91影视

Consider a matrix$A=\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}\right]$ .

To find$ker\left(A^T\right)$ determine the solution of linear system$A^T\stackrel{鈫�}{x}=\stackrel{鈫�}{0}$ .

Substitute the values$\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}\right]$ for A and A for$\stackrel{鈫�}{x}$ in the equation as follows.

$$\begin{gathered} A^T\stackrel{鈫�}{x}=\stackrel{鈫�}{0} \\ {\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}\right]}^T\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

Simplify the equation${\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}\right]}^T\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ as follows.

$$\begin{gathered} {\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}\right]}^T\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\end{array}\right]\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left|\begin{array}{c}{\stackrel{鈫�}{x}}_1+{\stackrel{鈫�}{x}}_2+{\stackrel{鈫�}{x}}_3=0\\ {\stackrel{鈫�}{x}}_1+2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3=0\end{array}\right| \end{gathered}$$

Subtract the equation${\stackrel{鈫�}{x}}_1+{\stackrel{鈫�}{x}}_2+{\stackrel{鈫�}{x}}_3=0$ from${\stackrel{鈫�}{x}}_1+2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3=0$ as follows.

$$\begin{gathered} \left\{{\stackrel{鈫�}{x}}_1+{\stackrel{鈫�}{x}}_2+{\stackrel{鈫�}{x}}_3\right\}-\left\{{\stackrel{鈫�}{x}}_1+2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3=0\right\} \\ {\stackrel{鈫�}{x}}_1+{\stackrel{鈫�}{x}}_2+{\stackrel{鈫�}{x}}_3-{\stackrel{鈫�}{x}}_1+2{\stackrel{鈫�}{x}}_2+3{\stackrel{鈫�}{x}}_3=0 \\ -{\stackrel{鈫�}{x}}_2-2{\stackrel{鈫�}{x}}_3=0 \\ {\stackrel{鈫�}{x}}_2=-2{\stackrel{鈫�}{x}}_3 \end{gathered}$$

Substitute the value$-2{\stackrel{鈫�}{x}}_3$ for${\stackrel{鈫�}{x}}_2$ in the equation${\stackrel{鈫�}{x}}_1+{\stackrel{鈫�}{x}}_2+{\stackrel{鈫�}{x}}_3=0$ as follows.

$$\begin{gathered} {\stackrel{鈫�}{x}}_1+{\stackrel{鈫�}{x}}_2+{\stackrel{鈫�}{x}}_3=0 \\ {\stackrel{鈫�}{x}}_1-2{\stackrel{鈫�}{x}}_3+{\stackrel{鈫�}{x}}_3=0 \\ {\stackrel{鈫�}{x}}_1-{\stackrel{鈫�}{x}}_3=0 \\ {\stackrel{鈫�}{x}}_1={\stackrel{鈫�}{x}}_3 \end{gathered}$$

Substitute the values$-2{\stackrel{鈫�}{x}}_3$ for${\stackrel{鈫�}{x}}_2$ and${\stackrel{鈫�}{x}}_3$ for${\stackrel{鈫�}{x}}_1$ in the equation$\stackrel{鈫�}{x}=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\end{array}\right]$ as follows.

$$\begin{gathered} \stackrel{鈫�}{x}=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_3\end{array}\right] \\ \stackrel{鈫�}{x}=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_3\\ -2{\stackrel{鈫�}{x}}_3\\ {\stackrel{鈫�}{x}}_3\end{array}\right] \\ \stackrel{鈫�}{x}={\stackrel{鈫�}{x}}_3\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right] \end{gathered}$$

As$ker\left(A^T\right)=im{\left(A\right)}^{鈯�}$ , therefore $im{\left(A\right)}^{鈯�}=\left\{t\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]\overline{)t}鈭�\mathrm{鈩�}\right\}$is a line spanned by$\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$ .

As $im\left(A\right)=\left\{\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}\right]\stackrel{鈫�}{x}\overline{)\stackrel{鈫�}{x}鈭�\mathrm{鈩�}}\right\}$, simplify $A\stackrel{鈫�}{x}$as follows.

$$\begin{gathered} A\stackrel{鈫�}{x}=\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 3\end{array}\right]\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right] \\ A\stackrel{鈫�}{x}={\stackrel{鈫�}{x}}_1\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+{\stackrel{鈫�}{x}}_2\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \end{gathered}$$

Therefore, the value $im\left(A\right)$is spanned by the vector${\stackrel{鈫�}{v}}_1=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$and${\stackrel{鈫�}{v}}_2=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$.

Draw the graph of the equation$ker\left(A^T\right)=im{\left(A\right)}^{鈯�}$as follows.

Hence, the $ker\left(A^T\right)$is a line spanned by $\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$and graph of the equation$ker\left(A^T\right)=im{\left(A\right)}^{鈯�}$is sketched.