectors $x_{1}, \dots, x_{m}$. Hence the space generated by the images $Az$ is contained in the span of the vectors $x_{1}, \dots, x_{m}$, so has dimension at most $m$.