魅影直播

Consider a matrix $A=\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]$.

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

Substitute the values $\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]$for A and A for $\stackrel{鈫�}{x}$in the equation as follows.

role="math" localid="1660110351584" $$\begin{gathered} A^T\stackrel{鈫�}{x}=0 \\ {\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]}^T\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}\stackrel{鈫�}{0}\\ \stackrel{鈫�}{0}\end{array}\right] \end{gathered}$$

Simplify the equation ${\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]}^T\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}\stackrel{鈫�}{0}\\ \stackrel{鈫�}{0}\end{array}\right]$as follows.

$$\begin{gathered} {\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]}^T\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}\stackrel{鈫�}{0}\\ \stackrel{鈫�}{0}\end{array}\right] \\ \left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right]=\left[\begin{array}{c}\stackrel{鈫�}{0}\\ \stackrel{鈫�}{0}\end{array}\right] \\ \left|\begin{array}{c}2{\stackrel{鈫�}{x}}_1+3{\stackrel{鈫�}{x}}_2=0\\ 4{\stackrel{鈫�}{x}}_1+6{\stackrel{鈫�}{x}}_2=0\end{array}\right| \end{gathered}$$

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

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

Substitute the value $-\frac{3}{2}{\stackrel{鈫�}{x}}_2$for ${\stackrel{鈫�}{x}}_1$in the equation ${\stackrel{鈫�}{x}}_1=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right]$as follows.

$$\begin{gathered} {\stackrel{鈫�}{x}}_1=\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right] \\ {\stackrel{鈫�}{x}}_1=\left[\begin{array}{c}-\frac{3}{2}{\stackrel{鈫�}{x}}_2\\ {\stackrel{鈫�}{x}}_2\end{array}\right] \\ \stackrel{鈫�}{x}={\stackrel{鈫�}{x}}_2\left[\begin{array}{c}-\frac{3}{2}\\ 1\end{array}\right] \end{gathered}$$

As $ker\left(A^T\right)=im\left(A^T\right)$, therefore $im\left(A^T\right)=\left\{t\left[\begin{array}{c}-\frac{3}{2}\\ 1\end{array}\right]\overline{)t}鈭�\mathrm{鈩�}\right\}$is a line spanned by $\left[\begin{array}{c}-\frac{3}{2}\\ 1\end{array}\right]$.

As $im\left(A\right)=\left\{\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]\stackrel{鈫�}{x}\overline{)\stackrel{鈫�}{x}}鈭�\mathrm{鈩�}\right\}$, simplify $A\stackrel{鈫�}{x}$as follows.

$$\begin{gathered} A\stackrel{鈫�}{x}=\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]\left[\begin{array}{c}{\stackrel{鈫�}{x}}_1\\ {\stackrel{鈫�}{x}}_2\end{array}\right] \\ ={\stackrel{鈫�}{x}}_1\left[\begin{array}{c}2\\ 3\end{array}\right]+{\stackrel{鈫�}{x}}_2\left[\begin{array}{c}6\\ 4\end{array}\right] \\ ={\stackrel{鈫�}{x}}_1\left[\begin{array}{c}2\\ 3\end{array}\right]+2{\stackrel{鈫�}{x}}_2\left[\begin{array}{c}2\\ 3\end{array}\right] \\ A\stackrel{鈫�}{x}=\left({\stackrel{鈫�}{x}}_1+2{\stackrel{鈫�}{x}}_2\right)\left[\begin{array}{c}2\\ 3\end{array}\right] \end{gathered}$$

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

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 $\left[\begin{array}{c}-\frac{3}{2}\\ 1\end{array}\right]$and graph of the equation$ker\left(A^T\right)=im{\left(A\right)}^{鈯�}$is sketched.