Let $A \in \mathbb R^{10\times10}$, $x_{1},x_{2},...,x_{7}\in \mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=\cdots=Ax_{7}$. Prove that $ \operatorname{rank} (A) \le 4$.
I know that it is truth because if I have seven linearly independent vectors and I multiply a matrix with these vectors I must have at least $6$ zero rows because I have $Ax_{1}=Ax_{2}=\cdots=Ax_{7}$. For example if I have $5$ zero rows then this vectors are not linearly independent.
However I don't know how to prove it in an elegant way.
You can find 6 linearly independent vectors in the kernel of $A$ by considering $y_1=x_2-x_1$, $y_2=x_3-x_1,\ldots,y_6=x_7-x_1$. Indeed, you can check that $Ay_i=0$ for all $1\geq i \geq 6$.
So $\dim \ker(A) \geq 6$. By the Rank-Nullity Theorem, $$\operatorname{rank}(T) = 10 - \dim \ker(T) \leq 10-6=4$$