Chapter 3.4, Problem 27PS

To determine

The six linearly independent $3$by $3$ echelon matrices $U_1,\mathrm{...},U_6$.

   $I,\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]$.

Given:

Linearly independent $3$by $3$ echelon matrices $U_1,\mathrm{...},U_6$.

Calculation:

It is noted that echelon matrices do not form a subspace, and therefore, they span the upper triangular matrices.

The sequence of vectors $v_1,\mathrm{...},v_n$is linearly independent if the only combination that gives the zero vector is $x_1{v}_1+x_2{v}_2+\mathrm{...}+x_n{v}_n$

Therefore, linear independence can be summarized as:

   $x_1{v}_1+x_2{v}_2+\mathrm{...}{x}_n{v}_n=0$

Will only happen when all x's are zero. So, the combination will be:

   $I,\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]$.