Question

Consider the vector

$\overline{\mathbf{v}}\mathbf{=}\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$in ${\mathbf{R}}^{\mathbf{4}}$

Find a basis of the subspace of ${\mathbf{R}}^{\mathbf{4}}$consisting of all vectors perpendicular to $\overrightarrow{\mathbf{v}}$.

Short answer

Any vector in the form $x=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$is perpendicular to $\overrightarrow{v}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$which is spanned by $\left\{\right(-2100),(-3010),(-4001\left)\right\}$.

Step-by-step solution

Determine the orthogonal vector

Consider the given vector $\overrightarrow{v}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$and localid="1659429971902" $\overrightarrow{x}=\left[\begin{array}{c}{\overrightarrow{x}}_1\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$.

If the vector$\overrightarrow{\mathbf{x}}$is perpendicular then .

As$\overrightarrow{v}$is perpendicular to $\overrightarrow{x}$, by the definition of orthogonal $\overrightarrow{x}.\overrightarrow{v}=0$.

Simplify the equation$\overrightarrow{x}.\overrightarrow{v}=0$as follows.

$$\begin{gathered} {\overrightarrow{x}}_1.{\overrightarrow{v}}_1=0 \\ \left[{\overrightarrow{x}}_1,{\overrightarrow{x}}_2,{\overrightarrow{x}}_3,{\overrightarrow{x}}_4\right]\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=0 \\ {\overrightarrow{x}}_1+2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4=0 \\ {\overrightarrow{x}}_1=0\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\} \end{gathered}$$

Substitute the value $-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}$for ${\overrightarrow{x}}_1$in the equation $\overrightarrow{x}=\left[\begin{array}{c}{\overrightarrow{x}}_1\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$as follows.

$$\begin{gathered} \overrightarrow{x}=\left[\begin{array}{c}{\overrightarrow{x}}_1\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right] \\ \overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right] \end{gathered}$$

Therefore, any vector in the form $\overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$is perpendicular to $\overrightarrow{v}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]=$.

Determine basis

Simplify the equation $\overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right]$as follows.

$$\begin{gathered} \overrightarrow{x}=\left[\begin{array}{c}-\left\{2{\overrightarrow{x}}_2+3{\overrightarrow{x}}_3+4{\overrightarrow{x}}_4\right\}\\ {\overrightarrow{x}}_2\\ {\overrightarrow{x}}_3\\ {\overrightarrow{x}}_4\end{array}\right] \\ =\left[\begin{array}{c}-2{\overrightarrow{x}}_2\\ {\overrightarrow{x}}_2\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}-3{\overrightarrow{x}}_2\\ 0\\ {\overrightarrow{x}}_3\\ 0\end{array}\right]+\left[\begin{array}{c}-4{\overrightarrow{x}}_4\\ 0\\ 0\\ {\overrightarrow{x}}_4\end{array}\right] \\ ={\overrightarrow{x}}_2\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+{\overrightarrow{x}}_3\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\end{array}\right]+{\overrightarrow{x}}_4\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right] \end{gathered}$$

Therefore, the vector $\overrightarrow{x}\in span\left\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 0\\ 1\end{array}\right]\right\}$.

Hence, the vector$\overrightarrow{x}$ is spanned by$\left\{\left(\begin{array}{cccc}-2& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}-3& 0& 1& 0\end{array}\right),\left(\begin{array}{cccc}-4& 0& 0& 1\end{array}\right)\right\}$ which is perpendicular to $\overrightarrow{v}=\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right]$.