Question

Consider the subspace$\mathit{i}\mathit{m}\left(A\right)$ of ${\mathbf{ℝ}}^{\mathbf{2}}$ . Where $\mathit{A}\mathbf{=}\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]$ . Find a basis of $\mathit{k}\mathit{e}\mathit{r}\left(A^T\right)$ , and draw a sketch illustrating the formula$\left(im\left(A^T\right)\right)\mathbf{=}\mathit{k}\mathit{e}\mathit{r}\left(A^T\right)$in this case.

Short answer

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 $ker\left(A^T\right)=im{\left(A\right)}^{\perp }$is

Step-by-step solution

Determine the value of ker(AT) .

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 $A^T\overrightarrow{x}=\overrightarrow{0}$.

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

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

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

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

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

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

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

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

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

Draw the graph of ker(AT)=im(A)⊥ .

As $im\left(A\right)=\left\{\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right]\overrightarrow{x}\overline{)\overrightarrow{x}}\in \mathrm{ℝ}\right\}$, simplify $A\overrightarrow{x}$as follows.

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

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

Draw the graph of the equation $ker\left(A^T\right)=im{\left(A\right)}^{\perp }$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$ker\left(A^T\right)=im{\left(A\right)}^{\perp }$is sketched.