This will be a proof by contradiction. Assume that the
\(i\)th row is a linear combination of the other rows of the matrix that is it can be written:
\begin{equation}
\alpha_i = c_1 \alpha_1 + c_2 \alpha_2 + \cdots + c_{i-1} \alpha_{i-1} + c_{i+1}
\alpha_{i+1} + \cdots + c_m \alpha_m\tag{1.3.2}
\end{equation}
The matrix
\(A\) can be written as
\begin{equation*}
A =
\begin{bmatrix}
0 \amp \cdots \amp 0 \amp \alpha_{1,\ell_1} \amp \alpha_{1, \ell_1+1} \amp \cdots \amp
\alpha_{1,\ell_i} \amp \cdots \amp \alpha_{1,\ell_m} \\
0 \amp \cdots \amp 0 \amp 0 \amp \alpha_{2,\ell_2} \amp \cdots \amp \alpha_{2,\ell_i}
\amp
\cdots \amp \alpha_{2, \ell_m} \\
\amp \amp \amp \amp \amp \amp \vdots \amp \amp \amp \vdots \\
\vdots \amp \amp \vdots \amp \amp 0 \amp \amp \alpha_{i,\ell_i} \amp \amp \vdots \\
0 \amp \cdots \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp \\
0 \amp \cdots \amp \amp \amp \amp \amp0 \amp0 \amp \alpha_{m,\ell_m}
\end{bmatrix}
\end{equation*}
To find the values of
\(c_j\) in (\ref{eq:row:linear:comb}), evaluate that equation for any column. To begin, consider
\(\ell_1\text{,}\) the column of the leading coefficient of the first row. Because the matrix is in echelon form,
\(\alpha_{2,\ell_1},
\alpha_{3,\ell_1},
\ldots, \alpha_{m,\ell_1}\) are all zero. Thus, examining the
\(\ell_1\)th element of
(1.3.2), one gets:
\begin{equation*}
0 = c_1 \alpha_{1,\ell_1} + c_2 \cdot 0 + \cdots + c_m \cdot 0
\end{equation*}
and because
\(\alpha_{1,\ell_1}\neq 0\) (it is a leading element),
\(c_1=0\text{.}\)
Inductively, assume that
\(c_1,c_2, \ldots, c_{k-1}=0\text{.}\) Let
\(\ell_k\) be the column with the leading element in row
\(k\text{.}\) This means that
\(\alpha_{k+1,\ell_k},
\alpha_{k+2,\ell_k}, \ldots, \alpha_{m,\ell_k}\) are all zero. If we examine the
\(\ell_k\)th component of
(1.3.2), then
Let
\(\ell_i\) be the column with the leading element in row
\(i\text{.}\) If the
\(\ell_i\)th element of (\ref{eq:row:linear:comb}) is extracted, one gets
\begin{align*}
\alpha_{i,\ell_k} \amp = (0) \alpha_{1,\ell_k} + (0) \alpha_{2,\ell_k} + \cdots +
(0) \alpha_{i-1,\ell_k} + (0) \alpha_{i+1,\ell_k} \\
\amp \qquad + \cdots + c_k\alpha_{k,\ell_k} + c_{k+1} \cdot 0 + \cdots + c_m \cdot
0
\end{align*}
If
\(i>k\text{,}\) then
\(\alpha_{i,\ell_k}=0\text{,}\) and this implied that
\(c_k=0\text{.}\) If
\(i
\lt k\text{,}\) then
