If $\pmb{u}$ is the matrix that row reduces $\pmb{a}$. Then the bottom rows of $u$ spans the annihilator of the range of $\pmb{a}$

This question is taken from the textbook "advanced calculus by lynn loomis and shlomo sternberg". (Question 6.10 of chapter 2):

Let $\pmb{a}$ be an $m\times n$ matrix, and let $\pmb{u}$ be the nonsingular matrix that row reduces $\pmb{a}$, so that $\pmb{r}=\pmb{u}\cdot\pmb{a}$ is the row-reduced echelon matrix obtained from $\pmb{a}$. Suppose that $\pmb{r}$ has $m-k>0$ zero rows at the bottom(the kth row being nonzero). Show that the bottom $m-k$ rows of $\pmb{u}$ span the annihilator $(range A)^\circ$ of the range of $A$. That is, $\pmb{y}=\pmb{a}\pmb{x}$ for some $\pmb{x}$ if and only if $$\sum_1^m c_iy_i = 0$$ for each $m$-tuple $\pmb{c}$ in the bottom $m-k$ rows of $\pmb{u}$. [Hint: The bottom row of $\pmb{r}$ is obtained by applying the bottom row of $\pmb{u}$ to the columns of $\pmb{a}$]

I have managed to show one direction of the question but not the converse.

My attempt:
Suppose that $\pmb{y}=\pmb{a}\pmb{x}$ for some $\pmb{x}$
We know that $\pmb{r}=\pmb{u}\pmb{a}$ therefore $$(\pmb{r})_{ij}=(\pmb{u}\pmb{a})_{ij}=\sum_{t=1}^m u_{it}a_{tj}$$ For $m-k<i\leq m$ We have $\pmb{r}_{ij} = 0$ and $u_{it} = c_t$ thus $$\sum_{t=1}^m c_ta_{tj} = 0 \tag{1}$$ for any $\pmb{c}$ in the bottom $m-k$ rows of $\pmb{u}$
$\pmb{y}=\pmb{a}\pmb{x}$ therefore $$y_j=\sum_{l=1}^n a_{jl}x_l$$ It follows from equation (1) that $$\sum_{t=1}^m c_t y_t = \sum_{t=1}^m c_t \sum_{l=1}^n a_{tl}x_l = \sum_{l=1}^n\underbrace{\sum_{t=1}^mc_t a_{tl}}_0 x_l = 0$$

I can't show the other direction. I need to probably use the fact that $\pmb{r}$ is row reduced but I don't know how.
Can someone please help me?

Solution 1:

Near the end you say

$$\sum_{t=1}^m c_t y_t = \sum_{t=1}^m c_t \sum_{l=1}^n a_{tl}x_l = \sum_{t=1}^m\sum_{l=1}^n \underbrace{c_t a_{tl}}_0 x_l = 0$$

But this doesn't quite make sense since the expression you have underlined is not equal to 0. You should explain your reasoning more clearly.

For the other direction consider the process of solving the linear system $\mathbf{a}\cdot \mathbf{x} = \mathbf{y}$ through Gaussian elimination. We form the augmented matrix $[\mathbf{a}~\mathbf{y}]$ and then row reduce. How can you use $\mathbf{r}$ to represent the row reduced form of $[\mathbf{a}~\mathbf{y}]$? We have that

$$\sum_{t =1}^{m} c_ty_t = 0$$

for each of the bottom $m-k$ rows $\mathbf{c}$ of $\mathbf{r}$. How can you use this to show that the system will be consistent?