# Echelon-form Matrix Lemma Proof

I couldn't follow the proof for the following lemma in the textbook, so I decided to try to prove it for myself. It is a little different from the proof in the textbook. I'd like to receive feedback on the following

I) Is the proof correct?

II) Are there any 'holes' or logical leaps or mistakes in it?

II) Is it okay for a first/second-year undergraduate?

(I've done an intro to proofs course but I'm still inexperienced in proofwriting.)

Lemma. In an echelon-form matrix, no nonzero row is a linear combination of other nonzero rows.

Proof. Let $$\mathbf{R}$$ be an echelon-form matrix with $$m$$ nonzero rows and $$n$$ columns. Suppose $$\mathbf{R}$$ has a nonzero row that is the linear combination of other rows. In other words, $$\vec{\rho_i} = c_1 \vec{\rho}_1 + \cdots + c_{i-1} \vec{\rho} _{i-1}+ c_{i+1} \vec{\rho}_{i+1} + \cdots + c_m \vec{\rho}_m$$

where $$1 \leq i \leq m$$, $$\vec{\rho}_k$$ is the $$k$$-th row of matrix $$\mathbf{R}$$, and $$c_1, \dots, c_m \in \mathbb{R}$$.

Now we use induction on the nonzero rows of the matrix to show that $$c_k = 0$$ for all $$1 \leq k \leq i-1$$.

Base case: Let $$k=1$$. Let the $$l$$-th column be the column with the leading variable of the $$k$$-th row. Since $$p_i$$ is a linear combination of other nonzero rows, we have $$\rho_{i, l} = c_1 \rho_{1, l} + \cdots + c_{i-1, l} \rho_{i-1, l} + c_{i+1, l} \rho_{i+1, l} + \cdots + c_m \rho_{m, l}$$

Since $$\mathbf{R}$$ is in echelon form, we have $$0 = c_1 \rho_{1, l} + c_2 \cdot 0 + + \cdots + c_{i-1, l} \cdot 0 + c_{i+1, l} \cdot 0 + \cdots + + c_m \cdot 0$$ and $$\rho_{1, l}\neq 0$$. It follows that $$c_1 = c_k = 0$$.

Induction step: Let $$1 \leq k \leq i-1$$. Again, let the $$l$$-th column be the column with the leading variable of the $$k$$-th row. Suppose $$c_j = 0$$ for all $$1 \leq j < k$$. Then we have $$\rho_{i, l} = 0 \cdot \rho_{1, l} + 0 \cdot \rho_{2, l} + \cdots + c_k \rho_{k, l} + \cdots + c_{i-1, l} \rho_{i-1, l} + c_{i+1, l} \rho_{i+1, l} + \cdots + c_m \rho_{m, l}$$

Since $$\mathbf{R}$$ is in echelon form, we have $$0 = c_k \rho_{k, l} + c_{k+1} \cdot 0 + \cdots + c_{i-1, l} \cdot 0 + c_{i+1, l} \cdot 0 + \cdots + + c_m \cdot 0$$ and $$\rho_{k, l}\neq 0$$. It follows that $$c_k = 0$$.

Thus we have shown that $$c_k = 0$$ for all $$1 \leq k \leq i-1$$. Now let the $$l$$-th column be the column with the leading variable of the $$i$$-th row. Then we have $$\rho_{i, l} = 0 \cdot \rho_{1, l} + 0 \cdot \rho_{2, l} + \cdots + 0 \cdot \rho_{i-1, l} + c_{i+1} \rho_{i+1, l} + \cdots + c_m \rho_{m, l}$$ But since $$\mathbf{R}$$ is in echelon form, we have $$\rho_{i, l} = c_{i+1} \cdot 0 + \cdots + c_m \cdot 0$$ and $$\rho_{i, l}\neq 0$$. Therefore, we have a contradiction. $$\square$$