This is in reference to Theorem 4.8 in the pictures below. I am facing difficulty in understanding the proof of this theorem.
I am able to understand the proof upto the second last statement. What I am not able to understand is the logic behind the last sentence. I am facing problem to bridge the gap. I am fine with the statement that $\textbf{u}$ has to be independent of the row vectors because it is non-null and orthogonal to the row vectors. But how does this imply that the number of independent row vectors in $\textbf{A}$ is less than $n$?
A non-null solution $u$ is orthogonal to all rows. The rows are in $\mathbb R^n$. If the row-rank was $n$, the rows would span $\mathbb R^n$ and there couldn't be any non-null vector orthogonal to them (because it would be orthogonal to $\mathbb R^n$). So, for the existence of a non-null solution it is necessary that the row-rank is $<n$. The claimed sufficiency in the theorem is not proved, though. But this is also clear: if the row rank is $<n$, the rows do not span $\mathbb R^n$ and you find a vector that is orthogonal to all rows, which will then be a non-null solution.