91影视

Bases of${\mathbf{鈩�}}^{\mathbf{n}}$

The vectors ${\stackrel{\mathbf{鈫�}}{\mathbf{蠀}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{鈫�}}{\mathbf{蠀}}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{鈫�}}{\mathbf{蠀}}}_{\mathbf{n}}$in${\mathbf{鈩�}}^{\mathbf{n}}$form a basis of if the matrix is

$\mathbf{\left[}\begin{array}{ccc}\overline{)}& & \overline{)}\\ {\stackrel{\mathbf{鈫�}}{\mathbf{蠀}}}_{\mathbf{1}}& \mathbf{鈥�}& {\stackrel{\mathbf{鈫�}}{\mathbf{蠀}}}_{\mathbf{n}}\\ \overline{)}& & \overline{)}\end{array}\mathbf{\right]}$is invertible.

The two independent vectors are given. To form a basis of ${\mathrm{鈩�}}^4$, two more vectors needed from ${\stackrel{鈫�}{e}}_1,{\stackrel{鈫�}{e}}_2,{\stackrel{鈫�}{e}}_3$and ${\stackrel{鈫�}{e}}_4$.

Let the vectors localid="1664286730397" ${\stackrel{鈫�}{e}}_1,{\stackrel{鈫�}{e}}_2,{\stackrel{鈫�}{e}}_3\&{\stackrel{鈫�}{e}}_4=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right]\&\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]$respectively.

So, the matrix can be written as:

$A=\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\\ 2& 4& 0& 0& 0& 0\\ 3& 6& 0& 1& 1& 0\\ 4& 8& 0& 0& 0& 1\end{array}\right]$

Using the transformation to reduce the matrix $A$:

$$\begin{gathered} R_2鈫�R_2-2R_1,R_3鈫�R_3-3R_1\text{\&}{R}_4鈫�R_4-4R_1 \\ 鈬�\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\\ 0& 2& -2& 1& 0& 0\\ 0& 3& -3& 0& 1& 0\\ 0& 4& -4& 0& 0& 1\end{array}\right] \\ {R}_3鈫�R_3-\frac{3}{2}{R}_2,R_4鈫�R_4-2R_2\text{\&}{R}_4鈫�R_4-\frac{4}{3}{R}_3 \\ 鈬�\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\\ 0& 2& -2& 1& 1& 0\\ 0& 0& 0& -\frac{3}{2}& -\frac{4}{3}& 1\\ 0& 0& 0& -2& 0& 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} R_2鈫�\frac{R_2}{2},R_3鈫�-\frac{2R_3}{3}\text{\&}{R}_4鈫�-\frac{3R_4}{4} \\ 鈬�\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\\ 0& 1& -1& \frac{1}{2}& 0& 0\\ 0& 0& 0& 0& -\frac{2}{3}& 0\\ 0& 0& 0& 0& 1& -\frac{3}{4}\end{array}\right] \\ {R}_3鈫�R_3+\frac{2}{3}{R}_4,R_2鈫�R_2-\frac{1}{2}{R}_1\text{\&}{R}_1鈫�R_1-R_2 \\ 鈬�\left[\begin{array}{cccccc}1& 0& 2& 0& 0& -\frac{1}{4}\\ 0& 1& -1& 0& 0& \frac{1}{4}\\ 0& 0& 0& 1& 0& -\frac{1}{2}\\ 0& 0& 0& 0& 1& -\frac{3}{4}\end{array}\right] \end{gathered}$$

$鈭�rrefA=\left[\begin{array}{cccccc}1& 0& 2& 0& 0& -\frac{1}{4}\\ 0& 1& -1& 0& 0& \frac{1}{4}\\ 0& 0& 0& 1& 0& -\frac{1}{2}\\ 0& 0& 0& 0& 1& -\frac{3}{4}\end{array}\right]$

From the matrix the redundant vectors are${\stackrel{鈫�}{e}}_1$and ${\stackrel{鈫�}{e}}_4$. The linearly independent vectors are${\stackrel{鈫�}{e}}_2$and ${\stackrel{鈫�}{e}}_3$.

Since, thelocalid="1664286903676" ${\stackrel{鈫�}{e}}_2$andlocalid="1664286847012">e鈫�3arelocalid="1664286811026" $\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right]\&\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right]$respectively. Therefore, the basis oflocalid="1664286834433">鈩�4is $\left\{\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],\left[\begin{array}{c}1\\ 4\\ 6\\ 8\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right]\right\}$.