Question

Let be the solution of the linear system

$\left|\begin{array}{c}{x}_1+x_2+x_3+x_4=0\\ x_1+2x_2+5x_3+4x_4=0\end{array}\right|$.

Find the basis of$\overline{){\mathbf{V}}^{\mathbf{\perp }}}$.

Short answer

The set $\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_1,....,{\overrightarrow{v}}_p,{\overrightarrow{w}}_1,{\overrightarrow{w}}_1,...{\overrightarrow{w}}_q\right\}$spanned${\mathrm{ℝ}}^n$ .

Step-by-step solution

Determine the Matrix.

Consider the linear system$\left|\begin{array}{c}{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=0\\ {\mathrm{x}}_1+2{\mathrm{x}}_2+5{\mathrm{x}}_3+4{\mathrm{x}}_4=0\end{array}\right|$ .

Simplify$\left|\begin{array}{c}{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=0\\ {\mathrm{x}}_1+2{\mathrm{x}}_2+5{\mathrm{x}}_3+4{\mathrm{x}}_4=0\end{array}\right|$ as follows.

$$\begin{gathered} \left|\begin{array}{c}{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4=0\\ {\mathrm{x}}_1+2{\mathrm{x}}_2+5{\mathrm{x}}_3+4{\mathrm{x}}_4=0\end{array}\right| \\ \left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 5& 4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=0 \end{gathered}$$

Assume that$A=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 5& 4\end{array}\right]$ .

Determine the kernel.

As the set V is set of all solution of the linear system implies $V=ker\left(A\right)$.

Assume$B=\left\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_1,....,{\overrightarrow{v}}_p,{\overrightarrow{w}}_1,{\overrightarrow{w}}_1,...{\overrightarrow{w}}_q\right\}$and$\overrightarrow{v}\in {\mathrm{ℝ}}^n$.

Theorem: For any matrix B,$im{\left(\mathrm{B}\right)}^{\perp }=ker{B}^{\mathrm{T}}$

By the theorem$im{\left(A^T\right)}^{\perp }=ker{A}^{T^T}$

As $A^{T^T}=A$and $A^{{\perp }^{\perp }}=A$, take $\perp$both side in the equation $im{\left(A^T\right)}^{\perp }=ker{A}^{T^T}$as follows.

$$\begin{gathered} im{\left(A^T\right)}^{\perp }=ker{A}^{T^T} \\ im{\left(A^T\right)}^{\perp }=kerA \\ {\left\{im{\left(A^T\right)}^{\perp }\right\}}^{\perp }={\left(kerA\right)}^{\perp } \\ im{\left(A^T\right)}^{\perp }={\left(kerA\right)}^{\perp } \end{gathered}$$

Therefore, the value$V^{\perp }=im\left(A^T\right)$ .

Determine the matrix AT .

Simplify$A^T$ as follows.

$$\begin{gathered} A^T={\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 5& 4\end{array}\right]}^T \\ {A}^T=\left[\begin{array}{cc}1& 1\\ 1& 2\\ 1& 5\\ 1& 4\end{array}\right] \end{gathered}$$

By the definition of $im\left(A\right)$, the basis for $im\left(A^T\right)$is $\left\{\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 2\\ 5\\ 4\end{array}\right]\right\}$.

Hence, the basis for$V^{\perp }$ is$\left\{\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 2\\ 5\\ 4\end{array}\right]\right\}$ .