Exercise 3.4.15

We call a column whose corresponding column in one fixed reduced echelon form contains a pivot (see Exercise 3.4.7) a pivotal column. Now we induct on the number of columns of a matrix. For matrix contains only one column $u_1$, the reduced echelon form of it would be the column $e_1$ if $u_1$ is nonzero (and hence $\{u_1\}$ is independent) and that of it would be the zero column if $u_1$ is zero (and hence $\{u_1\}$ is dependent). Suppose that the reduced echelon form of a matrix with $k$ columns is unique. Now consider a matrix $A$ with $k+1$ columns, say $u_1,u_2,\dots ,u_{k+1}$. Let $A^{\prime }$ be the matrix by deleting the final column of $A$. So we can write $A=(A^{\prime }|u_{k+1})$. Say $(R^{\prime }|b)$ is a reduced echelon form of $A$. By the previous exercise we know that $R^{\prime }$ is a reduced echelon form of $A^{\prime }$. And $R^{\prime }$ is unique by our induction hypothesis. So the set of pivotal columns $P^{\prime }$ in $A^{\prime }$ is also unique. By Theorem 3.16(c) and Theorem 3.16(d) we know that the set $P^{\prime }$ in $A^{\prime }$ is a maximal independent set of $\{u_1,u_2,\dots ,u_k\}$. Now if $P^{\prime }\cup \{u_k\}$ is linearly independent, this means

say the value is $r$. By Theorem 3.16(a) and Theorem 3.16(b), $b$ is the only column who can be $e_r$ and so we know that $b=e_r$. On the other hand, if $P^{\prime }\cup \{u_k\}$ is linearly dependent. The vector $u_k$ cannot be a pivotal column of $A$ since $P^{\prime }\cup \{u_k\}$ is the set of a pivotal column of $A$ and it must be linearly independent. Furthermore, $u_k$ has an unique representation with respect to the set $P^{\prime }$. By Theorem 3.16(d), the column vector $d$ must be the representation of $u_k$. In all cases, we know that $(R^{\prime }|b)$ is also unique. And by induction, we get the desired conclusion.